26
Apr 23

## The magic of the distribution function

Let $X$ be a random variable. The function $F_{X}\left( x\right) =P\left(X\leq x\right) ,$ where $x$ runs over real numbers, is called a distribution function of $X.$ In statistics, many formulas are derived with the help of $F_{X}\left( x\right) .$ The motivation and properties are given here

Oftentimes, working with the distribution function is an intermediate step to obtain a density $f_{X}$ using the link

$F_{X}\left( x\right) =\int_{-\infty }^{x}f_{X}\left( t\right) dt.$

A series of exercises below show just how useful the distribution function is.

Exercise 1. Let $Y$ be a linear transformation of $X,$ that is, $Y=\sigma X+\mu ,$ where $\sigma >0$ and $\mu \in R.$ Find the link between $F_{X}$ and $F_{Y}.$ Find the link between $f_{X}$ and $f_{Y}.$
The solution is here.

The more general case of a nonlinear transformation can also be handled:

Exercise 2. Let $Y=g\left( X\right)$ where $g$ is a deterministic function. Suppose that $g$ is strictly  monotone, differentiable and $g^{-1}$ exists. Find the link between $F_{X}$ and $F_{Y}.$ Find the link between $f_{X}$ and $f_{Y}.$

Solution. The result differs depending on whether $g$ is increasing or decreasing. Let's assume the latter, so that $x_{1}\leq x_{2}$ is equivalent to $g\left( x_{1}\right) \geq g\left( x_{2}\right) .$ Also for simplicity suppose that $P\left( X=c\right) =0$ for any $c\in R.$ Then

$F_{Y}\left( y\right) =P\left( g\left( X\right) \leq y\right) =P\left( X\geq g^{-1}\left( y\right) \right) =1-P\left( X\leq g^{-1}\left( y\right) \right)=1-F_{X}\left( g^{-1}\left( y\right) \right) .$

Differentiation of this equation produces

$f_{Y}\left( y\right) =-f_{X}\left( g^{-1}\left( y\right) \right) \left( g^{-1}\left( y\right) \right) ^{\prime }=f_{X}\left( g^{-1}\left( y\right) \right) \left\vert \left( g^{-1}\left( y\right) \right) ^{\prime }\right\vert$

(the derivative of $g^{-1}$ is negative).

For an example when $g$ is not invertible see the post about the chi-squared distribution.

Exercise 3. Suppose $T=X/Y$ where $X$ and $Y$ are independent, have densities $f_{X},f_{Y}$ and $Y>0.$ What are the distribution function and density of $T?$

Solution. The joint density $f_{X,Y}$ equals $f_{X}f_{Y},$ so

$F_{T}\left( t\right) =P\left( T\leq t\right) =P\left( X\leq tY\right) =\underset{x\leq ty}{\int \int }f_{X}\left( x\right) f_{Y}\left( y\right) dxdy$

(converting a double integral to an iterated integral and remembering that $f_{Y}$ is zero on the left half-axis)

$=\int_{0}^{\infty }\left( \int_{-\infty }^{ty}f_{X}\left( x\right) dx\right) f_{Y}\left( y\right) dy=\int_{0}^{\infty }F_{X}\left( ty\right) f_{Y}\left( y\right) dy.$

Now by the Leibniz integral rule

(1) $f_{T}\left( t\right) =\int_{0}^{\infty }f_{X}\left( ty\right) yf_{Y}\left( y\right) dy.$

A different method is indicated in Activity 4.11, p.207 of J. Abdey, Guide ST2133.

Exercise 4. Let $X,Y$ be two independent random variables with densities $f_{X},f_{Y}$. Find $F_{X+Y}$ and $f_{X+Y}.$

See this post.

Exercise 5. Let $X,Y$ be two independent random variables. Find $F_{\max\left\{ X_{1},X_{2}\right\} }$ and $F_{\min \left\{ X_{1},X_{2}\right\} }.$

Solution. The inequality $\max \left\{ X_{1},X_{2}\right\} \leq x$ holds if and only if both $X_{1}\leq x$ and $X_{2}\leq x$ hold. This means that the event $\left\{ \max \left\{ X_{1},X_{2}\right\} \leq x\right\}$ coincides with the event $\left\{ X_{1}\leq x\right\} \cap \left\{ X_{2}\leq x\right\} .$ It follows by independence that

$F_{\max \left\{ X_{1},X_{2}\right\}}\left( x\right) =P\left( \max \left\{ X_{1},X_{2}\right\} \leq x\right) =P\left( \left\{ X_{1}\leq x\right\} \cap \left\{ X_{2}\leq x\right\}\right)$

$=PX_{1}\leq x)P\left( X_{2}\leq x\right) =F_{X_{1}}\left( x\right)F_{X_{2}}\left( x\right) .$

For $\min \left\{ X_{1},X_{2}\right\}$ we need one more trick - pass to the complementary event by writing $F_{\min \left\{ X_{1},X_{2}\right\} }\left( x\right) =P\left( \min \left\{ X_{1},X_{2}\right\} \leq x\right) =1-P\left(\min \left\{ X_{1},X_{2}\right\} >x\right) .$ Now we can use the fact that the event $\left\{ \min \left\{ X_{1},X_{2}\right\} >x\right\}$ coincides with the event $\left\{ X_{1}>x\right\} \cap \left\{ X_{2}>x\right\} .$ Hence, by independence

$F_{\min \left\{ X_{1},X_{2}\right\} }\left( x\right) =1-P\left( \left\{ X_{1}>x\right\} \cap \left\{ X_{2}>x\right\} \right) =1-P\left( X_{1}>x\right) P\left( X_{2}>x\right)$

$=1-\left[ 1-P\left( X_1\leq x\right) \right] \left[ 1-P\left( X_2\leq x\right) \right] =1-\left( 1-F_{X_1}\left( x\right) \right) \left( 1-F_{X_2}\left( x\right) \right) .$

27
Dec 22

## Final exam in Advanced Statistics ST2133, 2022

Unlike most UoL exams, here I tried to relate the theory to practical issues.

KBTU International School of Economics

Compiled by Kairat Mynbaev

The total for this exam is 41 points. You have two hours.

Everywhere provide detailed explanations. When answering please clearly indicate question numbers. You don’t need a calculator. As long as the formula you provide is correct, the numerical value does not matter.

Question 1. (12 points)

a) (2 points) At a casino, two players are playing on slot machines. Their payoffs $X,Y$ are standard normal and independent. Find the joint density of the payoffs.

b) (4 points) Two other players watch the first two players and start to argue what will be larger: the sum $U = X + Y$ or the difference $V = X - Y$. Find the joint density. Are variables $U,V$ independent? Find their marginal densities.

c) (2 points) Are $U,V$ normal? Why? What are their means and variances?

d) (2 points) Which probability is larger: $P(U > V)$ or $P\left( {U < V} \right)$?

e) (2 points) In this context interpret the conditional expectation $E\left( {U|V = v} \right)$. How much is it?

