Full solution to Example 2.15 from the guide ST2134
Students tend to miss the ideas needed for this example for two reasons: in the guide there is a reference to ST104b Statistics 2, which nobody consults, and the short notation of the statistic conceals the essence.
Recommendation: for the statistic always use rather than Similarly, if is the observed sample, use rather than
Example 2.15. Let be a random sample (meaning i.i.d. variables) from a distribution, and let Show that is a sufficient statistic.
Solution. There are two ways to solve this problem. One is to use the definition of a sufficient statistic, and the other is to apply the sufficiency principle. It is a good idea to announce which way you go in the very beginning. We apply the definition, and we have to show that the density of conditional on does not depend on
Step 1. The density of is and by independence
Step 2. We need to characterize the distribution of and this is accomplished with the MGF. For one Poisson variable we have
Here we used the completeness axiom with
Step 3. This result for the sum implies (by independence) (since have identical distribution) (by Step 2). By Step 2 we know that implies As we just showed, the MGF of is the same. The uniqueness theorem says:
if random variables and have the same MGF's then their distributions are the same.
It follows that and that
(which is written as in the guide, and to me this is not transparent).
Step 4. To check that the conditional density does not depend on the parameter, we recall that the conditional density along the level set simplifies to (see Guide, p.30) (no joint density in the numerator). In our situation the full expression for the ratio on the right is
As there is no in the result, is sufficient for
Exercise. Do Example 2.16 from the Guide following this format.
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 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 or the difference . Find the joint density. Are variables independent? Find their marginal densities.
c) (2 points) Are normal? Why? What are their means and variances?
d) (2 points) Which probability is larger: or ?
e) (2 points) In this context interpret the conditional expectation . How much is it?
Reminder. The density of a normal variable is .
Question 2. (9 points) The distribution of a call duration of one Kcell [largest mobile operator in KZ] customer is exponential: The number of customers making calls simultaneously is distributed as Poisson: Thus the total call duration for all customers is for . We put . Assume that customers make their decisions about calling independently.
a) (3 points) Find the general formula (when are identically distributed and are independent but not necessarily exponential and Poisson, as above) for the moment generating function of explaining all steps.
b) (3 points) Find the moment generating functions of , and for your particular distributions.
c) (3 points) Find the mean and variance of . Based on the equations you obtained, can you suggest estimators of parameters ?
Remark. Direct observations on the exponential and Poisson distributions are not available. We have to infer their parameters by observing . This explains the importance of the technique used in Question 2.
Question 3. (8 points)
a) (2 points) For a non-negative random variable prove the Markov inequality
b) (2 points) Prove the Chebyshev inequality for an arbitrary random variable .
c) (4 points) We say that the sequence of random variables converges in probability to a random variable if as for any . Suppose that for all and that as . Prove that then converges in probability to .
Remark. Question 3 leads to the simplest example of a law of large numbers: if 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?
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 is the Poisson variable.
Define and derive the population mean and population variance of the sum where is an i.i.d. sample from .
Answer. is defined by Using and (ST2133 p.80) we have
(by independence and identical distribution). [Some students derived instead of respective equations for sample means].
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 Then so standardizing and gives the same result.
b) The MGF of is expressed through the MGF of :
(independence) (identical distribution)
c) If is a linear transformation of then
When answering the question we assume any i.i.d. sample from a population with mean and population variance :
Putting in c) and using a) we get
(using b) and )
This is a general result which for the Poisson distribution can be specified as follows. From ST2133, example 3.38 we know that . Therefore, we obtain
[Instead of some students gave .]
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 leads to
where are moments of and Differentiating this equation yields
and setting gives the rule for finding moments from MGF:
II) Decomposition of the cumulant generating function around zero. can also be decomposed into its Taylor series:
where the coefficients are called cumulants and can be found using . Since
Thus, for any random variable with mean and variance we have
terms of higher order for small.
III) If then by c)
IV) By b)
Using III), and then IV) we have
For the last term on the right we use the approximation around zero from II):
[Important. Why the above steps are necessary? Passing from the series to the series for is not straightforward and can easily lead to errors. It is not advisable in case of the Poisson to derive from .]
Prove the central limit theorem using the cumulant generating function you obtained.
Answer. In the previous question we proved that around zero
This implies that
(1) for each around zero.
But we know that for a standard normal its MGF is (ST2133 example 3.42) and hence for the standard normal
Theorem (link between pointwise convergence of MGFs of and convergence in distribution of ) Let be a sequence of random variables and let be some random variable. If converges for each from a neighborhood of zero to , then converges in distribution to
Using (1), (2) and this theorem we finish the proof that converges in distribution to the standard normal, which is the central limit theorem.
State the factorization theorem and apply it to show that is a sufficient statistic.
Answer. The solution is given on p.180 of ST2134. For the joint density is
The expression on the right does not depend on if and only of The last condition describes level sets of Thus it is minimal sufficient.
Find the Method of Moments estimator of the population mean.
Answer. The idea of the method is to take some populational property (for example, ) and replace the population characteristic (in this case ) by its sample analog () to obtain a MM estimator. In our case [Try to do this for the Gamma distribution].
Find the Fisher information.
Answer. From Problem 5 the log-likelihood is
Hence the score function is (see Example 2.30 in ST2134)
and the Fisher information is
Derive the Cramer-Rao lower bound for for a random sample.
Answer. (See Example 3.17 in ST2134) Since is an unbiased estimator of by Problem 1, from the Cramer-Rao theorem we know that
and in fact by Problem 1 this lower bound is attained.