Raising the bar Quantitative Finance Archives

Apr 19

Checklist for Quantitative Finance FN3142

Checklist for Quantitative Finance FN3142

Students of FN3142 often think that they can get by by picking a few technical tricks. The questions below are mostly about intuition that helps to understand and apply those tricks.

Everywhere we assume that $...,Y_{t-1},Y_t,Y_{t+1},...$ is a time series and $...,I_{t-1},I_t,I_{t+1},...$ is a sequence of corresponding information sets. It is natural to assume that $I_t\subset I_{t+1}$ for all $t.$ We use the short conditional expectation notation: $E_tX=E(X|I_t)$ .

Questions

Question 1. How do you calculate conditional expectation in practice?

Question 2. How do you explain $E_t(E_tX)=E_tX$ ?

Question 3. Simplify each of $E_tE_{t+1}X$ and $E_{t+1}E_tX$ and explain intuitively.

Question 4. $\varepsilon _t$ is a shock at time $t$ . Positive and negative shocks are equally likely. What is your best prediction now for tomorrow's shock? What is your best prediction now for the shock that will happen the day after tomorrow?

Question 5. How and why do you predict $Y_{t+1}$ at time $t$ ? What is the conditional mean of your prediction?

Question 6. What is the error of such a prediction? What is its conditional mean?

Question 7. Answer the previous two questions replacing $Y_{t+1}$ by $Y_{t+p}$ .

Question 8. What is the mean-plus-deviation-from-mean representation (conditional version)?

Question 9. How is the representation from Q.8 reflected in variance decomposition?

Question 10. What is a canonical form? State and prove all properties of its parts.

Question 11. Define conditional variance for white noise process and establish its link with the unconditional one.

Question 12. How do you define the conditional density in case of two variables, when one of them serves as the condition? Use it to prove the LIE.

Question 13. Write down the joint distribution function for a) independent observations and b) for serially dependent observations.

Question 14. If one variable is a linear function of another, what is the relationship between their densities?

Question 15. What can you say about the relationship between $a,b$ if $f(a)=f(b)$ ? Explain geometrically the definition of the quasi-inverse function.

Answers

Answer 1. Conditional expectation is a complex notion. There are several definitions of differing levels of generality and complexity. See one of them here and another in Answer 12.

The point of this exercise is that any definition requires a lot of information and in practice there is no way to apply any of them to actually calculate conditional expectation. Then why do they juggle conditional expectation in theory? The efficient market hypothesis comes to rescue: it is posited that all observed market data incorporate all available information, and, in particular, stock prices are already conditioned on $I_t.$

Answers 2 and 3. This is the best explanation I have.

Answer 4. Since positive and negative shocks are equally likely, the best prediction is $E_t\varepsilon _{t+1}=0$ (I call this equation a martingale condition). Similarly, $E_t\varepsilon _{t+2}=0$ but in this case I prefer to see an application of the LIE: $E_{t}\varepsilon _{t+2}=E_t(E_{t+1}\varepsilon _{t+2})=E_t0=0.$

Answer 5. The best prediction is $\hat{Y}_{t+1}=E_tY_{t+1}$ because it minimizes $E_t(Y_{t+1}-f(I_t))^2$ among all functions $f$ of current information $I_t.$ Formally, you can use the first order condition

$\frac{d}{df(I_t)}E_t(Y_{t+1}-f(I_t))^2=-2E_t(Y_{t+1}-f(I_t))=0$

to find that $f(I_t)=E_tf(I_t)=E_tY_{t+1}$ is the minimizing function. By the projector property
$E_t\hat{Y}_{t+1}=E_tE_tY_{t+1}=E_tY_{t+1}=\hat{Y}_{t+1}.$

Answer 6. It is natural to define the prediction error by

$\hat{\varepsilon}_{t+1}=Y_{t+1}-\hat{Y}_{t+1}=Y_{t+1}-E_tY_{t+1}.$

By the projector property $E_t\hat{\varepsilon}_{t+1}=E_tY_{t+1}-E_tY_{t+1}=0$ .

Answer 7. To generalize, just change the subscripts. For the prediction we have to use two subscripts: the notation $\hat{Y}_{t,t+p}$ means that we are trying to predict what happens at a future date $t+p$ based on info set $I_t$ (time $t$ is like today). Then by definition $\hat{Y} _{t,t+p}=E_tY_{t+p},$ $\hat{\varepsilon}_{t,t+p}=Y_{t+p}-E_tY_{t+p}.$

Answer 8. Answer 7, obviously, implies $Y_{t+p}=\hat{Y}_{t,t+p}+\hat{\varepsilon}_{t,t+p}.$ The simple case is here.

Answer 9. See the law of total variance and change it to reflect conditioning on $I_t.$

Answer 10. See canonical form.

Answer 11. Combine conditional variance definition with white noise definition.

Answer 12. The conditional density is defined similarly to the conditional probability. Let $X,Y$ be two random variables. Denote $p_X$ the density of $X$ and $p_{X,Y}$ the joint density. Then the conditional density of $Y$ conditional on $X$ is defined as $p_{Y|X}(y|x)=\frac{p_{X,Y}(x,y)}{p_X(x)}.$ After this we can define the conditional expectation $E(Y|X)=\int yp_{Y|X}(y|x)dy.$ With these definitions one can prove the Law of Iterated Expectations:

$E[E(Y|X)]=\int E(Y|x)p_X(x)dx=\int \left( \int yp_{Y|X}(y|x)dy\right) p_X(x)dx$

$=\int \int y\frac{p_{X,Y}(x,y)}{p_X(x)}p_X(x)dxdy=\int \int yp_{X,Y}(x,y)dxdy=EY.$

This is an illustration to Answer 1 and a prelim to Answer 13.