Reminder. The density of a normal variable $X \sim N\left( {\mu ,{\sigma ^2}} \right)$ is ${f_X}\left( x \right) = \frac{1}{{\sqrt {2\pi {\sigma ^2}} }}{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}$.

Question 2. (9 points) The distribution of a call duration $X$ of one Kcell [largest mobile operator in KZ] customer is exponential: ${f_X}\left( x \right) = \lambda {e^{ - \lambda x}},\,\,x \ge 0,\,\,{f_X}\left( x \right) = 0,\,\,x < 0.$ The number $N$ of customers making calls simultaneously is distributed as Poisson: $P\left( {N = n} \right) = {e^{ - \mu }}\frac{{{\mu ^n}}}{{n!}},\,\,n = 0,1,2,...$ Thus the total call duration for all customers is ${S_N} = {X_1} + ... + {X_N}$ for $N \ge 1$. We put ${S_0} = 0$. Assume that customers make their decisions about calling independently.

a) (3 points) Find the general formula (when ${X_1},...,{X_n}$ are identically distributed and $X,N$ are independent but not necessarily exponential and Poisson, as above) for the moment generating function of $S_N$ explaining all steps.

b) (3 points) Find the moment generating functions of $X$, $N$ and ${S_N}$ for your particular distributions.

c) (3 points) Find the mean and variance of ${S_N}$. Based on the equations you obtained, can you suggest estimators of parameters $\lambda ,\mu$?

Remark. Direct observations on the exponential and Poisson distributions are not available. We have to infer their parameters by observing ${S_N}$. This explains the importance of the technique used in Question 2.

Question 3. (8 points)

a) (2 points) For a non-negative random variable $X$ prove the Markov inequality $P\left( {X > c} \right) \le \frac{1}{c}EX,\,\,\,c > 0.$

b) (2 points) Prove the Chebyshev inequality $P\left( {|X - EX| > c} \right) \le \frac{1}{c^2}Var\left( X \right)$ for an arbitrary random variable $X$.

c) (4 points) We say that the sequence of random variables $\left\{ X_n \right\}$ converges in probability to a random variable $X$ if $P\left( {|{X_n} - X| > \varepsilon } \right) \to 0$ as $n \to \infty$ for any $\varepsilon > 0$.  Suppose that $E{X_n} = \mu$ for all $n$ and that $Var\left(X_n \right) \to 0$ as $n \to \infty$. Prove that then $\left\{X_n\right\}$ converges in probability to $\mu$.

Remark. Question 3 leads to the simplest example of a law of large numbers: if $\left\{ X_n \right\}$ are i.i.d. with finite variance, then their sample mean converges to their population mean in probability.

Question 4. (8 points)

a) (4 points) Define a distribution function. Give its properties, with intuitive explanations.

b) (4 points) Is a sum of two distribution functions a distribution function? Is a product of two distribution functions a distribution function?

Remark. The answer for part a) is here and the one for part b) is based on it.

Question 5. (4 points) The Rakhat factory prepares prizes for kids for the upcoming New Year event. Each prize contains one type of chocolates and one type of candies. The chocolates and candies are chosen randomly from two production lines, the total number of items is always 10 and all selections are equally likely.

a) (2 points) What proportion of prepared prizes contains three or more chocolates?

b) (2 points) 100 prizes have been sent to an orphanage. What is the probability that 50 of those prizes contain no more than two chocolates?

24
Oct 22

## A problem to do once and never come back

There is a problem I gave on the midterm that does not require much imagination. Just know the definitions and do the technical work, so I was hoping we could put this behind us. Turned out we could not and thus you see this post.

Problem. Suppose the joint density of variables $X,Y$ is given by

$f_{X,Y}(x,y)=\left\{\begin{array}{c}k\left( e^{x}+e^{y}\right) \text{ for }0\le y\le x\le 1, \\ 0\text{ otherwise.}\end{array}\right.$

I. Find $k$.

II. Find marginal densities of $X,Y$. Are $X,Y$ independent?

III. Find conditional densities $f_{X|Y},\ f_{Y|X}$.

IV. Find $EX,\ EY$.

When solving a problem like this, the first thing to do is to give the theory. You may not be able to finish without errors the long calculations but your grade will be determined by the beginning theoretical remarks.

### I. Finding the normalizing constant

Any density should satisfy the completeness axiom: the area under the density curve (or in this case the volume under the density surface) must be equal to one: $\int \int f_{X,Y}(x,y)dxdy=1.$ The constant $k$ chosen to satisfy this condition is called a normalizing constant. The integration in general is over the whole plain $R^{2}$ and the first task is to express the above integral as an iterated integral. This is where the domain where the density is not zero should be taken into account. There is little you can do without geometry. One example of how to do this is here.

The shape of the area $A=\left\{ (x,y):0 is determined by a) the extreme values of $x,y$ and b) the relationship between them. The extreme values are 0 and 1 for both $x$ and $y$, meaning that $A$ is contained in the square $\left\{ (x,y):0 The inequality $y means that we cut out of this square the triangle below the line $y=x$ (it is really the lower triangle because if from a point on the line $y=x$ we move down vertically, $x$ will stay the same and $y$ will become smaller than $x$).

In the iterated integral:

a) the lower and upper limits of integration for the inner integral are the boundaries for the inner variable; they may depend on the outer variable but not on the inner variable.

b) the lower and upper limits of integration for the outer integral are the extreme values for the outer variable; they must be constant.

This is illustrated in Pane A of Figure 1.

Figure 1. Integration order

Always take the inner integral in parentheses to show that you are dealing with an iterated integral.

a) In the inner integral integrating over $x$ means moving along blue arrows from the boundary $x=y$ to the boundary $x=1.$ The boundaries may depend on $y$ but not on $x$ because the outer integral is over $y.$

b) In the outer integral put the extreme values for the outer variable. Thus,

$\underset{A}{\int \int }f_{X,Y}(x,y)dxdy=\int_{0}^{1}\left(\int_{y}^{1}f_{X,Y}(x,y)dx\right) dy.$

Check that if we first integrate over $y$ (vertically along red arrows, see Pane B in Figure 1) then the equation

$\underset{A}{\int \int }f_{X,Y}(x,y)dxdy=\int_{0}^{1}\left(\int_{0}^{x}f_{X,Y}(x,y)dy\right) dx$

results.

In fact, from the definition $A=\left\{ (x,y):0 one can see that the inner interval for $x$ is $\left[ y,1\right]$ and for $y$ it is $\left[ 0,x\right] .$

### II. Marginal densities

The condition for independence of $X,Y$ is

$f_{X,Y}\left(x,y\right)=f_{X}\left( x\right) f_{Y}\left(y\right)$

(this is a direct analog of the independence condition for events $P\left(A\cap B\right) =P\left(A\right) P\left(B\right)$). In words: the joint density decomposes into a product of individual densities.

