Final #1 - Questions and Solutions

1. A population is distributed according to the probability function

P (x; r) = \{ \begin{matrix}\frac{2x}{r(r + 3)}, & x = 1, 2, …, r, \\ \frac{2}{r(r + 3)}, & x = −1, −2, …, −r, \\ 0, & otherwise, \end{matrix}

where r is a positive integer unknown parameter. A random sample X_{1}, X_{2}, …, X_{n} is drawn from the population.

(a) Find the maximum likelihood estimator of r.

We may write the likelihood function as follows:

L(r; x⃗) = \{ \begin{matrix}(\frac{2}{r(r + 3)})^{n} ∏_{i:x_{i}>0} x_{i}, & r ≥ x_{i}, x_{i} = ±1, ±2, …, \\ & i = 1, 2 …, n, \\ 0, & otherwise. \end{matrix}

To maximize this expression with respect to r, we have to require first that r ≥ max_{1≤i≤n} |x_{i}|, as otherwise the value is 0. For r satisfying

this condition, the likelihood decreases as a function of r, and therefore the maximum likelihood estimator of r is

R̂ = max_{1≤i≤n} |X_{i}| .

(b) Is the statistic S = (|X_{1}| , |X_{2}| , …, |X_{n}|) a sufficient statistic?

Write the joint density function in the form:

PX⃗(x⃗; r) = (\frac{2}{r(r + 3)})^{n} · ∏_{i=1}^{n} 1_{\{1,2,... ,r\}}(|x_{i}|) ∏_{i:x_{i}>0} x_{i} .

From the representation it is clear that S is a sufficient statistic.

2. A population is distributed (discretely) uniformly U[1, a], where a is an unknown positive integer parameter. A random sample X_{1}, X_{2}, …, X_{n} is drawn from the population.

(a) Show that for n = 1 there exists an unbiased estimator for \frac{1}{a^{2}} .

Since a is a positive integer and the distribution is supported on the integers in the interval [1, a], finding an unbiased estimator

means finding a function ψ : N ⟶ R such that E(ψ(X_{1})) = \frac{1}{a^{2}} for every a ∈ N. Namely, we need to have

\frac{1}{1}ψ(1) = \frac{1}{1^{2}} ,

\frac{1}{2}ψ(1) + \frac{1}{2}ψ(2) = \frac{1}{2^{2}} ,

and in general

\frac{1}{a}ψ(1) + \frac{1}{a}ψ(2) + … + \frac{1}{a}ψ(a) = \frac{1}{a^{2}}

for every positive integer a. Now the first of these equations uniquely determines ψ(1), the second uniquely determines ψ(2), and so forth. Hence there exists an estimator as required. (By the way, working out the equations, one can verify that ψ(1) = 1 and

ψ(k) = \frac{−1}{k(k − 1)} for k ≥ 2.)

(b) Is the estimator found in (a) of uniformly minimal variance among all unbiased estimators?

Since all the values ψ(k) are determined uniquely, the estimator is unique, and therefore trivially the best.

(c) For n ≥ 2 start from the unbiased estimator T defined by

T (X_{1}, X_{2}, …, X_{n}) = 1_{{(1,1)}}(X_{1}, X_{2})

to obtain an unbiased estimator of uniformly minimal variance

among all unbiased estimators for \frac{1}{a^{2}} .

As in the case of continuous uniform distribution, the statistic S = max_{1≤i≤n} X_{i} is a complete sufficient statistic (see also Exercise 94).

Consequently, the required estimator is E(T |S). Now

P (S = s) = \frac{s^{n} − (s − 1)^{n}}{a^{n}} , s = 1, 2, …, a ,

and therefore

\begin{aligned}E(T |S = s) & = P (X_{1} = X_{2} = 1|S = s) \\ & = \frac{P (X_{1} = X_{2} = 1 ⋂ S = s)}{P (S = s)} . \end{aligned}

For n ≥ 3 we find that

\begin{aligned}P (X_{1} = X_{2} = 1 ⋂ S = s) & = P (X_{1} = X_{2} = 1 ⋂ max_{3≤i≤n} X_{i} = s) \\ & = \frac{1}{a} · \frac{1}{a} · \frac{s^{n−2} − (s − 1)^{n−2}}{a^{n−2}} , \end{aligned}

which yields

E(T |S = s) = \frac{s^{n−2} − (s − 1)^{n−2}}{s^{n} − (s − 1)^{n}}

and consequently

E(T |S) = \frac{S^{n−2} − (S − 1)^{n−2}}{S^{n} − (S − 1)^{n}} .

For n = 2 the estimator T itself is a function of S, and therefore it is already an unbiased estimator of minimal variance. (We mention in passing that the expression given above for n ≥ 3 reduces to T for n = 2.)