Answer 13. Understanding this answer is essential for Section 8.6 on maximum likelihood of Patton's guide.

a) In case of independent observations $X_1,...,X_n$ the joint density of the vector $X=(X_1,...,X_n)$ is a product of individual densities:

$p_X(x_1,...,x_n)=p_{X_1}(x_1)...p_{X_n}(x_n).$

b) In the time series context it is natural to assume that the next observation depends on the previous ones, that is, for each $t,$ $X_t$ depends on $X_1,...,X_{t-1}$ (serially dependent observations). Therefore we should work with conditional densities $p_{X_1,...,X_t|X_1,...,X_{t-1}}.$ From Answer 12 we can guess how to make conditional densities appear:

$p_{X_1,...,X_n}(x_1,...,x_n)=\frac{p_{X_1,...,X_n}(x_1,...,x_n)}{ p_{X_1,...,X_{n-1}}(x_1,...,x_{n-1})}\frac{p_{X_1,...,X_{n-1}}(x_1,...,x_{n-1})}{ p_{X_1,...,X_{n-2}}(x_1,...,x_{n-2})}...\frac{p_{X_1,X_2}(x_1,x_2)}{p_{X_1}(x_1)}p_{X_1}(x_1).$

The fractions on the right are recognized as conditional probabilities. The resulting expression is pretty awkward:

$p_{X_1,...,X_n}(x_1,...,x_n)=p_{X_1,...,X_n|X_1,...,X_n-1}(x_1,...,x_n|x_1,...,x_{n-1})\times$

$\times p_{X_1,...,X_{n-1}|X_1,...,X_{n-2}}(x_1,...,x_{n-1}|x_1,...,x_{n-2})...\times$

$p_{X_1,X_2|X_1}(x_1,x_2|x_1)p_{X_1}(x_1).$

Answer 14. The answer given here helps one understand how to pass from the density of the standard normal to that of the general normal.

Answer 15. This elementary explanation of the function definition can be used in the fifth grade. Note that conditions sufficient for existence of the inverse are not satisfied in a case as simple as the distribution function of the Bernoulli variable (when the graph of the function has flat pieces and is not continuous). Therefore we need a more general definition of an inverse. Those who think that this question is too abstract can check out UoL exams, where examinees are required to find Value at Risk when the distribution function is a step function. To understand the idea, do the following:

a) Draw a graph of a good function $f$ (continuous and increasing).

b) Fix some value $y_0$ in the range of this function and identify the region $\{y:y\ge y_0\}$ .

c) Find the solution $x_0$ of the equation $f(x)=y_0$ . By definition, $x_0=f^{-1}(y_o).$ Identify the region $\{x:f(x)\ge y_0\}$ .

d) Note that $x_0=\min\{x:f(x)\ge y_0\}$ . In general, for bad functions the minimum here may not exist. Therefore minimum is replaced by infimum, which gives us the definition of the quasi-inverse:

$x_0=\inf\{x:f(x)\ge y_0\}$ .

Tags: best prediction, conditional density, conditional expectation, martingale condition, prediction error, quasi-inverse, serially dependent observations, University of London International Programmes, uol affiliate centre

Oct 18

Law of iterated expectations: geometric aspect

category: FN3142 Quantitative Finance, Quantitative Finance no comments

Law of iterated expectations: geometric aspect

There will be a separate post on projectors. In the meantime, we'll have a look at simple examples that explain a lot about conditional expectations.

Examples of projectors

The name "projector" is almost self-explanatory. Imagine a point and a plane in the three-dimensional space. Draw a perpendicular from the point to the plane. The intersection of the perpendicular with the plane is the points's projection onto that plane. Note that if the point already belongs to the plane, its projection equals the point itself. Besides, instead of projecting onto a plane we can project onto a straight line.

The above description translates into the following equations. For any $x\in R^3$ define

(1) $P_2x=(x_1,x_2,0)$ and $P_1x=(x_1,0,0).$

$P_2$ projects $R^3$ onto the plane $L_2=\{(x_1,x_2,0):x_1,x_2\in R\}$ (which is two-dimensional) and $P_1$ projects $R^3$ onto the straight line $L_1=\{(x_1,0,0):x_1\in R\}$ (which is one-dimensional).

Property 1. Double application of a projector amounts to single application.

Proof. We do this just for one of the projectors. Using (1) three times we get

(1) $P_2[P_2x]=P_2(x_1,x_2,0)=(x_1,x_2,0)=P_2x.$

Property 2. A successive application of two projectors yields the projection onto a subspace of a smaller dimension.

Proof. If we apply first $P_2$ and then $P_1$ , the result is

(2) $P_1[P_2x]=P_1(x_1,x_2,0)=(x_1,0,0)=P_1x.$

If we change the order of projectors, we have

(3) $P_2[P_1x]=P_2(x_1,0,0)=(x_1,0,0)=P_1x.$

Exercise 1. Show that both projectors are linear.

Exercise 2. Like any other linear operator in a Euclidean space, these projectors are given by some matrices. What are they?

The simple truth about conditional expectation

In the time series setup, we have a sequence of information sets $...\subset I_t\subset I_{t+1}\subset...$ (it's natural to assume that with time the amount of available information increases). Denote

$E_tX=E(X|I_t)$

the expectation of $X$ conditional on $I_t$ . For each $t$ ,

$E_t$ is a projector onto the space of random functions that depend only on the information set $I_t$ .

Property 1. Double application of conditional expectation gives the same result as single application:

(4) $E_t(E_tX)=E_tX$

( $E_tX$ is already a function of $I_t$ , so conditioning it on $I_t$ doesn't change it).

Property 2. A successive conditioning on two different information sets is the same as conditioning on the smaller one:

(5) $E_tE_{t+1}X=E_tX,$

(6) $E_{t+1}E_tX=E_tX.$

Property 3. Conditional expectation is a linear operator: for any variables $X,Y$ and numbers $a,b$

$E_t(aX+bY)=aE_tX+bE_tY.$

It's easy to see that (4)-(6) are similar to (1)-(3), respectively, but I prefer to use different names for (4)-(6). I call (4) a projector property. (5) is known as the Law of Iterated Expectations, see my post on the informational aspect for more intuition. (6) holds simply because at time $t+1$ the expectation $E_tX$ is known and behaves like a constant.

