Mathematics 376 – Probability and Statistics II Problem Set 5 Solutions March 19, 2004

Extra Problem A. 1) Recall that a random variable Y has an F (n, m) distribution if and only if Y has the form Y = \frac{W_{1}/n}{W_{2}/m}, where W_{1} has a χ^{2}(n) distribution and W_{2} has a χ^{2}(m) distribution. Hence 1/Y = Z has an F (m, n) distribution, and hence taking reciprocals, then reversing the inequality (replacing α by 1 − α) we get:

\begin{aligned}P (Y ≥ f_{α}(n, m)) & = α \\ ⇒ P (\frac{1}{Y} ≤ \frac{1}{f_{α}(n, m)}) & = α \\ ⇒ P (Z ≥ \frac{1}{f_{α}(n, m)}) & = 1 − α \end{aligned}

By definition, this means

\frac{1}{f_{α}(n, m)} = f_{1−α}(m, n)

as claimed.

2) By part 1, and the F-table in the book’s appendix,

f_{.99}(7, 10) = \frac{1}{f_{1−.99}(10, 7)} = \frac{1}{f_{.01}(10, 7)} = \frac{1}{6.62} ≐ .15106

8.73/ Since the sample sizes n_{1} = 16 and n_{2} = 20 are relatively small, we use the tdistribution formula to find the confidence interval. Using the given information, the pooled estimator for the variance (assumed equal) is

S_{p}^{2} = \frac{15 · 36 + 19 · 64}{34} ≐ 51.647

so the 95% confidence interval is

μ_{1} − μ_{2} = \overline{X}_{1} − \overline{X}_{2} ± t_{.025}(34)\sqrt{51.647}\sqrt{\frac{1}{16} + \frac{1}{20}} = −1 ± 4.7245

(using t_{.025}(34) = 1.96 from the t-table in the book).

8.76 a) Similar to 8.73: The interval is

446 − 534 ± t_{.025}(28)\sqrt{\frac{14 · 42^{2} + 14 · 45^{2}}{28}} · \sqrt{\frac{2}{15}} = −88 ± 32.55

b) Also similar:

548 − 517 ± t_{.025}(28)\sqrt{\frac{14 · 57^{2} + 14 · 52^{2}}{28}} · \sqrt{\frac{2}{15}} = 31 ± 40.79

c) The data supports the conclusion that the verbal scores of language/literature students are higher than the verbal scores of engineering students (the interval in part a consists entirely of negative numbers). But the data does not support the conclusion that the math scores are higher for engineering students than for language/literature students.

d) We assume that the scores are independently and randomly selected from the two groups, and that the scores in each case have normal distributions with equal variances.

8.81/Computing \overline{X} and S^{2} from the given data, we get

\overline{X} = 85.73 S^{2} ≐ .5027

Then the 90% confidence interval for σ^{2} is (since n = 6),

[\frac{5S^{2}}{χ_{.05}^{2}(5)}, \frac{5S^{2}}{χ_{.95}^{2}(5)}] = [\frac{5 · .5027}{11.0705}, \frac{5 · .5047}{1.145476}] = [.2276, 2.194]

8.83/a) Since \frac{(n−1)S^{2}}{σ^{2}} has a χ^{2}(n − 1) distribution, we have

P (\frac{(n − 1)S^{2}}{σ^{2}} ≥ χ_{1−α}^{2}) = 1 − α,

so multiplying by σ^{2}, dividing by χ_{1−α}^{2}, we get

P (σ^{2} ≤ \frac{(n − 1)S^{2}}{χ_{1−α}^{2}} ) = 1 − α.

So the (1 − α) × 100% upper confidence bound is

\frac{(n − 1)S^{2}}{χ_{1−α}^{2}} ,

(using n − 1 degrees of freedom). b) Similarly, to get a lower confidence bound,

P (\frac{(n − 1)S^{2}}{σ^{2}} ≤ χ_{α}^{2}) = 1 − α,

so multiplying by σ^{2}, dividing by χ_{1−α}^{2}, we get

P (σ^{2} ≥ \frac{(n − 1)S^{2}}{χ_{α}^{2}} ) = 1 − α.