### III. Conditional densities

In this case the easiest is to recall the definition of conditional probability $P\left( A|B\right) =\frac{P\left( A\cap B\right) }{P\left(B\right) }.$ The definition of conditional densities $f_{X|Y},\ f_{Y|X}$ is quite similar:

(2) $f_{X|Y}\left( x|y\right) =\frac{f_{X,Y}\left( x,y\right) }{f_{Y}\left(y\right) },\ f_{Y|X}\left( y|x\right) =\frac{f_{X,Y}\left( x,y\right) }{f_{X}\left( x\right) }$.

Of course, $f_{Y}\left( y\right) ,f_{X}\left( x\right)$ here can be replaced by their marginal equivalents.

### IV. Finding expected values of $X,Y$$X,Y$

The usual definition $EX=\int xf_{X}\left( x\right) dx$ takes an equivalent form using the marginal density:

$EX=\int x\left( \int f_{X,Y}\left( x,y\right) dy\right) dx=\int \int xf_{X,Y}\left( x,y\right) dydx.$

Which equation to use is a matter of convenience.

Another replacement in the usual definition gives the definition of conditional expectations:

$E\left( X|Y\right) =\int xf_{X|Y}\left( x|y\right) dx,$ $E\left( Y|X\right) =\int yf_{Y|X}\left( y|x\right) dx.$

Note that these are random variables: $E\left( X|Y=y\right)$ depends in $y$ and $E\left( Y|X=x\right)$ depends on $x.$

### Solution to the problem

Being a lazy guy, for the problem this post is about I provide answers found in Mathematica:

I. $k=0.581977$

II. $f_{X}\left( x\right) =-1+e^{x}\left( 1+x\right) ,$ for $x\in[ 0,1],$ $f_{Y}\left( y\right) =e-e^{y}y,$ for $y\in \left[ 0,1\right] .$

It is readily seen that the independence condition is not satisfied.

III. $f_{X|Y}\left( x|y\right) =\frac{k\left( e^{x}+e^{y}\right) }{e-e^{y}y}$ for $0

$f_{Y|X}\left(y|x\right) =\frac{k\left(e^x+e^y\right) }{-1+e^x\left( 1+x\right) }$ for $0

IV. $EX=0.709012,$ $EY=0.372965.$

24
Oct 22

## Marginal probabilities and densities

This is to help everybody, from those who study Basic Statistics up to Advanced Statistics ST2133.

### Discrete case

Suppose in a box we have coins and banknotes of only two denominations: $1 and$5 (see Figure 1).

Figure 1. Illustration of two variables

We pull one out randomly. The division of cash by type (coin or banknote) divides the sample space (shown as a square, lower left picture) with probabilities $p_{c}$ and $p_{b}$ (they sum to one). The division by denomination ($1 or$5) divides the same sample space differently, see the lower right picture, with the probabilities to pull out $1 and$5 equal to $p_{1}$ and $p_{5}$, resp. (they also sum to one). This is summarized in the tables

 Variable 1: Cash type Prob coin $p_{c}$$p_{c}$ banknote $p_{b}$$p_{b}$
 Variable 2: Denomination Prob $1 $p_{1}$$p_{1}$$5 $p_{5}$$p_{5}$

Now we can consider joint events and probabilities (see Figure 2, where the two divisions are combined).

Figure 2. Joint probabilities

For example, if we pull out a random $item$ it can be a $coin$ and \$1 and the corresponding probability is $P\left(item=coin,\ item\ value=\1\right) =p_{c1}.$ The two divisions of the sample space generate a new division into four parts. Then geometrically it is obvious that we have four identities:

Adding over denominations: $p_{c1}+p_{c5}=p_{c},$ $p_{b1}+p_{b5}=p_{b},$

Adding over cash types: $p_{c1}+p_{b1}=p_{1},$ $p_{c5}+p_{b5}=p_{5}.$

Formally, here we use additivity of probability for disjoint events

$P\left( A\cup B\right) =P\left( A\right) +P\left( B\right) .$

In words: we can recover own probabilities of variables 1,2 from joint probabilities.

### Generalization

Suppose we have two discrete random variables $X,Y$ taking values $x_{1},...,x_{n}$ and $y_{1},...,y_{m},$ resp., and their own probabilities are $P\left( X=x_{i}\right) =p_{i}^{X},$ $P\left(Y=y_{j}\right) =p_{j}^{Y}.$ Denote the joint probabilities $P\left(X=x_{i},Y=y_{j}\right) =p_{ij}.$ Then we have the identities

(1) $\sum_{j=1}^mp_{ij}=p_{i}^{X},$ $\sum_{i=1}^np_{ij}=p_{j}^{Y}$ ($n+m$ equations).

In words: to obtain the marginal probability of one variable (say, $Y$) sum over the values of the other variable (in this case, $X$).

The name marginal probabilities is used for $p_{i}^{X},p_{j}^{Y}$ because in the two-dimensional table they arise as a result of summing table entries along columns or rows and are displayed in the margins.

### Analogs for continuous variables with densities

Suppose we have two continuous random variables $X,Y$ and their own densities are $f_{X}$ and $f_{Y}.$ Denote the joint density $f_{X,Y}$. Then replacing in (1) sums by integrals and probabilities by densities we get

(2) $\int_R f_{X,Y}\left( x,y\right) dy=f_{X}\left( x\right) ,\ \int_R f_{X,Y}\left( x,y\right) dx=f_{Y}\left( y\right) .$

In words: to obtain one marginal density (say, $f_{Y}$) integrate out the other variable (in this case, $x$).

22
Mar 22

## Blueprint for exam versions

This is the exam I administered in my class in Spring 2022. By replacing the Poisson distribution with other random variables the UoL examiners can obtain a large variety of versions with which to torture Advanced Statistics students. On the other hand, for the students the answers below can be a blueprint to fend off any assaults.

During the semester my students were encouraged to analyze and collect information in documents typed in Scientific Word or LyX. The exam was an open-book online assessment. Papers typed in Scientific Word or LyX were preferred and copying from previous analysis was welcomed. This policy would be my preference if I were to study a subject as complex as Advanced Statistics. The students were given just two hours on the assumption that they had done the preparations diligently. Below I give the model answers right after the questions.

## Midterm Spring 2022

You have to clearly state all required theoretical facts. Number all equations that you need to use in later calculations and reference them as necessary. Answer the questions in the order they are asked. When you don't know the answer, leave some space. For each unexplained fact I subtract one point. Put your name in the file name.

In questions 1-9 $X$ is the Poisson variable.

### Question 1

Define $X$ and derive the population mean and population variance of the sum $S_{n}=\sum_{i=1}^{n}X_{i}$ where $X_{i}$ is an i.i.d. sample from $X$.