Summary. (4)-(6) are easy to remember as one property. The smaller information set wins: $E_sE_tX=E_{\min\{s,t\}}X.$

Tags: conditional expectation, conditioning as projector, Law of iterated expectations, projector, time series setup, University of London International Programmes, uol affiliate centre

Oct 18

Law of iterated expectations: informational aspect

category: FN3142 Quantitative Finance, Quantitative Finance no comments

Law of iterated expectations: informational aspect

The notion of Brownian motion will help us. Suppose we observe a particle that moves back and forth randomly along a straight line. The particle starts at zero at time zero. The movement can be visualized by plotting on the horizontal axis time and on the vertical axis - the position of the particle. $W(t)$ denotes the random position of the particle at time $t$ .

Figure 1. Unconditional expectation

In Figure 1, various paths starting at the origin are shown in different colors. The intersections of the paths with vertical lines at times 0.5, 1 and 1.5 show the positions of the particle at these times. The deviations of those positions from $y=0$ to the upside and downside are assumed to be equally likely (more precisely, they are normal variables with mean zero and variance $t$ ).

Unconditional expectation

“In the beginning there was nothing, which exploded.” ― Terry Pratchett, Lords and Ladies

If we are at the origin (like the Big Bang), nothing has happened yet and $EW(t)=0$ is the best prediction for any moment $t>0$ we can make (shown by the blue horizontal line in Figure 1). The usual, unconditional expectation $EX$ corresponds to the empty information set.

Conditional expectation

Figure 2. Conditional expectation

In Figure 2, suppose we are at $t=2.$ The dark blue path between $t=0$ and $t=2$ has been realized. We know that the particle has reached the point $W(2)$ at that time. With this knowledge, we see that the paths starting at this point will have the average

(1) $E(W(t)|W(2))=W(2),$ $t>2.$

This is because the particle will continue moving randomly, with the up and down moves being equally likely. Prediction (1) is shown by the horizontal light blue line between $t=2$ and $t=4.$ In general, this prediction is better than $EW(t)=0$ .

Note that for different realized paths, $W(2)$ takes different values. Therefore $E(W(t)|W(2))$ , for $t<2$ , is a random variable. It is a function of the event we condition the expectation on.

Law of iterated expectations

Figure 3. Law of iterated expectations

Suppose you are at time $t=2$ (see Figure 3). You send many agents to the future $t=3$ to fetch the information about what will happen. They bring you the data on the means $E(W(t)|W(3))$ they see (shown by horizontal lines between $t=3$ and $t=4).$ Since there are many possible future realizations, you have to average the future means. For this, you will use the distributional belief you have at time $t=2.$ The result is $E[E(W(t)|W(3))|W(2)].$ Since the up and down moves are equally likely, your distribution at time $t=2$ is symmetric around $W(2).$ Therefore the above average will be equal to $E(W(t)|W(2)).$ This is the Law of Iterated Expectations, also called the tower property:

(2) $E[E(W(t)|W(3))|W(2)]=E(W(t)|W(2)).$

The knowledge of all of the future predictions $E(W(t)|W(3))$ , upon averaging, does not improve or change our current prediction $E(W(t)|W(2))$ .

For a full mathematical treatment of conditional expectation see Lecture 10 by Gordan Zitkovic.

Tags: Brownian motion, conditional expectation, Law of Iterated Expectations intuition, tower property, University of London International Programmes, uol affiliate centre

Sep 18

Portfolio analysis: return on portfolio

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Portfolio analysis: return on portfolio

Exercise 1. Suppose a portfolio contains $n_1$ shares of stock 1 whose price is $S_1$ and $n_2$ shares of stock 2 whose price is $S_2$ . Stock prices fluctuate and are random variables. Numbers of shares are assumed fixed and are deterministic. What is the expected value of the portfolio?

Solution. The portfolio value is its market price $V=n_1S_1+n_2S_2$ . Since this is a linear combination, the expected value is $EV=n_1ES_1+n_2ES_2$ .

In fact, the portfolio analysis is a little bit different than suggested by Exercise 1. To explain the difference, we start with fixing two points of view.

View 1. I hold a portfolio of stocks. I may have inherited it, and it does not matter how much it cost at the moment it was formed. If I want to sell it, I am interested in knowing its market value. In this situation the numbers of shares in my portfolio, which are constant, and the market prices of stocks, which are random, determine the market value of the portfolio, defined in Exercise 1. The value of the portfolio is a linear combination of stock prices.

View 2. I have a certain amount of money $M^0$ to invest. Being a gambler, I am not interested in holding a portfolio forever. I am thinking about buying a portfolio of stocks now and selling it, say, in a year at price $M^1$ . In this case I am interested in the rate of return defined by $r=\frac{M^1-M^0}{M^0}.$ $M^0$ is considered deterministic (current prices are certain) and $M^1$ is random (future prices are unpredictable). Thus the rate of return is random.

We pursue the second view (prevalent in finance). As it often happens in economics and finance, the result depends on how one understands the things. Suppose the initial amount $M^0$ is invested in $n$ assets. Denoting $M_i^0$ the amount invested in asset $i$ , we have $M^0=\sum\limits_{i = 1}^nM_i^0$ . Denoting $s_i=M_i^0/{M^0}$ the share (percentage) of $M_i^0$ in the total investment $M^0$ , we have

(1) $M_i^0=s_iM^0,\ M^0=\sum\limits_{i = 1}^ns_iM^0.$

The initial shares $s_i$ are deterministic.

Let $M_i^1$ be what becomes of $M_i^0$ in one year and let $M^1=\sum\limits_{i = 1}^nM_i^1$ be the total value of the investment at the end of the year. Since different assets grow at different rates, generally it is not true that $M_i^1 =s_iM^1$ . Denote $r_i=\frac{M_i^1-M_i^0}{M_i^0}$ the rate of return on asset $i$ . Then

(2) $M_i^1=(1+r_i)M_i^0,$ $M^1=\sum\limits_{i = 1}^n(1+r_i)M_i^0.$

Exercise 2. The rate of return on the portfolio is a linear combination of the rates of return on separate assets, the coefficients being the initial shares of investment.