3. A population has a density function f(x; b), which is linear in each of the intervals [−1, 0] and [0, 1], assumes the values f(−1) = f(1) = b, f(0) = 1 − b at the endpoints of these intervals, and vanishes outside [−1, 1], where 0 < b < 1 is an unknown parameter. A random sample X_{1}, X_{2}, …, X_{n} is drawn from the population. Consider the statistics:

T_{1} = ∑_{i=1}^{n} X_{i},

T_{2} = ∑_{i=1}^{n} (X_{i} − X̄_{n})^{2},

T_{3} = ∑_{i=1}^{n} 1_{[0,1]}(X_{i}).

(a) Find an unbiased estimator of b based on one of the T_{i}’s.

Since the density of the population is symmetric around 0, we have E(X) = 0. A routine calculation gives the variance:

V (X) = E (X^{2})−E^{2}(X) = 2 ∫_{0}^{1} (1 − b + (2b − 1)x) x^{2} dx = \frac{1 + 2b}{6} .

One can use either one of the first two statistics to obtain an unbiased estimator of b. In fact

E (T_{1}^{2}) = V (T_{1}) = nV (X) = \frac{(1 + 2b)n}{6} ,

so that

E (\frac{3T_{1}^{2}}{n} − \frac{1}{2}) = b .

Alternatively

E (T_{2}) = (n − 1)V (X) = \frac{(1 + 2b)(n − 1)}{6} ,

which gives

E (\frac{3T_{2}}{n − 1} − \frac{1}{2}) = b .

(b) Find a confidence interval of confidence level 1 − α for b which is based on one of the other two T_{i}’s. (Assume n to be large.)

Again, either T_{1} or T_{2} may be used. Take, for example, T_{1}. The random variable

Z = \frac{T_{1} − nE(X)}{\sqrt{nV (X)}} = \frac{T_{1}\sqrt{6}}{\sqrt{n(1 + 2b)}}

is distributed approximately N(0, 1). Thus (approximately):

P (−z_{α/2} ≤ \frac{T_{1}\sqrt{6}}{\sqrt{n(1 + 2b)}} ≤ z_{α/2}) = 1 − α .

Solving for b we obtain the confidence interval

I = [\frac{3T_{1}^{2}}{nz_{α/2}^{2}} − \frac{1}{2}, ∞) .

To obtain a finite confidence interval, one should start with limiting Z to the complement of the interval (−z_{(1+α)/2}, z_{(1+α)/2}).

(c) Do you think we can use the remaining T_{i} in any way to estimate b? If yes – explain intuitively how, if not – why not?

The probability for any X_{i} belonging to [0, 1] is 1/2, whatever the value of b is. Hence T_{3} is distributed B(n, 1/2). As the distribution does not depend on b, the value T_{3} assumes cannot provide any indication as to the value of b.

(d) Show that at least one of the T_{i}’s is not a complete statistic.

T_{1} is not complete since E(T_{1}) = 0, and T_{3} is not complete since E(2T_{3} − n) = 0.

4. A population has the density function

f(x; r) = \{ \begin{matrix}\frac{2}{π} sin^{2} rx, & 0 ≤ x ≤ π, \\ 0, & otherwise, \end{matrix}

with an unknown positive integer parameter r. A random sample X_{1}, X_{2}, …, X_{n} is drawn from the population. Consider the statistics:

T_{1} = ∑_{i=1}^{n} X_{i},

T_{2} = ∑_{i=1}^{n} 1_{[π/4,3π/4]}(X_{i}) .

Suggest a test based on one of these statistics for testing

\begin{matrix}H_{0} : & r = 1, \\ H_{1} : & r = 2, \end{matrix}

with significance level α. (Assume n to be large.)

The function f is symmetric with respect to the point π/2 for any value of r, so that E(T_{1}) = 0. Hence the size of T_{1} does not give an immediate indication on the value of b (although of course T_{1} may well serve to estimate b). It is reasonable therefore to use T_{2}. We have

P (\frac{π}{4} ≤ X ≤ \frac{3π}{4} ) = ∫_{π/4}^{3π/4} \frac{2}{π} sin^{2} rx dx = \frac{2}{π} ∫_{π/4}^{3π/4} \frac{1 − cos 2rx}{2} dx

= \frac{1}{π} [x − \frac{sin 2rx}{2r}]_{π/4}^{3π/4} = \frac{1}{2} − \frac{sin 3rπ/2 − sin rπ/2}{2rπ}

= \frac{1}{2} − \frac{1}{rπ} · sin \frac{rπ}{2} cos rπ = \frac{1}{2} + \frac{(−1)^{r−1}}{rπ} sin \frac{rπ}{2} .

In particular, for r = 1 this probability is 1/2 + 1/π, while for r = 2 it is 1/2. Thus, 1_{[π/4,3π/4]}(X_{i}) is distributed B(1, p), where the original H_{0} and H_{1} may be rewritten as

\begin{matrix}H′_{0} : & p = \frac{1}{2} + \frac{1}{π} , \\ H′_{1} : & p = \frac{1}{2}, \end{matrix}

respectively. Using the general test for proportions we find in our case

R = \{T_{2} ≥ n (\frac{1}{2} + \frac{1}{π} ) − z_{α}\sqrt{(\frac{1}{2} + \frac{1}{π} ) (\frac{1}{2} − \frac{1}{π} ) n}\} .