13
Mar 24

## My book on Early Education and Brainbuilding

has been published on Apple books.

One of the main ideas concerns what I call an internal vision.

If you are interested in reading the book, send me a message, and I will give you a promo code. Note: as of now, the book is not available in Kazakhstan.

## Publisher Description

The book contains recommendations for early education (preschool and primary school) and brainbuilding (adults). The emphasis is on intensive methods that give strong and lasting results and, correspondingly, require a considerable effort. Most advice will be useful also to school teachers as it explains how to develop basic supporting skills, such as memory, logic and imagination.

The author, Kairat Mynbaev, is a mathematician and university professor in Kazakhstan. He did research in mathematics, statistics, economics and finance. Two of his Math books have been published in the USA. The book reflects his personal experience in brainbuilding and teaching kids.

Book cover

4
Mar 24

## AR(1) model: Tesla stock versus Tesla return

Question. You run two AR(1) regressions: 1) for Tesla stock price $Y_{t}$, $Y_{t}=\alpha +\beta Y_{t-1}+\varepsilon _{t}$, and 2) for its return $R_{t}$$R_{t}=\phi +\psi R_{t-1}+\delta _{t}$. Here the errors $\varepsilon _{t},\delta _{t}$ are i.i.d. normal with mean $0$ and variance $\sigma ^{2}.$ Based on the 5-year chart of the stock price below (see Chart 1), what would be your expectations about the coefficients $\alpha ,$ $\beta ,$ $\phi ,$ and $\psi ?$

Chart 1. 5-year chart of TSLA stock. Source: barchart.com

Answer. Suppose that instead of the time series model $Y_{t}=\alpha +\beta Y_{t-1}+\varepsilon _{t}$ we have a simple regression $Y_{t}=\alpha +\beta X_{t}+\varepsilon _{t}$ and on the stock chart we have the values of $X_{t}$ on the horizontal axis and the values of $Y_{t}$ on the vertical axis. Then instead of the time series chart we would have a scatterplot. Drawing a straight line to approximate the cloud of observed pairs $\left(X_{t},Y_{t}\right)$, we can see that both $\alpha$ and $\beta$ must be positive (see Chart 2). The same intuition applies to the time series model $Y_{t}=\alpha +\beta Y_{t-1}+\varepsilon _{t}.$

Chart 2. Same chart of Tesla stock viewed as a scatterplot with fitted line

Table 1 contains estimation results for the first model.

 Coefficient Estimate p-value $\alpha$ 152.282 0.023 $\beta$ 0.9973 0.000

The fundamental difference between the stock and its return is that the return cannot be trending for extended periods of time. The intuition is that if, for example, the return for some stock is persistently positive, then everybody starts investing in it and seeing sizable profits. However, the paper profits must be realized sooner or later, which means investors at some point will start selling the stock and the return becomes negative. As a result, the return must oscillate around zero. This intuition is confirmed in Chart 3, which displays the return for Tesla stock, and in Chart 4, which is a nonparametric estimation of the density of that return.

Chart 3. Chart for return on Tesla, from Stata

The straight line that approximates the cloud of observed pairs $\left(R_{t-1},R_{t}\right)$ should be very close to the $x$ axis. That is, both $\phi$ and $\psi$ should be very close to zero.

Chart 4. The density of return on Tesla is centered almost at zero

See estimation results in Table 2.

 Coefficient Estimate p-value $\phi$ 0.0018 0.106 $\psi$ -0.0056 0.795

22
Nov 23

## Simple tools for combinatorial problems

Before solving the problem, it is useful to compare the case of independent events with that of dependent events.

Suppose the events $A_{1},A_{2},...,A_{n}$ are independent (in the context of the problem, it will be drawings with replacement). Then by definition the joint probability is the product of individual probabilities:

(1) $P\left( A_{1}\cap ... \cap A_{n}\right) =P\left( A_{1}\right) ... P\left(A_{n}\right) .$

Now assume that the event $A_{1}$ occurs first, $A_{2}$ occurs second, ...., $A_{n}$ occurs last and each subsequent event depends on the previous one (as in the case of drawings without replacement). Then

$P\left( A_{1}\cap A_{2}\right) =\frac{P\left( A_{1}\cap A_{2}\right) }{P\left( A_{1}\right) }P\left( A_{1}\right) =P\left( A_{2}|A_{1}\right)P\left( A_{1}\right) .$

Similarly, by multiplying and dividing many times, we get

(2) $P\left( A_{1}\cap ... \cap A_{n}\right) =P\left(A_{n}|A_{1},...,A_{n-1}\right) P\left( A_{n-1}|A_{1},...,A_{n-2}\right)... P\left( A_{1}\right) .$

Equation (2) is called a chain rule for probability. Several of my students have been able to solve the problem without explicitly using (2). It is advisable to use (2) or other relevant theoretical properties to achieve clarity and avoid errors.