Answer. $X$ is defined by $P\left( X=x\right) =e^{-\lambda }\frac{\lambda ^{x}}{x!},\ x=0,1,...$ Using $EX=\lambda$ and $Var\left( X\right) =\lambda$ (ST2133 p.80) we have

$ES_{n}=\sum EX_{i}=n\lambda ,$ $Var\left( S_{n}\right) =\sum V\left(X_{i}\right) =n\lambda$

(by independence and identical distribution). [Some students derived $EX=\lambda ,$ $Var\left( X\right) =\lambda$ instead of respective equations for sample means].

### Question 2

Derive the MGF of the standardized sample mean.

Answer. Knowing this derivation is a must because it is a combination of three important facts.

a) Let $z_{n}=\frac{\bar{X}-E\bar{X}}{\sigma \left( \bar{X}\right) }.$ Then $z_{n}=\frac{nS_{n}-EnS_{n}}{\sigma \left( nS_{n}\right) }=\frac{S_{n}-ES_{n}}{\sigma \left( S_{n}\right) },$ so standardizing $\bar{X}$ and $S_{n}$ gives the same result.

b) The MGF of $S_{n}$ is expressed through the MGF of $X$:

$M_{S_{n}}\left( t\right)=Ee^{S_{n}t}=Ee^{X_{1}t+...+X_{n}t}=Ee^{X_{1}t}...e^{X_{n}t}=$

(independence) $=Ee^{X_{1}t}...Ee^{X_{n}t}=$ (identical distribution) $=\left[ M_{X}\left( t\right) \right] ^{n}.$

c) If $X$ is a linear transformation of $Y,$ $X=a+bY,$ then

$M_{X}\left( t\right) =Ee^{X}=Ee^{\left( a+bY\right) t}=e^{at}Ee^{Y\left(bt\right) }=e^{at}M_{Y}\left( bt\right) .$

When answering the question we assume any i.i.d. sample from a population with mean $\mu$ and population variance $\sigma ^{2}$:

Putting in c) $a=-\frac{ES_{n}}{\sigma \left( S_{n}\right) },$ $b=\frac{1}{\sigma \left( S_{n}\right) }$ and using a) we get

$M_{z_{n}}\left( t\right) =E\exp \left( \frac{S_{n}-ES_{n}}{\sigma \left(S_{n}\right) }t\right) =e^{-ES_{n}t/\sigma \left( S_{n}\right)}M_{S_{n}}\left( t/\sigma \left( S_{n}\right) \right)$

(using b) and $ES_{n}=n\mu ,$ $Var\left( S_{n}\right) =n\sigma ^{2}$)

$=e^{-ES_{n}t/\sigma \left( S_{n}\right) }\left[ M_{X}\left( t/\sigma \left(S_{n}\right) \right) \right] ^{n}=e^{-n\mu t/\left( \sqrt{n}\sigma \right) }\left[ M_{X}\left( t/\left( \sqrt{n}\sigma \right) \right) \right] ^{n}.$

This is a general result which for the Poisson distribution can be specified as follows. From ST2133, example 3.38 we know that $M_{X}\left( t\right)=\exp \left( \lambda \left( e^{t}-1\right) \right)$. Therefore, we obtain

$M_{z_{n}}\left( t\right) =e^{-\sqrt{\lambda }t}\left[ \exp \left( \lambda\left( e^{t/\left( n\sqrt{\lambda }\right) }-1\right) \right) \right] ^{n}=e^{-t\sqrt{\lambda }+n\lambda \left( e^{t/\left( n\sqrt{\lambda }\right)}-1\right) }.$

[Instead of $M_{z_n}$ some students gave $M_X$.]

### Question 3

Derive the cumulant generating function of the standardized sample mean.

Answer. Again, there are a couple of useful general facts.

I) Decomposition of MGF around zero. The series $e^{x}=\sum_{i=0}^{\infty }\frac{x^{i}}{i!}$ leads to

$M_{X}\left( t\right) =Ee^{tX}=E\left( \sum_{i=0}^{\infty }\frac{t^{i}X^{i}}{i!}\right) =\sum_{i=0}^{\infty }\frac{t^{i}}{i!}E\left( X^{i}\right)=\sum_{i=0}^{\infty }\frac{t^{i}}{i!}\mu _{i}$

where $\mu _{i}=E\left( X^{i}\right)$ are moments of $X$ and $\mu_{0}=EX^{0}=1.$ Differentiating this equation yields

$M_{X}^{(k)}\left( t\right) =\sum_{i=k}^{\infty }\frac{t^{i-k}}{\left(i-k\right) !}\mu _{i}$

and setting $t=0$ gives the rule for finding moments from MGF: $\mu_{k}=M_{X}^{(k)}\left( 0\right) .$

II) Decomposition of the cumulant generating function around zero. $K_{X}\left( t\right) =\log M_{X}\left( t\right)$ can also be decomposed into its Taylor series:

$K_{X}\left( t\right) =\sum_{i=0}^{\infty }\frac{t^{i}}{i!}\kappa _{i}$

where the coefficients $\kappa _{i}$ are called cumulants and can be found using $\kappa _{k}=K_{X}^{(k)}\left( 0\right)$. Since

$K_{X}^{\prime }\left( t\right) =\frac{M_{X}^{\prime }\left( t\right) }{M_{X}\left( t\right) }$ and $K_{X}^{\prime \prime }\left( t\right) =\frac{M_{X}^{\prime \prime }\left( t\right) M_{X}\left( t\right) -\left(M_{X}^{\prime }\left( t\right) \right) ^{2}}{M_{X}^{2}\left( t\right) }$

we have

$\kappa _{0}=\log M_{X}\left( 0\right) =0,$ $\kappa _{1}=\frac{M_{X}^{\prime}\left( 0\right) }{M_{X}\left( 0\right) }=\mu _{1},$

$\kappa _{2}=\mu_{2}-\mu _{1}^{2}=EX^{2}-\left( EX\right) ^{2}=Var\left( X\right) .$

Thus, for any random variable $X$ with mean $\mu$ and variance $\sigma ^{2}$ we have

$K_{X}\left( t\right) =\mu t+\frac{\sigma ^{2}t^{2}}{2}+$ terms of higher order for $t$ small.