Solution. Using Equations (1) and (2) we get

(3) $r=\frac{M^1-M^0}{M^0}=\frac{\sum(1+r_i)M_i^0-\sum M_i^0}{M^0}=\frac{\sum r_iM_i^0}{M^0}=\frac{\sum r_is_iM^0}{M^0}=\sum s_ir_i .$

Once you know this equation you can find the mean and variance of the rate of return on the portfolio in terms of investment shares and rates of return on assets.

Tags: Portfolio analysis, portfolio value, rate of return, return on portfolio, University of London International Programmes, uol affiliate centre

Sep 18

Applications of the diagonal representation IV

category: FN3142 Quantitative Finance, Matrix algebra, MT2175 Further linear algebra, Quantitative Finance no comments

Applications of the diagonal representation IV

Principal component analysis is a general method based on diagonalization of the variance matrix. We consider it in a financial context. The variance matrix measures riskiness of the portfolio. We want to see which stocks contribute most to the portfolio risk. The surprise is that the answer is given not in terms of the vector of returns but in terms of its linear transformation.

8. Principal component analysis (PCA)

Let $R$ be a column-vector of returns on $p$ stocks with the variance matrix $V(R)=E(R-ER)(R-ER)^{T}$ . The idea is to find an orthogonal matrix $W$ such that $W^{-1}V(R)W=D$ is a diagonal matrix $D=diag[\lambda_1,...,\lambda_p]$ with $\lambda_1\geq...\geq\lambda_p.$

With such a matrix, instead of $R$ we can consider its transformation $Y=W^{-1}R$ for which

$V(Y)=W^{-1}V(R)(W^{-1})^T=W^{-1}V(R)W=D.$

We know that $V(Y)$ has variances $V(Y_1),...,V(Y_p)$ on the main diagonal. It follows that $V(Y_i)=\lambda_i$ for all $i.$ Variance is a measure of riskiness. Thus, the transformed variables $Y_1,...,Y_p$ are put in the order of declining risk. What follows is the realization of this idea using sample data.

In a sampling context, all population means shoud be replaced by their sample counterparts. Let $R^{(t)}$ be a $p\times 1$ vector of observations on $R$ at time $t.$ These observations are put side by side into a matrix $\mathbb{R}=(R^{(1)},...,R^{(n)})$ where $n$ is the number of moments in time. The population mean $ER$ is estimated by the sample mean

$\bar{\mathbb{R}}=\frac{1}{n}\sum_{t=1}^nR^{(t)}.$

The variance matrix $V(R)$ is estimated by

$\hat{V}=\frac{1}{n-1}(\mathbb{R}-\bar{\mathbb{R}}l)(\mathbb{R}-\bar{\mathbb{R}}l)^T$

where $l$ is a $1\times n$ vector of ones. It is this matrix that is diagonalized: $W^{-1}\hat{V}W=D.$

In general, the eigenvalues in $D$ are not ordered. Ordering them and at the same time changing places of the rows of $W^{-1}$ correspondingly we get a new orthogonal matrix $W_1$ (this requires a small proof) such that the eigenvalues in $W_1^{-1}\hat{V}W_1=D_1$ will be ordered. There is a lot more to say about the method and its applications.

Tags: Principal component analysis, University of London International Programmes, uol affiliate centre

May 18

Efficient market hypothesis is subject to interpretation

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Efficient market hypothesis is subject to interpretation

The formulation on Investopedia seems to me the best:

The efficient market hypothesis (EMH) is an investment theory that states it is impossible to "beat the market" because stock market efficiency causes existing share prices to always incorporate and reflect all relevant information. According to the EMH, stocks always trade at their fair value on stock exchanges, making it impossible for investors to either purchase undervalued stocks or sell stocks for inflated prices. As such, it should be impossible to outperform the overall market through expert stock selection or market timing, and the only way an investor can possibly obtain higher returns is by purchasing riskier investments.

This is not Math, and the EMH interpretation is subjective. My purpose is not to discuss the advantages and drawbacks of various versions of the EMH but indicate some errors students make on exams.

Best(?) way to answer questions related to EMH

Since there is a lot of talking, the best is to use the appropriate key words.

Start with "The EMH states that it is impossible to make economic profit".

Then explain why: The stock market is efficient in the sense that stocks trade at their fair value, so that undervalued or overvalued stocks don't exist.

Then specify that "to obtain economic profits, from the revenues we subtract opportunity (hidden) costs, in addition to direct costs, such as transaction fees". What on the surface seems to be a profitable activity may in fact be balancing at break-even.

Next is to address the specification by Malkiel that the EMH depends on the information set $\Omega_t$ available at time $t$ .

Weak form of EMH. The information set $\Omega_t^1$ contains only historical values of asset prices, dividends (and possibly volume) up until time $t$ . This is basically what an investor sees on a stock price chart. Many students say "historical information" but fail to mention that it is about prices of financial assets. The birthdays of celebrities are also historical information but they are not in this info set.

Semi-strong form of EMH. The info set $\Omega_t^2$ is all publicly available information. Some students don't realize that it includes $\Omega_t^1$ . The risk-free rate is in $\Omega_t^2$ but not in $\Omega_t^1$ because 1) it is publicly known and 2) it is not traded (it is fixed by the central bank for extended periods of time).

Strong form of EMH. The info set $\Omega_t^3$ includes all publicly available info plus private company information. Firstly, this info set includes the previous two: $\Omega_t^1\subset\Omega_t^2\subset\Omega_t^3$ . Secondly, whether a certain piece of information belongs to $\Omega_t^2$ or $\Omega_t^3$ depends on time. For example, the number of shares Warren Buffett purchased today of the stock is in $\Omega_t^3$ but over time it becomes a part of $\Omega_t^2$ because large holdings must be reported within 45 days of the end of a calendar quarter. If there are nuances like this you have to explain them.

Implications for time series analysis

Conditional expectation is a relatively complex mathematical construct. The simplest definition is accessible to basic statistics students. The mid-level definition in case of conditioning on a set of positive probability already raises questions about practical calculation. The most general definition is based on Radon-Nikodym derivatives. Moreover, nobody knows exactly any of those $\Omega_t$ . So how do you apply time series models which depend so heavily on conditioning? The answer is simple: since by the EMH the stock price "reflects all relevant information", that price is already conditioned on that information, and you don't need to worry about theoretical complexities of conditioning in applications.