## Problem statement and solution

Suppose there are $n\geq 2$ red balls and $3$ green balls in a bag. All balls with the same color are indistinguishable.

### Part i.

Suppose one ball is drawn at a time at random with replacement from the bag. Let $X$ be the number of balls drawn until a red ball is obtained (including the red ball). Write down the probability mass function of $X$.

Solution. Most students answer that this is a hypergeometric distribution with probabilities given by $q^{x-1}p,$ $x=1,2,...$ where $p$ is the probability of success. Without specifying $p$ (the probability of drawing a red ball) the answer is incomplete. Since $p=\frac{n}{n+3},$ we have

$q^{x-1}p=\left( \frac{3}{n+3}\right) ^{x-1}\frac{n}{n+3}=\frac{n3^{x-1}}{\left( n+3\right) ^{x}}.$

### Part ii.

Now suppose one ball is drawn at random at a time without replacement from the bag. Let $Y$ be the number of balls drawn until a red ball is obtained (including the red ball). Write down the probability mass function of $Y$.

Solution. Let us denote $R_{i}$ the event that the $i$th ball is red and $G_{i}$ the event that the $i$th ball is green, respectively. Note that the only way $R_{3}$ appears is by obtaining $G_{1},G_{2}$ before it. Hence, $R_{3}$ equals $\left( G_{1},G_{2},R_{3}\right) .$ Besides, $R_{1},...,R_{4}$ are the only (mutually exclusive) possibilities and it remains to find their probabilities.

Obviously, $P\left( R_{1}\right) =\frac{n}{n+3}.$

Next, using (2)

$P\left( R_{2}\right) =P\left( G_{1},R_{2}\right) =P\left( G_{1}\right)P\left( R_{2}|G_{1}\right) =\frac{3}{n+3}\frac{n}{n+2}.$

Further,

$P\left( R_{3}\right) =P\left( G_{1},G_{2},R_{3}\right) =P\left(G_{1}\right) P\left( G_{2}|G_{1}\right) P\left( R_{3}|G_{1},G_{2}\right) =\frac{3}{n+3}\frac{2}{n+2}\frac{n}{n+1}.$

Finally,

$P\left( R_{4}\right) =P\left( G_{1},G_{2},G_{3},R_{4}\right)$

$=P\left( G_{1}\right) P\left( G_{2}|G_{1}\right) P\left(G_{3}|G_{1},G_{2}\right) P\left( R_{4}|G_{1},G_{2},G_{3}\right) =\frac{3}{n+3}\frac{2}{n+2}\frac{1}{n+1}\frac{n}{n}.$

The results can be summarized in a table:

$\begin{array}{cc} \textrm{Values} & \textrm{Prob}\\R_{1} & \frac{n}{n+3} \\R_{2} & \frac{3}{n+3}\frac{n}{n+2} \\R_{3} & \frac{3}{n+3}\frac{2}{n+2}\frac{n}{n+1} \\R_{4} & \frac{3}{n+3}\frac{2}{n+2}\frac{1}{n+1}\frac{n}{n}\end{array}$

This distribution is not of one of standard types.