III) If $X=a+bY$ then by c)

$K_{X}\left( t\right) =K_{a+bY}\left( t\right) =\log \left[e^{at}M_{Y}\left( bt\right) \right] =at+K_{X}\left( bt\right) .$

IV) By b)

$K_{S_{n}}\left( t\right) =\log \left[ M_{X}\left( t\right) \right]^{n}=nK_{X}\left( t\right) .$

Using III), $z_{n}=\frac{S_{n}-ES_{n}}{\sigma \left( S_{n}\right) }$ and then IV) we have

$K_{z_{n}}\left( t\right) =\frac{-ES_{n}}{\sigma \left( S_{n}\right) }t+K_{S_{n}}\left( \frac{t}{\sigma \left( S_{n}\right) }\right) =\frac{-ES_{n}}{\sigma \left( S_{n}\right) }t+nK_{X}\left( \frac{t}{\sigma \left(S_{n}\right) }\right) .$

For the last term on the right we use the approximation around zero from II):

$K_{z_{n}}\left( t\right) =\frac{-ES_{n}}{\sigma \left( S_{n}\right) }t+nK_{X}\left( \frac{t}{\sigma \left( S_{n}\right) }\right) \approx \frac{-ES_{n}}{\sigma \left( S_{n}\right) }t+n\mu \frac{t}{\sigma \left(S_{n}\right) }+n\frac{\sigma ^{2}}{2}\left( \frac{t}{\sigma \left(S_{n}\right) }\right) ^{2}$ $=-\frac{n\mu }{\sqrt{n}\sigma }t+n\mu \frac{t}{\sqrt{n}\sigma }+n\frac{\sigma ^{2}}{2}\left( \frac{t}{\sigma \left( S_{n}\right) }\right)^{2}=t^{2}/2.$

[Important. Why the above steps are necessary? Passing from the series $M_{X}\left( t\right) =\sum_{i=0}^{\infty }\frac{t^{i}}{i!}\mu _{i}$ to the series for $K_{X}\left( t\right) =\log M_{X}\left( t\right)$ is not straightforward and can easily lead to errors. It is not advisable in case of the Poisson to derive $K_{z_{n}}$ from $M_{z_{n}}\left( t\right)=$ $e^{-t\sqrt{\lambda }+n\lambda \left( e^{t/\left( n\sqrt{\lambda }\right)}-1\right) }$.]

### Question 4

Prove the central limit theorem using the cumulant generating function you obtained.

Answer. In the previous question we proved that around zero

$K_{z_{n}}\left( t\right) \rightarrow \frac{t^{2}}{2}.$

This implies that

(1) $M_{z_{n}}\left( t\right) \rightarrow e^{t^{2}/2}$ for each $t$ around zero.

But we know that for a standard normal $X$ its MGF is $M_{X}\left( t\right)=\exp \left( \mu t+\frac{\sigma ^{2}t^{2}}{2}\right)$ (ST2133 example 3.42) and hence for the standard normal

(2) $M_{z}\left( t\right) =e^{t^{2}/2}.$

Theorem (link between pointwise convergence of MGFs of $\left\{X_{n}\right\}$ and convergence in distribution of $\left\{ X_{n}\right\}$) Let $\left\{ X_{n}\right\}$ be a sequence of random variables and let $X$ be some random variable. If $M_{X_{n}}\left( t\right)$ converges for each $t$ from a neighborhood of zero to $M_{X}\left( t\right)$, then $X_{n}$ converges in distribution to $X.$

Using (1), (2) and this theorem we finish the proof that $z_{n}$ converges in distribution to the standard normal, which is the central limit theorem.

### Question 5

State the factorization theorem and apply it to show that $U=\sum_{i=1}^{n}X_{i}$ is a sufficient statistic.

Answer. The solution is given on p.180 of ST2134. For $x_{i}=1,...,n$ the joint density is

(3) $f_{X}\left( x,\lambda \right) =\prod\limits_{i=1}^{n}e^{-\lambda }\frac{\lambda ^{x_{i}}}{x_{i}!}=\frac{\lambda ^{\Sigma x_{i}}e^{-n\lambda }}{\Pi _{i=1}^{n}x_{i}!}.$

To satisfy the Fisher-Neyman factorization theorem set

$g\left( \sum x_{i},\lambda \right) =\lambda ^{\Sigma x_{i}e^{-n\lambda }},\ h\left( x\right) =\frac{1}{\Pi _{i=1}^{n}x_{i}!}$

and then we see that $\sum x_{i}$ is a sufficient statistic for $\lambda .$

### Question 6

Find a minimal sufficient statistic for $\lambda$ stating all necessary theoretical facts.

AnswerCharacterization of minimal sufficiency A statistic $T\left(X\right)$ is minimal sufficient if and only if level sets of $T$ coincide with sets on which the ratio $f_{X}\left( x,\theta \right) /f_{X}\left(y,\theta \right)$ does not depend on $\theta .$

From (3)

$f_{X}\left( x,\lambda \right) /f_{X}\left( y,\lambda \right) =\frac{\lambda^{\Sigma x_{i}}e^{-n\lambda }}{\Pi _{i=1}^{n}x_{i}!}\left[ \frac{\lambda^{\Sigma y_{i}}e^{-n\lambda }}{\Pi _{i=1}^{n}y_{i}!}\right] ^{-1}$ $=\lambda ^{\left[ \Sigma x_{i}-\Sigma y_{i}\right] }\frac{\Pi _{i=1}^{n}y_{i}!}{\Pi_{i=1}^{n}x_{i}!}.$

The expression on the right does not depend on $\lambda$ if and only of $\Sigma x_{i}=\Sigma y_{i}.$ The last condition describes level sets of $T\left( X\right) =\sum X_{i}.$ Thus it is minimal sufficient.

### Question 7

Find the Method of Moments estimator of the population mean.

Answer. The idea of the method is to take some populational property (for example, $EX=\lambda$) and replace the population characteristic (in this case $EX$) by its sample analog ($\bar{X}$) to obtain a MM estimator. In our case $\hat{\lambda}_{MM}=\bar{X}.$ [Try to do this for the Gamma distribution].

### Question 8

Find the Fisher information.

Answer. From Problem 5 the log-likelihood is

$l_{X}\left( \lambda ,x\right) =-n\lambda +\sum x_{i}\log \lambda -\sum \log\left( x_{i}!\right) .$

Hence the score function is (see Example 2.30 in ST2134)

$s_{X}\left( \lambda ,x\right) =\frac{\partial }{\partial \lambda }l_{X}\left( \lambda ,x\right) =-n+\frac{1}{\lambda }\sum x_{i}.$

Then

$\frac{\partial ^{2}}{\partial \lambda ^{2}}l_{X}\left( \lambda ,x\right) =-\frac{1}{\lambda ^{2}}\sum x_{i}$

and the Fisher information is

$I_{X}\left( \lambda \right) =-E\left( \frac{\partial ^{2}}{\partial \lambda^{2}}l_{X}\left( \lambda ,x\right) \right) =\frac{1}{\lambda ^{2}}E\sum X_{i}=\frac{n\lambda }{\lambda ^{2}}=\frac{n}{\lambda }.$

### Question 9

Derive the Cramer-Rao lower bound for $V\left( \bar{X}\right)$ for a random sample.

Answer. (See Example 3.17 in ST2134) Since $\bar{X}$ is an unbiased estimator of $\lambda$ by Problem 1, from the Cramer-Rao theorem we know that

$V\left( \bar{X}\right) \geq \frac{1}{I_{X}\left( \lambda \right) }=\frac{\lambda }{n}$

and in fact by Problem 1 this lower bound is attained.