Tags: efficient market hypothesis, market efficiency, University of London International Programmes, uol affiliate centre

May 18

The Newey-West estimator: uncorrelated and correlated data

category: FN3142 Quantitative Finance, Quantitative Finance no comments

The Newey-West estimator: uncorrelated and correlated data

I hate long posts but here we by necessity have to go through all ideas and calculations, to understand what is going on. One page of formulas in A. Patton's guide to FN3142 Quantitative Finance in my rendition becomes three posts.

Preliminaries and autocovariance function

Let $X_1,...,X_n$ be random variables. We need to recall that the variance of the vector $X=(X_1,...,X_n)^T$ is

(1) $V(X)=\left(\begin{array}{cccc}V(X_1)&Cov(X_1,X_2)&...&Cov(X_1,X_n)\\Cov(X_2,X_1)&V(X_2)&...&Cov(X_2,X_n)\\...&...&...&...\\Cov(X_n,X_1)&Cov(X_n,X_2)&...&V(X_n)\end{array}\right).$

With the help of this matrix we derived two expressions for variance of a linear combination:

(2) $V\left(\sum_{i=1}^na_iX_i\right)=\sum_{i=1}^na_i^2V(X_i)$

for uncorrelated variables and

(3) $V\left(\sum_{i=1}^na_iX_i\right)=\sum_{i=1}^na_i^2V(X_i)+2\sum_{i=1}^{n-1}\sum_{j=i+1}^na_ia_jCov(X_i,X_j)$

when there is autocorrelation.

In a time series context $X_1,...,X_n$ are observations along time. $1,...,n$ stand for moments in time and the sequence $X_1,...,X_n$ is called a time series. We need to recall the definition of a stationary process. Of that definition, we will use only the part about covariances: $Cov(X_i,X_j)$ depends only on the distance $|i-j|$ between the time moments $i,j.$ For example, in the top right corner of (1) we have $Cov(X_1,X_n),$ which depends only on $n-1.$

Preamble. Let $X_1,...,X_n$ be a stationary times series. Firstly, $Cov(X_i,X_i+k)$ depends only on $k.$ Secondly, for all integer $k=0,\pm 1,\pm 2,...$ denoting $j=i+k$ we have

(4) $Cov(X_i,X_{i+k})=Cov(X_{j-k},X_j)=Cov(X_j,X_{j-k}).$

Definition. The autocovariance function is defined by

(5) $\gamma_k=Cov(X_i,X_{k+i})$ for all integer $k=0,\pm 1,\pm 2,...$

In particular,

(6) $\gamma_0=Cov(X_i,X_i)=V(X_i)$ for all $i.$

The preamble shows that definition (5) is correct (the right side in (5) depends only on $k$ and not on $i$ ). Because of (4) we have symmetry $\gamma_{-k}=\gamma _k,$ so negative $k$ can be excluded from consideration.

With (5) and (6) for a stationary series (1) becomes

(7) $V(X)=\left(\begin{array}{cccc}\gamma_0&\gamma_1&...&\gamma_{n-1}\\ \gamma_1&\gamma_0&...&\gamma_{n-2}\\...&...&...&...\\ \gamma_{n-1}&\gamma_{n-2}&...&\gamma_0\end{array}\right).$

Estimating variance of a sample mean

Uncorrelated observations. Suppose $X_1,...,X_n$ are uncorrelated observations from the same population with variance $\sigma^2.$ From (2)
we get

(8) $V\left(\frac{1}{n}\sum_{i=1}^nX_i\right) =\frac{1}{n^2}\sum_{i=1}^nV(X_i)=\frac{n\sigma^2}{n^2}=\frac{\sigma^2}{n}.$

This is a theoretical relationship. To actually obtain an estimator of the sample variance, we need to replace $\sigma^2$ by some estimator. It is known that

(9) $s^2=\frac{1}{n}\sum_{i=1}^n(X_i-\bar{X})^2$

consistently estimates $\sigma^2.$ Plugging it in (8) we see that variance of the sample mean is consistently estimated by

$\hat{V}=\frac{1}{n}s^2=\frac{1}{n^2}\sum_{i=1}^n(X_i-\bar{X})^2.$

This is the estimator derived on p.151 of Patton's guide.

Correlated observations. In this case we use (3):

$V\left( \frac{1}{n}\sum_{i=1}^nX_i\right) =\frac{1}{n^{2}}\left[\sum_{i=1}^nV(X_i)+2\sum_{i=1}^{n-1}\sum_{j=i+1}^nCov(X_i,X_j)\right]$ .

Here visualization comes in handy. The sums in the square brackets include all terms on the main diagonal of (7) and above it. That is, we have $n$ copies of $\gamma_0,$ $n-1$ copies of $\gamma_{1}$ ,..., 2 copies of $\gamma _{n-2}$ and 1 copy of $\gamma _{n-1}.$ The sum in the brackets is

$\sum_{i=1}^nV(X_i)+2\sum_{i=1}^{n-1}\sum_{j=i+1}^nCov(X_i,X_j)=n\gamma_0+2[(n-1)\gamma_1+...+2\gamma_{n-2}+\gamma _{n-1}]=n\gamma _0+2\sum_{k=1}^{n-1}(n-k)\gamma_k.$

Thus we obtain the first equation on p.152 of Patton's guide (it's up to you to match the notation):

(10) $V(\bar{X})=\frac{1}{n}\gamma_0+\frac{2}{n}\sum_{k=1}^{n-1}(1-\frac{k}{n})\gamma_k.$

As above, this is just a theoretical relationship. $\gamma_0=V(X_i)=\sigma^2$ is estimated by (9). Ideally, the estimator of $\gamma_k=Cov(X_i,X_{k+i})$ is obtained by replacing all population means by sample means:

(11) $\hat{\gamma}_k=\frac{1}{n}\sum_{i=1}^n(X_i-\bar{X})(X_{k+i}-\bar{X}).$

There are two problems with this estimator, though. The first problem is that when $i$ runs from $1$ to $n,$ $k+i$ runs from $k+1$ to $k+n.$ To exclude out-of-sample values, the summation in (11) is reduced:

(12) $\hat{\gamma}_k=\frac{1}{n-k}\sum_{i=1}^{n-k}(X_i-\bar{X})(X_{k+i}-\bar{X}).$

The second problem is that the sum in (12) becomes too small when $k$ is close to $n.$ For example, for $k=n-1$ (12) contains just one term (there is no averaging). Therefore the upper limit of summation $n-1$ in (10) is replaced by some function $M(n)$ that tends to infinity slower than $n.$ The result is the estimator

$\hat{V}=\frac{1}{n}\hat{\gamma}_0+\frac{2}{n}\sum_{k=1}^{M(n)}(1-\frac{k}{M(n)})\hat{\gamma}_k$

where $\hat{\gamma}_0$ is given by (9) and $\hat{\gamma}_k$ is given by (12). This is almost the Newey-West estimator from p.152. The only difference is that instead of $\frac{k}{M(n)}$ they use $\frac{k}{M(n)+1}$ , and I have no idea why. One explanation is that for low $n$ , $M(n)$ can be zero, so they just wanted to avoid division by zero.

Tags: autocovariance function, correlated observations, Estimating variance of a sample mean, Newey-West estimator, Uncorrelated observations, University of London International Programmes, uol affiliate centre, Variance of a linear combination

May 18

Different faces of vector variance: again visualization helps

category: EC2020 Elements of econometrics, Econometrics, FN3142 Quantitative Finance, Quantitative Finance no comments

Different faces of vector variance: again visualization helps

In the previous post we defined variance of a column vector $X$ with $n$ components by

$V(X)=E(X-EX)(X-EX)^T.$

In terms of elements this is the same as:

(1) $V(X)=\left(\begin{array}{cccc}V(X_1)&Cov(X_1,X_2)&...&Cov(X_1,X_n)\\Cov(X_2,X_1)&V(X_2)&...&Cov(X_2,X_n)\\...&...&...&...\\Cov(X_n,X_1)&Cov(X_n,X_2)&...&V(X_n)\end{array}\right).$

So why knowing the structure of this matrix is so important?

Let $X_1,...,X_n$ be random variables and let $a_1,...,a_n$ be numbers. In the derivation of the variance of the slope estimator for simple regression we have to deal with the expression of type

(2) $V\left(\sum_{i=1}^na_iX_i\right).$

Question 1. How do you multiply a sum by a sum? I mean, how do you use summation signs to find the product $\left(\sum_{i=1}^na_i\right)\left(\sum_{i=1}^nb_i\right)$ ?

Answer 1. Whenever you have problems with summation signs, try to do without them. The product

$\left(a_1+...+a_n\right)\left(b_1+...+b_n\right)=a_1b_1+...+a_1b_n+...+a_nb_1+...+a_nb_n$

should contain ALL products $a_ib_j.$ Again, a matrix visualization will help:

$\left(\begin{array}{ccc}a_1b_1&...&a_1b_n\\...&...&...\\a_nb_1&...&a_nb_n\end{array}\right).$

The product we are looking for should contain all elements of this matrix. So the answer is

(3) $\left(\sum_{i=1}^na_i\right)\left(\sum_{i=1}^nb_i\right)=\sum_{i=1}^n\sum_{j=1}^na_ib_j.$

Formally, we can write $\sum_{i=1}^nb_i=\sum_{j=1}^nb_j$ (the sum does not depend on the index of summation, this is another point many students don't understand) and then perform the multiplication in (3).

Question 2. What is the expression for (2) in terms of covariances of components?

Answer 2. If you understand Answer 1 and know the relationship between variances and covariances, it should be clear that

(4) $V\left(\sum_{i=1}^na_iX_i\right)=Cov(\sum_{i=1}^na_iX_i,\sum_{i=1}^na_iX_i)$

$=Cov(\sum_{i=1}^na_iX_i,\sum_{j=1}^na_jX_j)=\sum_{i=1}^n\sum_{j=1}^na_ia_jCov(X_i,X_j).$

Question 3. In light of (1), separate variances from covariances in (4).

Answer 3. When $i=j,$ we have $Cov(X_i,X_j)=V(X_i),$ which are diagonal elements of (1). Otherwise, for $i\neq j$ we get off-diagonal elements of (1). So the answer is

(5) $V\left(\sum_{i=1}^na_iX_i\right)=\sum_{i=1}^na_i^2V(X_i)+\sum_{i\neq j}a_ia_jCov(X_i,X_j).$

Once again, in the first sum on the right we have only variances. In the second sum, the indices $i,j$ are assumed to run from $1$ to $n$ , excluding the diagonal $i=j.$

Corollary. If $X_{i}$ are uncorrelated, then the second sum in (5) disappears:

(6) $V\left(\sum_{i=1}^na_iX_i\right)=\sum_{i=1}^na_i^2V(X_i).$

This fact has been used (with a slightly different explanation) in the derivation of the variance of the slope estimator for simple regression.

Question 4. Note that the matrix (1) is symmetric (elements above the main diagonal equal their mirror siblings below that diagonal). This means that some terms in the second sum on the right of (5) are repeated twice. If you group equal terms in (5), what do you get?

Answer 4. The idea is to write

$a_ia_jCov(X_i,X_j)+a_ia_jCov(X_j,X_i)=2a_ia_jCov(X_i,X_j),$

that is, to join equal elements above and below the main diagonal in (1). For this, you need to figure out how to write a sum of the elements that are above the main diagonal. Make a bigger version of (1) (with more off-diagonal elements) to see that the elements that are above the main diagonal are listed in the sum $\sum_{i=1}^{n-1}\sum_{j=i+1}^n.$ This sum can also be written as $\sum_{1\leq i<j\leq n}.$ Hence, (5) is the same as

(7) $V\left(\sum_{i=1}^na_iX_i\right)=\sum_{i=1}^na_i^2V(X_i)+2\sum_{i=1}^{n-1}\sum_{j=i+1}^na_ia_jCov(X_i,X_j)$

$=\sum_{i=1}^na_i^2V(X_i)+2\sum_{1\leq i<j\leq n}a_ia_jCov(X_i,X_j).$

Unlike (6), this equation is applicable when there is autocorrelation.