### Part iii.

Suppose two balls are drawn at the same time at random with replacement from the bag. Let $Z$ denote the number of these double draws performed until two green balls are obtained. Show that the probability of drawing two green balls is

$\frac{6}{\left( n+2\right) \left( n+3\right) }.$

Hence, show that the probability mass function for $Z$ is

$P\left( Z=z\right) =\frac{6\left( n^{2}+5n\right) ^{z-1}}{\left( n+3\right)^{z}\left( n+2\right) ^{z}},$ $z=1,2,...$

Solution. Using the same notation as before,

$P\left( G_{1},G_{2}\right) =P\left( G_{2}|G_{1}\right) P\left( G_{1}\right)=\frac{2}{n+2}\frac{3}{n+3}.$

All other events (two red balls or one green and one red) are considered a failure. Thus we have a hypergeometric distribution with

$p=\frac{6}{\left( n+2\right) \left( n+3\right) },$

$q=1-\frac{6}{\left( n+2\right) \left( n+3\right) }=\frac{n^{2}+5n+6-6}{\left( n+2\right) \left( n+3\right) }=\frac{n^{2}+5n}{\left( n+2\right) \left( n+3\right) }$

and

$q^{z-1}p=\left[ \frac{n^{2}+5n}{\left( n+2\right) \left( n+3\right) }\right]^{z-1}\frac{6}{\left( n+2\right) \left( n+3\right) }=\frac{6\left( n^{2}+5n\right) ^{z-1}}{\left( n+2\right) ^{z}\left( n+3\right) ^{2}},$ $z=1,2,...$

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

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

I can't say about this more than I said here.

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 $X,Y$

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 $p_{c}$ banknote $p_{b}$
 Variable 2: Denomination Prob $1 $p_{1}$$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, $x$).

25
Dec 21

## Analysis of problems with conditioning

These problems are among the most difficult. It's important to work out a general approach to such problems. All references are to J. Abdey,  Advanced statistics: distribution theory, ST2133, University of London, 2021.

### General scheme

Step 1. Conditioning is usually suggested by the problem statement: $Y$ is conditioned on $X$.

Your life will be easier if you follow the notation used in the guide: use $p$ for probability mass functions (discrete variables) and $f$ for (probability) density functions (continuous variables).

a) If $Y|X$ and $X$ both are discrete (Example 5.1, Example 5.13, Example 5.18):

$p_{Y}\left( y\right) =\sum_{Set}p_{Y\vert X}\left( y\vert x\right) p_{X}\left(x\right) .$

b) If $Y|X$ and $X$ both are continuous (Activity 5.6):

$f_{Y}\left( y\right) =\int_{Set}f_{Y\vert X}\left( y\vert x\right) f_{X}\left(x\right) dx.$

c) If $Y|X$ is discrete, $X$ is continuous (Example 5.2, Activity 5.5):

$p_{Y}\left( y\right) =\int_{Set}p_{Y\vert X}\left( y\vert x\right) f_{X}\left(x\right) dx$

d) If $Y|X$ is continuous, $X$ is discrete (Activity 5.12):

$f_{Y}\left( y\right) =\sum_{Set}f_{Y\vert X}\left( y\vert x\right) p_{X}\left(x\right) .$

In all cases you need to figure out $Set$ over which to sum or integrate.

Step 2. Write out the conditional densities/probabilities with the same arguments
as in your conditional equation.

Step 3. Reduce the result to one of known distributions using the completeness
axiom.

### Example 5.1

Let $X$ denote the number of hurricanes which form in a given year, and let $Y$ denote the number of these which make landfall. Suppose each hurricane has a probability of $\pi$ making landfall independent of other hurricanes. Given the number of hurricanes $x$, then $Y$ can be thought of as the number of successes in $x$ independent and identically distributed Bernoulli trials. We can write this as $Y|X=x\sim Bin(x,\pi )$. Suppose we also have that $X\sim Pois(\lambda )$. Find the distribution of $Y$ (noting that $X\geq Y$ ).

### Solution

Step 1. The number of hurricanes $X$ takes values $0,1,2,...$ and is distributed as Poisson. The number of landfalls for a given $X=x$ is binomial with values $y=0,...,x$. It follows that $Set=\{x:x\ge y\}$.

Write the general formula for conditional probability:

$p_{Y}\left( y\right) =\sum_{x=y}^{\infty }p_{Y\vert X}\left( y\vert x\right)p_{X}\left( x\right) .$

Step 2. Specifying the distributions:

$p_{X}\left( x\right) =e^{-\mu }\frac{\mu ^{x}}{x!},$ where $x=0,1,2,...,$

and

$P\left( Bin\left( x,\pi \right) =y\right) =p_{Y\vert X}\left( y\vert x\right)=C_{x}^{y}\pi ^{y}\left( 1-\pi \right) ^{x-y}$ where $y\leq x.$

Step 3. Reduce the result to one of known distributions:

$p_{Y}\left( y\right) =\sum_{x=y}^{\infty }C_{x}^{y}\pi ^{y}\left( 1-\pi\right) ^{x-y}e^{-\mu }\frac{\mu ^{x}}{x!}$

(pull out of summation everything that does not depend on summation variable
$x$)

$=\frac{e^{-\mu }\mu ^{y}}{y!}\pi ^{y}\sum_{x=y}^{\infty }\frac{1}{\left(x-y\right) !}\left( \mu \left( 1-\pi \right) \right) ^{x-y}$

(replace $x-y=z$ to better see the structure)

$=\frac{e^{-\mu }\mu ^{y}}{y!}\pi ^{y}\sum_{z=0}^{\infty }\frac{1}{z!}\left(\mu \left( 1-\pi \right) \right) ^{z}$

(using the completeness axiom $\sum_{x=0}^{\infty }\frac{\mu ^{x}}{x!}=e^{\mu }$ for the Poisson variable)

$=\frac{e^{-\mu }}{y!}\left( \mu \pi \right) ^{y}e^{\mu \left( 1-\pi \right)}=\frac{e^{-\mu \pi }}{y!}\left( \mu \pi \right) ^{y}=p_{Pois(\mu \pi)}\left( y\right) .$

27
Jan 21

## AP Stats and Business Stats

Its content, organization and level justify its adoption as a textbook for introductory statistics for Econometrics in most American or European universities. The book's table of contents is somewhat standard, the innovation comes in a presentation that is crisp, concise, precise and directly relevant to the Econometrics course that will follow. I think instructors and students will appreciate the absence of unnecessary verbiage that permeates many existing textbooks.

Having read Professor Mynbaev's previous books and research articles I was not surprised with his clear writing and precision. However, I was surprised with an informal and almost conversational one-on-one style of writing which should please most students. The informality belies a careful presentation where great care has been taken to present the material in a pedagogical manner.

Carlos Martins-Filho
Professor of Economics
University of Colorado at Boulder
Boulder, USA

26
May 20

2
Mar 20

## Statistical calculator

In my book I explained how one can use Excel to do statistical simulations and replace statistical tables commonly used in statistics courses. Here I go one step further by providing a free statistical calculator that replaces the following tables from the book by Newbold et al.:

Table 1 Cumulative Distribution Function, F(z), of the Standard Normal Distribution Table

Table 2 Probability Function of the Binomial Distribution

Table 5 Individual Poisson Probabilities

Table 7a Upper Critical Values of Chi-Square Distribution with $\nu$ Degrees of Freedom

Table 8 Upper Critical Values of Student’s t Distribution with $\nu$ Degrees of Freedom

Tables 9a, 9b Upper Critical Values of the F Distribution

The calculator is just a Google sheet with statistical functions, see Picture 1:

Picture 1. Calculator using Google sheet

## How to use Calculator

1. Open an account at gmail.com, if you haven't already. Open Google Drive.
2. Install Google sheets on your phone.
3. Find the sheet on my Google drive and copy it to your Google drive (File/Make a copy). An icon of my calculator will appear in your drive. That's not the file, it's just a link to my file. To the right of it there are three dots indicating options. One of them is "Make a copy", so use that one. The copy will be in your drive. After that you can delete the link to my file. You might want to rename "Copy of Calculator" as "Calculator".
4. Open the file on your drive using Google sheets. Your Calculator is ready!
5. When you click a cell, you can enter what you need either in the formula bar at the bottom or directly in the cell. You can also see the functions I embedded in the sheet.
6. In cell A1, for example, you can enter any legitimate formula with numbers, arithmetic signs, and Google sheet functions. Be sure to start it with =,+ or - and to press the checkmark on the right of the formula bar after you finish.
7. The cells below A1 replace the tables listed above. Beside each function there is a verbal description and further to the right - a graphical illustration (which is not in Picture 1).
8. On the tab named Regression you can calculate the slope and intercept. The sample size must be 10.
9. Keep in mind that tables for continuous distributions need two functions. For example, in case of the standard normal distribution one function allows you to go from probability (area of the left tail) to the cutting value on the horizontal axis. The other function goes from the cutting value on the horizontal axis to probability.
10. Feel free to add new sheets or functions as you may need. You will have to do this on a tablet or computer.