19
Feb 22

## Estimation of parameters of a normal distribution

Here we show that the knowledge of the distribution of $s^{2}$ for linear regression allows one to do without long calculations contained in the guide ST 2134 by J. Abdey.

Theorem. Let $y_{1},...,y_{n}$ be independent observations from $N\left( \mu,\sigma ^{2}\right)$. 1) $s^{2}\left( n-1\right) /\sigma ^{2}$ is distributed as $\chi _{n-1}^{2}.$ 2) The estimators $\bar{y}$ and $s^{2}$ are independent. 3) $Es^{2}=\sigma ^{2},$ 4) $Var\left( s^{2}\right) =\frac{2\sigma ^{4}}{n-1},$ 5) $\frac{s^{2}-\sigma ^{2}}{\sqrt{2\sigma ^{4}/\left(n-1\right) }}$ converges in distribution to $N\left( 0,1\right) .$

Proof. We can write $y_{i}=\mu +e_{i}$ where $e_{i}$ is distributed as $N\left( 0,\sigma ^{2}\right) .$ Putting $\beta =\mu ,\ y=\left(y_{1},...,y_{n}\right) ^{T},$ $e=\left( e_{1},...,e_{n}\right) ^{T}$ and $X=\left( 1,...,1\right) ^{T}$ (a vector of ones) we satisfy (1) and (2). Since $X^{T}X=n,$ we have $\hat{\beta}=\bar{y}.$ Further,

$r\equiv y-X\hat{\beta}=\left( y_{1}-\bar{y},...,y_{n}-\bar{y}\right) ^{T}$

and

$s^{2}=\left\Vert r\right\Vert ^{2}/\left( n-1\right) =\sum_{i=1}^{n}\left(y_{i}-\bar{y}\right) ^{2}/\left( n-1\right) .$

Thus 1) and 2) follow from results for linear regression.

3) For a normal variable $X$ its moment generating function is $M_{X}\left( t\right) =\exp \left(\mu t+\frac{1}{2}\sigma ^{2}t^{2}\right)$ (see Guide ST2133, 2021, p.88). For the standard normal we get

$M_{z}^{\prime }\left( t\right) =\exp \left(\frac{1}{2}t^{2}\right) t,$ $M_{z}^{\prime \prime }\left( t\right) =\exp \left( \frac{1}{2}t^{2}\right) (t^{2}+1),$

$M_{z}^{\prime \prime \prime}\left( t\right) =\exp \left( \frac{1}{2}t^{2}\right) (t^{3}+2t+t),$ $M_{z}^{(4)}\left( t\right) =\exp \left( \frac{1}{2}t^{2}\right)(t^{4}+6t^{2}+3).$

Applying the general property $EX^{r}=M_{X}^{\left(r\right) }\left( 0\right)$ (same guide, p.84) we see that

$Ez=0,$ $Ez^{2}=1,$ $Ez^{3}=0,$ $Ez^{4}=3,$

$Var(z)=1,$ $Var\left( z^{2}\right) =Ez^{4}-\left( Ez^{2}\right)^{2}=3-1=2.$

Therefore

$Es^{2}=\frac{\sigma ^{2}}{n-1}E\left( z_{1}^{2}+...+z_{n-1}^{2}\right) =\frac{\sigma ^{2}}{n-1}\left( n-1\right)=\sigma ^{2}.$

4) By independence of standard normals

$Var\left( s^{2}\right) =$ $\left(\frac{\sigma ^{2}}{n-1}\right) ^{2}\left[ Var\left( z_{1}^{2}\right)+...+Var\left( z_{n-1}^{2}\right) \right] =\frac{\sigma ^{4}}{\left(n-1\right) ^{2}}2\left( n-1\right) =\frac{2\sigma ^{4}}{n-1}.$

5) By standardizing $s^{2}$ we have $\frac{s^{2}-Es^{2}}{\sigma \left(s^{2}\right) }=\frac{s^{2}-\sigma ^{2}}{\sqrt{2\sigma ^{4}/\left( n-1\right) }}$ and this converges in distribution to $N\left( 0,1\right)$ by the central limit theorem.

5
Feb 22

## Sufficiency and minimal sufficiency

### Sufficient statistic

I find that in the notation of a statistic it is better to reflect the dependence on the argument. So I write $T\left( X\right)$ for a statistic, where $X$ is a sample, instead of a faceless $U$ or $V.$

Definition 1. The statistic $T\left( X\right)$ is called sufficient for the parameter $\theta$ if the distribution of $X$ conditional on $T\left( X\right)$ does not depend on $\theta .$

The main results on sufficiency and minimal sufficiency become transparent if we look at them from the point of view of Maximum Likelihood (ML) estimation.

Let $f_{X}\left( x,\theta \right)$ be the joint density of the vector $X=\left( X_{1},...,X_{n}\right)$, where $\theta$ is a parameter (possibly a vector). The ML estimator is obtained by maximizing over $\theta$ the function $f_{X}\left( x,\theta \right)$ with $x=\left(x_{1},...,x_{n}\right)$ fixed at the observed data. The estimator depends on the data and can be denoted $\hat{\theta}_{ML}\left( x\right) .$

Fisher-Neyman theorem. $T\left( X\right)$ is sufficient for $\theta$ if and only if the joint density can be represented as

(1) $f_{X}\left( x,\theta \right) =g\left( T\left( x\right) ,\theta \right) k\left( x\right)$

where, as the notation suggests, $g$ depends on $x$ only through $T\left(x\right)$ and $k$ does not depend on $\theta .$

Maximizing the left side of (1) is the same thing as maximizing $g\left(T\left( x\right) ,\theta \right)$ because $k$ does not depend on $\theta .$ But this means that $\hat{\theta}_{ML}\left( x\right)$ depends on $x$ only through $T\left( x\right) .$ A sufficient statistic is all you need to find the ML estimator. This interpretation is easier to understand than the definition of sufficiency.

### Minimal sufficient statistic

Definition 2. A sufficient statistic $T\left( X\right)$ is called minimal sufficient if for any other statistic $S\left( X\right)$ there exists a function $g$ such that $T\left( X\right) =g\left( S\left( X\right) \right) .$

A level set is a set of type $\left\{ x:T\left( x\right) =c\right\} ,$ for a constant $c$ (which in general can be a constant vector). See the visualization of level sets.  A level set is also called a preimage and denoted $T^{-1}\left( c\right) =\left\{ x:T\left(x\right) =c\right\} .$ When $T$ is one-to-one the preimage contains just one point. When $T$ is not one-to-one the preimage contains more than one point. The wider it is the less information about the sample carries the statistic (because many data sets are mapped to a single point and you cannot tell one data set from another by looking at the statistic value). In the definition of the minimal sufficient statistic we have

$\left\{x:T\left( X\right) =c\right\} =\left\{ x:g\left( S\left( X\right) \right)=c\right\} =\left\{ x:S\left( X\right) \in g^{-1}\left( c\right) \right\} .$

Since $g^{-1}\left( c\right)$ generally contains more than one point, this shows that the level sets of $T\left( X\right)$ are generally wider than those of $S\left( X\right) .$ Since this is true for any $S\left( X\right) ,$ $T\left( X\right)$ carries less information about $X$ than any other statistic.

