Final #2 - Questions and Solutions

Let X_{1}, X_{2}, ..., X_{n} be a random sample from a U(a, 2a) population, where a > 0 is an unknown parameter.

(a) Estimate a by the method of moments.

Since E(X) = \frac{3}{2}a, the required estimator is Θ̂_{1} = \frac{2}{3} X̄_{n}.

(b) Find a sufficient two-dimensional statistic for the parameter a.

The joint density function of the sample X_{1}, X_{2}, ..., X_{n} is

\begin{aligned}f(x_{1}, x_{2}, ..., x_{n}; a) & = \frac{1}{a^{n}} · ∏_{i=1}^{n} 1_{[a,2a]}(x_{i}) \\ & = \frac{1}{a^{n}} 1_{[a,∞)} (min_{1≤i≤n}x_{i}) · 1_{[0,2a]} (max_{1≤i≤n}x_{i}) . \end{aligned}

(1)

By the factorization theorem

( min_{1≤i≤n} X_{i}, max_{1≤i≤n} X_{i})  (2)

is a sufficient statistic for a.

(c) Is the statistic found in the preceding part a complete statistic?

The statistic given by (2) is not complete, as the following example shows: Denote S = min_{1≤i≤n} X_{i} and T = max_{1≤i≤n} X_{i}. Here and later it will be convenient to observe that, if X ∼ U(a, 2a), then the random variable U defined by

U = \frac{X}{a} − 1

is distributed U(0, 1). Put U_{i} = \frac{X_{i}}{a} − 1, and denote S′ = min_{1≤i≤n} U_{i} and

T′ = max_{1≤i≤n} U_{i}. Calculating E(S′) and E(T′) we obtain:

\begin{aligned}E(S′) & = ∫_{0}^{1} n(1 − u)^{n−1} · udu = n ∫_{0}^{1} v^{n−1}(1 − v)dv \\ & = n (\frac{1}{n} − \frac{1}{n + 1}) = \frac{1}{n + 1} \end{aligned}

and

E(T′) = ∫_{0}^{1} nu^{n−1} · udu = \frac{n}{n + 1}.

Since X = aU + a, we obviously have

S = aS′ + a  (3)

and

T = aT′ + a.  (4)

Thus, calculating

E(S) = aE(S′) + a = \frac{a(n + 2)}{n + 1}  (5)

and

E(T) = aE(T′) + a = \frac{a(2n + 1)}{n + 1},  (6)

we obtain

E ((n + 2)T − (2n + 1)S) = 0, a > 0.

Hence (S, T) is not a complete statistic.

(d) Find the maximum likelihood estimator for a. Is it a sufficient statistic?

In the joint probability function (1), the factor \frac{1}{a^{n}} increases as a decreases. The factor 1_{[0,2a]}(t) vanishes unless a ≥ \frac{t}{2}, and the factor 1_{[a,∞)}(s) – unless a ≤ s. It follows that \frac{T}{2} is the maximum likelihood estimator for a.

The conditional distribution of (X_{1}, X_{2}, ... , X_{n}) given that T = t is a uniform distribution of the (n − 1)-dimensional set

⋃_{j=1}^{n}[a, t]^{j−1} × {t} × [a, t]^{n−j}.

Since it depends on a, the statistic T is not sufficient.

(e) Find three unbiased estimators for a, based on X̄_{n}, S and T. Rank them.

According to (a), we may simply take Θ̂_{1} = \frac{2}{3} X̄_{n}. By (5) and (6), both

Θ̂_{2} = \frac{n + 1}{n + 2}S and Θ̂_{3} = \frac{n + 1}{2n + 1}T are unbiased estimators for a.

To rank the estimators, consider their variances. We have:

V (Θ̂_{1}) = \frac{4}{9} · \frac{V (X)}{n} = \frac{4}{9} · \frac{a^{2}}{12n} = \frac{a^{2}}{27n}.  (7)

By (3) and (4) we have V (S) = a^{2}V (S′) and V (T) = a^{2}V (T′). Now

V (S′) = ∫_{0}^{1} n(1 − u)^{n−1} · u^{2}du − \frac{1}{(n + 1)^{2}}

= n ∫_{0}^{1} v^{n−1}(1 − v)^{2}dv − \frac{1}{(n + 1)^{2}}

= n (\frac{1}{n} − \frac{2}{n + 1} + \frac{1}{n + 2}) − \frac{1}{(n + 1)^{2}}

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

so that

V (S) = \frac{a^{2}n}{(n + 1)^{2}(n + 2)}

and

V (Θ̂_{2}) = V (\frac{n + 1}{n + 2}S) = \frac{a^{2}n}{(n + 2)^{3}} .  (8)

Similarly we obtain

V (T) = \frac{a^{2}n}{(n + 1)^{2}(n + 2)}

and

V (Θ̂_{3}) = V (\frac{n + 1}{2n + 1}S) = \frac{a^{2}n}{(n + 2)^{2}(2n + 1)}.  (9)

Comparing (7), (8) and (9), the estimators can be ranked as follows:

Θ̂_{3} ≻ Θ̂_{2} ≻ Θ̂_{1}.