Tags: autocorrelation, University of London International Programmes, uol affiliate centre, vector variance

May 18

Variance of a vector: motivation and visualization

category: EC2020 Elements of econometrics, FN3142 Quantitative Finance, Quantitative Finance no comments

Variance of a vector: motivation and visualization

I always show my students the definition of the variance of a vector, and they usually don't pay attention. You need to know what it is, already at the level of simple regression (to understand the derivation of the slope estimator variance), and even more so when you deal with time series. Since I know exactly where students usually stumble, this post is structured as a series of questions and answers.

Think about ideas: how would you define variance of a vector?

Question 1. We know that for a random variable $X$ , its variance is defined by

(1) $V(X)=E(X-EX)^{2}.$

Now let

$X=\left(\begin{array}{c}X_{1}\\...\\X_{n}\end{array}\right)$

be a vector with $n$ components, each of which is a random variable. How would you define its variance?

The answer is not straightforward because we don't know how to square a vector. Let $X^T=(\begin{array}{ccc}X_1& ...&X_n\end{array})$ denote the transposed vector. There are two ways to multiply a vector by itself: $X^TX$ and $XX^T.$

Question 2. Find the dimensions of $X^TX$ and $XX^T$ and their expressions in terms of coordinates of $X.$

Answer 2. For a product of matrices there is a compatibility rule that I write in the form

(2) $A_{n\times m}B_{m\times k}=C_{n\times k}.$

Recall that $n\times m$ in the notation $A_{n\times m}$ means that the matrix $A$ has $n$ rows and $m$ columns. For example, $X$ is of size $n\times 1.$ Verbally, the above rule says that the number of columns of $A$ should be equal to the number of rows of $B.$ In the product that common number $m$ disappears and the unique numbers ( $n$ and $k$ ) give, respectively, the number of rows and columns of $C.$ Isn't the the formula
easier to remember than the verbal statement? From (2) we see that $X_{1\times n}^TX_{n\times 1}$ is of dimension 1 (it is a scalar) and $X_{n\times 1}X_{1\times n}^T$ is an $n\times n$ matrix.

For actual multiplication of matrices I use the visualization

(3) $\left(\begin{array}{ccccc}&&&&\\&&&&\\a_{i1}&a_{i2}&...&a_{i,m-1}&a_{im}\\&&&&\\&&&&\end{array}\right) \left(\begin{array}{ccccc}&&b_{1j}&&\\&&b_{2j}&&\\&&...&&\\&&b_{m-1,j}&&\\&&b_{mj}&&\end{array}\right) =\left( \begin{array}{ccccc}&&&&\\&&&&\\&&c_{ij}&&\\&&&&\\&&&&\end{array}\right)$

Short formulation. Multiply rows from the first matrix by columns from the second one.

Long Formulation. To find the element $c_{ij}$ of $C,$ we find a scalar product of the $i$ th row of $A$ and $j$ th column of $B:$ $c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+...$ To find all elements in the $i$ th row of $C,$ we fix the $i$ th row in $A$ and move right the columns in $B.$ Alternatively, to find all elements in the $j$ th column of $C,$ we fix the $j$ th column in $B$ and move down the rows in $A$ . Using this rule, we have

(4) $X^TX=X_1^2+...+X_n^2,$ $XX^T=\left(\begin{array}{ccc}X_1^2&...&X_1X_n\\...&...&...\\X_nX_1&...&X_n^2 \end{array}\right).$

Usually students have problems with the second equation.

Based on (1) and (4), we have two candidates to define variance:

(5) $V(X)=E(X-EX)^T(X-EX)$

and

(6) $V(X)=E(X-EX)(X-EX)^T.$

Answer 1. The second definition contains more information, in the sense to be explained below, so we define variance of a vector by (6).

Question 3. Find the elements of this matrix.

Answer 3. Variance of a vector has variances of its components on the main diagonal and covariances outside it:

(7) $V(X)=\left(\begin{array}{cccc}V(X_1)&Cov(X_1,X_2)&...&Cov(X_1,X_n)\\Cov(X_2,X_1)&V(X_2)&...&Cov(X_2,X_n)\\...&...&...&...\\Cov(X_n,X_1)&Cov(X_n,X_2)&...&V(X_n)\end{array}\right).$

If you can't get this on your own, go back to Answer 2.

There is a matrix operation called trace and denoted $tr$ . It is defined only for square matrices and gives the sum of diagonal elements of a matrix.

Exercise 1. Show that $tr(V(X))=E(X-EX)^T(X-EX).$ In this sense definition (6) is more informative than (5).

Exercise 2. Show that if $EX_1=...=EX_n=0$ , then (7) becomes

$V(X)=\left(\begin{array}{cccc}EX^2_1&EX_1X_2&...&EX_1X_n\\EX_2X_1&EX^2_2&...&EX_2X_n\\...&...&...&...\\EX_nX_1&EX_nX_2&...&EX^2_n\end{array}\right).$

Tags: compatibility rule for matrix multiplication, multiplication of matrices, product of matrices, scalar product, trace of a matrix, transposed vector, University of London International Programmes, uol affiliate centre, variance of a vector

May 18

Solution to Question 2 from UoL exam 2016, zone A

category: FN3142 Quantitative Finance, Quantitative Finance no comments

Solution to Question 2 from UoL exam 2016, Zone A (FN3142)

This is another difficult question, and I don't think it will appear again in its entirety. However, the ideas applied here and here are worth repeating and will surely be on future UoL exams in some form.