Definition 2 is an existence statement and is difficult to verify directly as there are words "for any" and "exists". Again it's better to relate it to ML estimation.

Suppose for two sets of data $x,y$ there is a positive number $k\left(x,y\right)$ such that

(2) $f_{X}\left( x,\theta \right) =k\left( x,y\right) f_{X}\left( y,\theta\right) .$

Maximizing the left side we get the estimator $\hat{\theta}_{ML}\left(x\right) .$ Maximizing $f_{X}\left( y,\theta \right)$ we get $\hat{\theta}_{ML}\left( y\right) .$ Since $k\left( x,y\right)$ does not depend on $\theta ,$ (2) tells us that

$\hat{\theta}_{ML}\left( x\right) =\hat{\theta}_{ML}\left( y\right) .$

Thus, if two sets of data $x,y$ satisfy (2), the ML method cannot distinguish between $x$ and $y$ and supplies the same estimator. Let us call $x,y$ indistinguishable if there is a positive number $k\left( x,y\right)$ such that (2) is true.

An equation $T\left( x\right) =T\left( y\right)$ means that $x,y$ belong to the same level set.

Characterization of minimal sufficiency. A statistic $T\left( X\right)$ is minimal sufficient if and only if its level sets coincide with sets of indistinguishable $x,y.$

The advantage of this formulation is that it relates a geometric notion of level sets to the ML estimator properties. The formulation in the guide by J. Abdey is:

A statistic $T\left( X\right)$ is minimal sufficient if and only if the equality $T\left( x\right) =T\left( y\right)$ is equivalent to (2).

Rewriting (2) as

(3) $f_{X}\left( x,\theta \right) /f_{X}\left( y,\theta \right) =k\left(x,y\right)$

we get a practical way of finding a minimal sufficient statistic: form the ratio on the left of (3) and find the sets along which the ratio does not depend on $\theta .$ Those sets will be level sets of $T\left( X\right) .$

28
Dec 21

## Chi-squared distribution

This post is intended to close a gap in J. Abdey's guide ST2133, which is absence of distributions widely used in Econometrics.

### Chi-squared with one degree of freedom

Let $X$ be a random variable and let $Y=X^{2}.$

Question 1. What is the link between the distribution functions of $Y$ and $X?$

Chart 1. Inverting a square function

The start is simple: just follow the definitions. $F_{Y}\left( y\right)=P\left( Y\leq y\right) =P\left( X^{2}\leq y\right) .$ Assuming that $y>0$, on Chart 1 we see that $\left\{ x:x^{2}\leq y\right\} =\left\{x: -\sqrt{y}\leq x\leq \sqrt{y}\right\} .$ Hence, using additivity of probability,

(1) $F_{Y}\left( y\right) =P\left( -\sqrt{y}\leq X\leq \sqrt{y}\right) =P\left( X\leq \sqrt{y}\right) -P\left( X<-\sqrt{y}\right)$

$=F_{X}\left( \sqrt{y}\right) -F_{X}\left( -\sqrt{y}\right) .$

The last transition is based on the assumption that $P\left( X for all $x$, which is maintained for continuous random variables throughout the guide by Abdey.

Question 2. What is the link between the densities of $X$ and $Y=X^{2}?$ By the Leibniz integral rule (1) implies

(2) $f_{Y}\left( y\right) =f_{X}\left( \sqrt{y}\right) \frac{1}{2\sqrt{y}} +f_{X}\left( -\sqrt{y}\right) \frac{1}{2\sqrt{y}}.$

Exercise. Assuming that $g$ is an increasing differentiable function with the inverse $h$ and $Y=g(X)$ answer questions similar to 1 and 2.

See the definition of $\chi _{1}^{2}.$ Just applying (2) to $X=z$ and $Y=z^{2}=\chi _{1}^{2}$ we get

$f_{\chi _{1}^{2}}\left( y\right) =\frac{1}{\sqrt{2\pi }}e^{-y/2}\frac{1}{2 \sqrt{y}}+\frac{1}{\sqrt{2\pi }}e^{-y/2}\frac{1}{2\sqrt{y}}=\frac{1}{\sqrt{ 2\pi }}y^{1/2-1}e^{-y/2},\ y>0.$

Since $\Gamma \left( 1/2\right) =\sqrt{\pi },$ the procedure for identifying the gamma distribution gives

$f_{\chi _{1}^{2}}\left( x\right) =\frac{1}{\Gamma \left( 1/2\right) }\left( 1/2\right) ^{1/2}x^{1/2-1}e^{-x/2}=f_{1/2,1/2}\left( x\right) .$

We have derived the density of the chi-squared variable with one degree of freedom, see also Example 3.52, J. Abdey, Guide ST2133.

### General chi-squared

For $\chi _{n}^{2}=z_{1}^{2}+...+z_{n}^{2}$ with independent standard normals $z_{1},...,z_{n}$ we can write $\chi _{n}^{2}=\chi _{1}^{2}+...+\chi _{1}^{2}$ where the chi-squared variables on the right are independent and all have one degree of freedom. This is because deterministic (here quadratic) functions of independent variables are independent.

Recall that the gamma density is closed under convolutions with the same $\alpha .$ Then by the convolution theorem we get

$f_{\chi _n^2}=f_{\chi _1^2}\ast ... \ast f_{\chi_1^2}=f_{1/2,1/2} \ast ... \ast f_{1/2,1/2}$ $=f_{1/2,n/2}=\frac{1}{\Gamma \left( n/2\right) 2^{n/2}}x^{n/2-1}e^{-x/2}.$
27
Dec 21

## Gamma distribution

Definition. The gamma distribution $Gamma\left( \alpha ,\nu \right)$ is a two-parametric family of densities. For $\alpha >0,\nu >0$ the density is defined by

$f_{\alpha ,\nu }\left( x\right) =\frac{1}{\Gamma \left( \nu \right) }\alpha ^{\nu }x^{\nu -1}e^{-\alpha x},\ x>0;$ $f_{\alpha ,\nu }\left( x\right) =0,\ x<0.$

Obviously, you need to know what is a gamma function. My notation of the parameters follows Feller, W. An Introduction to Probability Theory and its Applications, Volume II, 2nd edition (1971). It is different from the one used by J. Abdey in his guide ST2133.