(f) On which of the following statistics can an unbiased estimator for a^{2} be based? Explain.

(i) \frac{T}{S} ,  (ii) T − S,  (iii) T − 2S,  (iv) \frac{T^{2}}{S} ,  (v) \frac{(S + T)^{2}}{ST} .

By (3) and (4) we have:

(i) \frac{T}{S} = \frac{T′ + 1}{S′ + 1}.

As the distribution of this statistic does not depend on a, \frac{S}{T} cannot be used as a basis for an unbiased estimator for a^{2}.

(ii) T − S = a(T′ − S′). Hence for an appropriate constant c = c(n), the statistic c(T − S)^{2} forms an unbiased estimator for a^{2}.

(iii) T − 2S = a(T′ − 2S′ − 1). Similarly to (ii), the statistic T − 2S can be used to produce an unbiased estimator for a^{2}.

(iv) \frac{T^{2}}{S} = a\frac{(T′ + 1)^{2}}{S′ + 1} .

Hence \frac{T^{2}}{S} can be used as a basis for an unbiased estimator for a^{2}.

(v) \frac{(T + S)^{2}}{ST} = (\sqrt{\frac{T}{S}} + \sqrt{\frac{S}{T}} )^{2} .

Thus, this statistic is a function of \frac{T}{S}. It follows from part (i) that we cannot base on it an estimator as required.

(g) Which of the following statistics are pivotal quantities?

(i) S − a,  (ii) \frac{S}{a} ,  (iii) \frac{ST}{a} ,  (iv) \frac{T^{2} − S^{2}}{a^{2}} ,  (v) e^{S} − e^{a}.

By (3) and (4) we have:

(i) S − a = aS′. Since the distribution depends on a, this is not a pivotal quantity.

(ii) \frac{S}{a} = S′ + 1.

The distribution does not depend on a, so that it is a pivotal quantity.

(iii) \frac{ST}{a} = a(S′ + 1)(T′ + 1).

Since the distribution depends on a, this is not a pivotal quantity.

(iv) \frac{T^{2} − S^{2}}{a^{2}} = (T′ + 1)^{2} − (S^{2} + 1)^{2},

The distribution does not depend on a, so that it is a pivotal quantity.

(v) The statistic e^{S} − e^{a} assumes values in the interval [0, e^{2a} − e^{a}], which depends non-trivially on a. Hence it is not a pivotal quantity.

(h) Find a confidence interval of confidence level 1 − α for a based on S and T.

Obviously, [\frac{T}{2} , S] is a confidence interval of level 1. Hence it is natural

to look for a confidence interval of the form

I_{1} = [\frac{T}{2} (1 + c), S(1 − d)] ,

where c, d > 0 are some constants. Let us find under what conditions on c and d, the level of the confidence interval I_{1} will be equal to 1 − α. Consider:

\begin{aligned}P (\frac{T}{2} (1 + c) ≤ a ≤ S(1 − d)) & = P (T ≤ \frac{2a}{1 + c}, S ≥ \frac{a}{1 − d}) \\ & = ( \frac{\frac{2a}{1+c} − \frac{a}{1−d}}{a} )^{n} = (\frac{1 − 2d − c}{(1 + c)(1 − d)})^{n} , \end{aligned}

which equals to 1 − α if and only if

\frac{1 − 2d − c}{(1 + c)(1 + d)} = \sqrt[n]{1 − α}.

(i) Find a confidence interval of confidence level 1 − α (approximately) for θ based on X̄_{n} for large n.

Since E(X̄_{n}) = \frac{3a}{2} and V (X̄_{n}) = \frac{a^{2}}{12n}, we obtain

P (−z_{α/2} ≤ \frac{X̄_{n} − \frac{3a}{2}}{\sqrt{\frac{a^{2}}{12n}}} ≤ z_{α/2}) = 1 − α.

A routine calculation gives the confidence interval:

I_{2} = [ \frac{X̄_{n}}{\frac{3}{2} + z_{α/2}\sqrt{\frac{1}{12n}}} , \frac{X̄_{n}}{\frac{3}{2} − z_{α/2}\sqrt{\frac{1}{12n}}} ] .

(j) Suggest a test with significance level α based on S to test:

\begin{matrix}H_{0} : & a = 10, \\ H_{1} : & a > 10. \end{matrix}

Find the power function.