Problem statement

You hold two different corporate bonds (bond A and bond B), each with a face value of $1 mln. The issuing firms have a 2% probability of defaulting on the bonds, and both the default events and the recovery values are independent of each other. Without default, the notional value is repaid, while in case of default, the recovery value is uniformly distributed between 0 and the notional value.
(a1) (20 points) Find the 1% VaR for bond A or bond B and report your calculations. (HINT: you will need the definition and properties of a uniform distribution $U[a,b]$ on $[a,b]$ .)
(a2) (60 points) Taking into account the formula for the distribution function of the sum of two independent uniformly distributed random variables on $[a,b]$ explain how you would find the 1% VaR for a portfolio combining the two bonds (A + B). Report your calculations without finding the actual value.
(b) (20 points) The 1% expected shortfall is defined as the expected loss given that the loss exceeds the 1% VaR. Find the 1% expected shortfall for bond A and report your calculations.

I made the statement shorter by referencing my post on the uniform distribution. The formulas for the triangular distribution given in the exam and examiner's comments are both wrong.

Solution

As with previous solutions related to VaR, I emphasize general methods that reduce the amount of guessing to a minimum. Besides, I work with the definition that gives a negative VaR in percents (that is, I work with returns) and then translate that value to positive dollar amounts. Translating the problem statement to returns: without default, the return on each bond is zero, while in case of default the loss is uniformly distributed between -1 and 0 (-1 corresponds to -100%). The recovery value is the notional value minus the dollar loss.

The method consists of two steps. Step 1. Derive the distribution function. Step 2. Use the usual or generalized inverse to find the VaR.

(a1) $R_A$ denotes the return on A, $D_A$ is the event that A defaults and $\bar{D}_A$ is the complement event of no default. Similar notation is employed for B. By the law of total probability for any real $x$

(1) $F_{R_A}(x)=P(R_A\le x)=P(R_A\le x|D_A)P(D_A)+P(R_A\le x|\bar{D}_A)P(\bar{D}_A).$

We want to show that for our purposes the second term can be made zero. We know that without default, return is zero. This means that conditionally we deal with a random variable that takes value 0 with probability 1. Hence, conditional on no default, by additivity of conditional probability for $x\ge 0$

(2) $P(R_A\le x|\bar{D}_A)=P(R_A<0|\bar{D}_A)+P(R_A=0|\bar{D}_A)+P(0<R_A\le x|\bar{D}_A)=1$

because the middle term on the right is 1 and the other two are zero. It follows that for $x\ge 0$ the second term in (1) is

(3) $P(R_A\le x|\bar{D}_A)P(\bar{D}_A)=1\times 0.98>0.01.$

To find the VaR, we need the whole expression in (1) to be 0.01, so we have to consider only $x<0$ , in which case the second term on the right of (1) is zero.

In the first term on the right of (1) $P(R_A\le x|D_A)$ is uniformly distributed on $[-1,0]$ , so from its distribution function it's clear that we need to consider only $-1<x<0$ :

$F_{R_A}(x)=P(R_A\le x|D_A)P(D_A)=(x+1)0.02$ .

As this should be equal to 0.01, we obtain $x=-0.5$ . The loss is $1 mln times 50% and the recovery value is half a million.

(a2) Denote $R=R_A+R_B$ . This allows us to use the formula for the convolution of two independent uniform distributions. We should remember that this is not a return on the total portfolio because it takes values between -2 and 0. By the law of total probability for any real $x$

(4) $F_R(x)=P(R\le x|D_A\cap D_B)P(D_A\cap D_B)+P(R\le x|D_A\cap \bar{D}_B)P(D_A\cap \bar{D}_B)$

$+P(R\le x|\bar{D}_A\cap D_B)P(\bar{D}_A\cap D_B)+P(R\le x|\bar{D}_A\cap \bar{D}_B)P(\bar{D}_A\cap \bar{D}_B)$ .

As above, we can show that the last term in (4) is large if $x\ge 0$ . Indeed, if no company defaults, the return is zero and, as in (2), for such $x$

$P(R\le x|\bar{D}_A\cap \bar{D}_B)=1$ .

Therefore, as in (3)

$P(R\le x|\bar{D}_A\cap \bar{D}_B)P(\bar{D}_A\cap \bar{D}_B)=1\times 0.98^2=0.9604>0.01$ .

It follows that we can restrict our attention to $x<0$ , in which case the last term on the right of (4) is zero, the second and third terms are the same and we get

(5) $F_R(x)=P(R\le x|D_A\cap D_B)0.02^2+2P(R\le x|D_A\cap \bar{D}_B)0.02\cdot 0.98$ .

In case of two defaults, $R_A,\ R_B$ are independent and distributed as $U[-1,0]$ . Hence, $R$ has a triangular distribution. Denoting its density $f_T$ , we have

(6) $P(R\le x|D_A\cap D_B)0.02^2=0.0004\int_{-2}^xf_T(x)dx,\ -2<x<0$ .

The integral here is

(7) $\int_{-2}^xf_T(x)dx=\left\{\begin{array}{ll}x^2/2+2x+2, &-2<x<-1\\1-x^2/2, &-1\le x<0\end{array}\right.$

In case of one default, we use the uniform distribution to find

(8) $2P(R\le x|D_A\cap \bar{D}_B)0.02\cdot 0.98=\left\{\begin{array}{ll}0.0392(x+1),&-1\le x<0\\0,&x<-1\end{array}\right.$ .

Plugging (6), (7), (8) in (5) and equating the result to 0.01, we get

(9) $0.0004(x^2/2+2x+2)=0.01, -2<x<-1$

and

(10) $0.0004(1-x^2/2)+0.0392(x+1)=0.01, -1\le x<0$ .

Mathematica gives two solutions for (9), none of which is in the interval (-2,-1). Of the two solutions of (10), the one which belongs to (-1,0) is -0.75. This is 37.5% of -2 and the loss is $750,000 out of $2 mln. The exam question doesn't require you to give numerical values.

(b) From the general definition of expected shortfall

$ES^{0.01}=\frac{ER_A1_{R_A\le -0.5}}{P(R_A\le -0.5)}=\frac{\int_{-1}^{-0.5}tdt}{\int_{-1}^{-0.5}dt}=\frac{-3/8}{1/2}=-0.75$

which means an expected loss of $750,000.

Tags: additivity of conditional probability, expected shortfall, recovery value of a bond, triangular distribution, University of London International Programmes, uol affiliate centre, Value at Risk

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Solution to Question 2 from UoL exam 2016, Zone A (FN3142)

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