27
Dec 22

Final exam in Advanced Statistics ST2133, 2022

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?

20
Sep 16

The pearls of AP Statistics 29

Normal distributions: sometimes it is useful to breast the current

The usual way of defining normal variables is to introduce the whole family of normal distributions and then to say that the standard normal is a special member of this family. Here I show that, for didactic purposes, it is better to do the opposite.

Standard normal distribution

The standard normal distribution z is defined by its probability density

p(x)=\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2}).

Usually students don't remember this equation, and they don't need to. The point is to emphasize that this is a specific density, not a generic "bell shape".

standard-normal

Figure 1. Standard normal density

From the plot of the density (Figure 1) they can guess that the mean of this variable is zero.

 

 

 

 

 

area-under-xpx

Figure 2. Plot of xp(x)

Alternatively, they can look at the definition of the mean of a continuous random variable Ez=\int_\infty^\infty xp(x)dx. Here the function f(x)=xp(x) has the shape given in Figure 2, where the positive area to the right of the origin exactly cancels out with the negative area to the left of the origin. Since an integral means the area under the function curve, it follows that

(1) Ez=0.

 

To find variance, we use the shortcut:

Var(z)=Ez^2-(Ez)^2=Ez^2=\int_{-\infty}^\infty x^2p(x)dx=2\int_0^\infty x^2p(x)dx=1.

plot-for-variance

Figure 3. Plot of x^2p(x)

 

The total area under the curve is twice the area to the right of the origin, see Figure 3. Here the last integral has been found using Mathematica. It follows that

(2) \sigma(z)=\sqrt{Var(z)}=1.

General normal distribution

linear-transformation

Figure $. Visualization of linear transformation - click to view video

Fix some positive \sigma and real \mu. A (general) normal variable \mu is defined as a linear transformation of z:

(3) X=\sigma z+\mu.

Changing \mu moves the density plot to the left (if \mu is negative) and to the right (if \mu is positive). Changing \sigma makes the density peaked or flat. See video. Enjoy the Mathematica file.

 

 

 

Properties follow like from the horn of plenty:

A) Using (1) and (3) we easily find the mean of X:

EX=\sigma Ez+\mu=\mu.

B) From (2) and (3) we have

Var(X)=Var(\sigma z)=\sigma^2Var(z)=\sigma^2

(the constant \mu does not affect variance and variance is homogeneous of degree 2).

C) Solving (3) for z gives us the z-score:

z=\frac{X-\mu}{\sigma}.

D) Moreover, we can prove that a linear transformation of a normal variable is normal. Indeed, let X be defined by (3) and let Y be its linear transformation: Y=\delta X+\nu. Then

Y=\delta (\sigma z+\mu)+\nu=\delta\sigma z+(\delta\mu+\nu)

is a linear transformation of the standard normal and is therefore normal.

Remarks. 1) In all of the above, no derivation is longer than one line. 2) Reliance on geometry improves understanding. 3) Only basic properties of means and variances are used. 4) With the traditional way of defining the normal distribution using the equation

p(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})

there are two problems. Nobody understands this formula and it is difficult to extract properties of the normal variable from it.

Compare the above exposition with that of Agresti and Franklin: a) The normal distribution is symmetric, bell-shaped, and characterized by its mean μ and standard deviation σ (p.277) and b) The Standard Normal Distribution has Mean = 0 and Standard Deviation = 1 (p.285). It is the same old routine: remember this, remember that.