We need to use the distribution function of S:

F_{S}(s) = \{ \begin{matrix}0, & s < a, \\ 1 − (\frac{2a − s}{a})^{n} , & a ≤ s ≤ 2a, \\ 1, & s > 2a. \end{matrix}  (10)

It is natural to consider a critical region of the form R = {S > b_{0}}. By (10):

P_{a=10}(S > b) = \{ \begin{matrix}1, & b < 10, \\ (\frac{20 − b}{10})^{n} , & 10 ≤ b ≤ 20, \\ 0, & b > 20. \end{matrix}

Now the equation

(\frac{20 − b}{10})^{n} = α

yields b_{0} = 20 − 10\sqrt[n]{α}.

To find the power, calculate:

1 − β = P_{a}(S > b_{0}) = \{ \begin{matrix}(\frac{2a − b_{0}}{a} )^{n} , & a ≤ b_{0} ≤ 2a, \\ 1, & a > b_{0}. \end{matrix}

(k) The same as in the previous part, with the test based on T.

Calculate the distribution function of T:

F_{T} (t) = \{ \begin{matrix}0, & t < a, \\ (\frac{t − a}{a})^{n}, & a ≤ t ≤ 2a, \\ 1, & t > 2a. \end{matrix}  (11)

Consider a critical region of the form R = {T > c_{0}}. By (11):

P_{a=10}(T > c) = \{ \begin{matrix}1, & c < 10, \\ 1 − (\frac{c − 10}{10})^{n} , & 10 ≤ c ≤ 20, \\ 0, & c > 20. \end{matrix}

Now the equation

1 − (\frac{c − 10}{10})^{n} = α

yields c_{0} = 10(1 + \sqrt[n]{1 − α}).

To find the power, calculate:

1 − β = P_{a}(T > c_{0}) = \{ \begin{matrix}1 − (\frac{c_{0} − a}{a})^{n} , & a ≤ c_{0}, \\ 1, & a > c_{0}. \end{matrix}

(l) One statistisian A, used 0.05 significance level test, based on the sample mean with n = 1200 to test the above hypothesis. Another statistisian, B, (who did not know that the sample was drawn from a U(a, 2a) population), used the sign test for testing:

\begin{matrix}H′_{0} : & m = 15, \\ H′_{1} : & m > 15, \end{matrix}

using the same observations and significance level. Assume that a = 10.1. Is it possible that A will reject H_{0}, but B will accept H′_{0}? Is the opposite possible? What is the probability for A to reject H_{0}? What is the probability for B to reject H′_{0}?

Let us start with the test used by A. The critical region of the test is

R_{1} = \{X̄_{1200} > \frac{3}{2} · 10 + z_{α}\sqrt{\frac{10^{2}}{12 · 1200}}\} ,

which, for α = 0.05, gives

R_{1} = {X̄_{1200} > 15.137}.  (12)

Now:

\begin{aligned}P_{a=10.1}(X̄_{1200} > 15.137) & ≈ P_{a=10.1} (Z > \frac{15.137 − 10.1 · 1.5}{\sqrt{\frac{10.1^{2}}{12·1200}}} ) \\ & = 1 − Φ(−0.155) = 1 − 0.4384 = 0.5616. \end{aligned}

To find the test used by B, denote by Y the number of observations bigger than 15. Under H′_{0} we have Y ∼ B(1200, 0.5). Consider a critical region of the form R_{2} = {Y > c_{0}}, where c_{0} is an integer. To obtain c_{0} use the normal approximation:

\begin{aligned}P_{m=15}(Y > c_{0}) & = P_{m=15} (\frac{Y − \frac{1200}{2}}{\sqrt{\frac{1200}{4}}} ≥ \frac{c_{0} + \frac{1}{2} − \frac{1200}{2}}{\sqrt{\frac{1200}{4}}} ) \\ & ≈ 1 − Φ (\frac{c_{0} − 599.5}{10\sqrt{3}}) . \end{aligned}

To obtain the required significance level α = 0.05 we find, using the normal distribution table, that

\frac{c_{0} − 599.5}{10\sqrt{3}} = 1.645,

which yields c_{0} = 628. Thus:

R_{2} = {Y ≥ 628}.  (13)

If a = 10.1 then Y ∼ B(1200, p_{0}), where

p_{0} = P (X_{1} > 15) = 1 − \frac{15 − 10.1}{10.1} = 0.5149.

Hence:

P_{m=15.15}(Y ≥ 628) = P_{m=15.15}(Y > 627.5)

≈ P (Z > \frac{627.5 − 1200 · 0.5149}{\sqrt{1200 · 0.4851 · 0.5149}})

= 1 − Φ (0.556) = 1 − 0.71 = 0.29.

To answer the remaining questions, by (12) and (13) we see that it is possible that A will reject H_{0} but B will accept H′_{0}. For example, that is the case if Y = 600 and the average of the observations bigger that 15 is 18, while the average of the observations smaller than 15 is 13. The opposite situation is also possible. This happens, for example, if Y = 700 and the average of the observations bigger that 15 is 17, while the average of the observations smaller than 15 is 12.