### Property 1

It is really a density because

$\frac{1}{\Gamma \left( \nu \right) }\alpha ^{\nu }\int_{0}^{\infty }x^{\nu -1}e^{-\alpha x}dx=$ (replace $\alpha x=t$)

$=\frac{1}{\Gamma \left( \nu \right) }\alpha ^{\nu }\int_{0}^{\infty }t^{\nu -1}\alpha ^{1-\nu -1}e^{-t}dt=1.$

Suppose you see an expression $x^{a}e^{-bx}$ and need to determine which gamma density this is. The power of the exponent gives you $\alpha =b$ and the power of $x$ gives you $\nu =a+1.$ It follows that the normalizing constant should be $\frac{1}{\Gamma \left( a+1\right) }b^{a+1}$ and the density is $\frac{1}{\Gamma \left( a+1\right) }b^{a+1}x^{a}e^{-bx},$ $x>0.$

### Property 2

The most important property is that the family of gamma densities with the same $\alpha$ is closed under convolutions. Because of the associativity property $f_{X}\ast f_{Y}\ast f_{Z}=\left( f_{X}\ast f_{Y}\right) \ast f_{Z}$ it is enough to prove this for the case of two gamma densities.

First we want to prove

(1) $\left( f_{\alpha ,\mu }\ast f_{\alpha ,\nu }\right) \left( x\right) = \frac{\Gamma \left( \mu +\nu \right) }{\Gamma \left( \mu \right) \Gamma \left( \nu \right) }\int_{0}^{1}\left( 1-t\right) ^{\mu -1}t^{\nu -1}dt\times f_{\alpha ,\mu +\nu }(x).$

Start with the general definition of convolution and recall where the density vanishes:

$\left( f_{\alpha ,\mu }\ast f_{\alpha ,\nu }\right) \left( x\right) =\int_{-\infty }^{\infty }f_{\alpha ,\mu }\left( x-y\right) f_{\alpha ,\nu }\left( y\right) dy=\int_{0}^{x}f_{\alpha ,\mu }\left( x-y\right) f_{\alpha ,\nu }\left( y\right) dy$

(plug the densities and take out the constants)

$=\int_{0}^{x}\left[ \frac{1}{\Gamma \left( \mu \right) }\alpha ^{\mu }\left( x-y\right) ^{\mu -1}e^{-\alpha \left( x-y\right) }\right] \left[ \frac{1}{\Gamma \left( \nu \right) }\alpha ^{\nu }y^{\nu -1}e^{-\alpha y} \right] dy$ $=\frac{\alpha ^{\mu +\nu }e^{-\alpha x}}{\Gamma \left( \mu \right) \Gamma \left( \nu \right) }\int_{0}^{x}\left( x-y\right) ^{\mu -1}y^{\nu -1}dy$

(replace $y=xt$)

$=\frac{\Gamma \left( \mu +\nu \right) }{\Gamma \left( \mu \right) \Gamma \left( \nu \right) }\frac{\alpha ^{\mu +\nu }x^{\mu +\nu -1}e^{-\alpha x}}{ \Gamma \left( \mu +\nu \right) }\int_{0}^{1}\left( 1-t\right) ^{\mu -1}t^{\nu -1}dt$ $=\frac{\Gamma \left( \mu +\nu \right) }{\Gamma \left( \mu \right) \Gamma \left( \nu \right) }\int_{0}^{1}\left( 1-t\right) ^{\mu -1}t^{\nu -1}dt\times f_{\alpha ,\mu +\nu }\left( x\right).$

Thus (1) is true. Integrating it we have

$\int_{R}\left( f_{\alpha ,\mu }\ast f_{\alpha ,\nu }\right) \left( x\right) dx=\frac{\Gamma \left( \mu +\nu \right) }{\Gamma \left( \mu \right) \Gamma \left( \nu \right) }\int_{0}^{1}\left( 1-t\right) ^{\mu -1}t^{\nu -1}dt\times \int_{R}f_{\alpha ,\mu +\nu }\left( x\right) dx.$

We know that the convolution of two densities is a density. Therefore the last equation implies

$\frac{\Gamma \left( \mu +\nu \right) }{\Gamma \left( \mu \right) \Gamma \left( \nu \right) }\int_{0}^{1}\left( 1-t\right) ^{\mu -1}t^{\nu -1}dt=1$

and

$f_{\alpha ,\mu }\ast f_{\alpha ,\nu }=f_{\alpha ,\mu +\nu },\ \mu ,\nu >0.$

Alternative proof. The moment generating function of a sum of two independent beta distributions with the same $\alpha$ shows that this sum is again a beta distribution with the same $\alpha$, see pp. 141, 209 in the guide ST2133.

26
Dec 21

## Gamma function

### Gamma function

The gamma function and gamma distribution are two different things. This post is about the former and is a preparatory step to study the latter.

Definition. The gamma function is defined by

$\Gamma \left( t\right) =\int_{0}^{\infty }x^{t-1}e^{-x}dx,\ t> 0.$

The integrand $f(t)=x^{t-1}e^{-x}$ is smooth on $\left( 0,\infty \right) ,$ so its integrability is determined by its behavior at $\infty$ and $0$. Because of the exponent, it is integrable in the neighborhood of $\infty .$ The singularity at $0$ is integrable if $t>0.$ In all calculations involving the gamma function one should remember that its argument should be positive.

## Properties

1) Factorial-like property. Integration by parts shows that

$\Gamma \left( t\right) =-\int_{0}^{\infty }x^{t-1}\left( e^{-x}\right) ^{\prime }dx=-x^{t-1}e^{-x}|_{0}^{\infty }+\left( t-1\right) \int_{0}^{\infty }x^{t-2}e^{-x}dx$

$=\left( t-1\right) \Gamma \left( t-1\right)$ if $t>1.$

2) $\Gamma \left( 1\right) =1$ because $\int_{0}^{\infty }e^{-x}dx=1.$

3) Combining the first two properties we see that for a natural $n$

$\Gamma \left( n+1\right) =n\Gamma ( n) =...=n\times \left( n-1\right) ...\times 1\times \Gamma \left( 1\right) =n!$

Thus the gamma function extends the factorial to non-integer $t>0.$

4) $\Gamma \left( 1/2\right) =\sqrt{\pi }.$

Indeed, using the density $f_{z}$ of the standard normal $z$ we see that

$\Gamma \left( 1/2\right) =\int_{0}^{\infty }x^{-1/2}e^{-x}dx=$

(replacing $x^{1/2}=u$)

$=\int_{0}^{\infty }\frac{1}{u}e^{-u^{2}}2udu=2\int_{0}^{\infty }e^{-u^{2}}du=\int_{-\infty }^{\infty }e^{-u^{2}}du=$

(replacing $u=z/\sqrt{2}$)

$=\frac{\sqrt{\pi }}{\sqrt{2\pi }}\int_{-\infty }^{\infty }e^{-z^{2}/2}dz= \sqrt{\pi }\int_{R}f_{z}\left( t\right) dt=\sqrt{\pi }.$

Many other properties are not required in this course.