So the (1 − α) × 100% upper confidence bound is

\frac{(n − 1)S^{2}}{χ_{α}^{2}} ,

(using n − 1 degrees of freedom again). 8.84/Since

P (\frac{(n − 1)S^{2}}{χ_{α/2}^{2}} ≤ σ^{2} ≤ \frac{(n − 1)S^{2}}{χ_{1−α/2}^{2}} ) = 1 − α,

if we take square roots everywhere on the “inside”, we get:

P (\sqrt{\frac{(n − 1)S^{2}}{χ_{α/2}^{2}}} ≤ σ ≤ \sqrt{\frac{(n − 1)S^{2}}{χ_{1−α/2}^{2}}} ) = 1 − α,

So the (1 − α) × 100% confidence interval for σ is

[\sqrt{\frac{(n − 1)S^{2}}{χ_{α/2}^{2}}} , \sqrt{\frac{(n − 1)S^{2}}{χ_{1−α/2}^{2}}} ] .

9.13/For each sample Y_{i}, we have

E(Y_{i}) = ∫_{0}^{1} y · θy^{θ−1} dy = θ ∫_{0}^{1} y^{θ} dy = \frac{θ}{θ + 1}

and

E(Y_{i}^{2}) = ∫_{0}^{1} y^{2} · θy^{θ−1} dy = θ ∫_{0}^{1} y^{θ+1} dy = \frac{θ}{θ + 2}

Hence

V (Y_{i}) = E(Y_{i}^{2}) − (E(Y_{i}))^{2} = \frac{θ}{θ + 2} − (\frac{θ}{θ + 1})^{2} = \frac{θ}{(θ + 1)^{2}(θ + 2)}.

Now, \overline{Y} = \frac{1}{n}(Y_{1} + ⋯ + Y_{n}) satisfies

E(\overline{Y}) = \frac{1}{n}(E(Y_{1}) + ⋯ + E(Y_{n})) = \frac{θ}{θ + 1}

so \overline{Y} is unbiased for \frac{θ}{θ+1}. Moreover, by independence,

V (\overline{Y}) = \frac{1}{n^{2}} (V (Y_{1}) + ⋯ + V (Y_{n})) = \frac{1}{n} · \frac{θ}{(θ + 1)^{2}(θ + 2)}

As n → ∞, this clearly shows

lim_{n→∞} V (\overline{Y}) = 0

so \overline{Y} is consistent by Theorem 9.1.

9.19/a) Since Y_{1} has an N(μ, 1) distribution, E(Y_{1}) = μ. Hence Y_{1} is unbiased for μ.

b) Using the standard normal table, P (|Y_{1}−μ| ≤ 1) = P (−1 ≤ Y_{1}−μ ≤ 1) = 1−2(.1587) = .6824.

c) No, Y_{1} is not a consistent estimator for μ. Let ϵ = 1. As n → ∞, the probability that P (|Y_{1} − μ| ≤ 1) does not change (since the estimator does not depend on the additional samples after the first). In particular it does not tend to 1.

9.61/To find the method of moments estimator, we first compute the expected value a random variable with this pdf:

μ′_{1} = E(Y) = ∫_{0}^{1} y · (θ + 1)y^{θ} dy = \frac{θ + 1}{θ + 2}

(similar to 9.13 above). Set μ′_{1} = \frac{θ+1}{θ+2} = \overline{Y} = m′_{1} and solve for θ:

θ + 1 = \overline{Y} θ + 2\overline{Y}

This is the method of moments estimator:

θ̂ = \frac{2\overline{Y} − 1}{1 − \overline{Y}}.

To see this is consistent, we can use Theorem 9.2. As in 9.13, n → ∞ lim_{n→∞} \overline{Y} = \frac{θ+1}{θ+2} “in probability” (i.e. intuitively, the probability that they are equal tends to 1, or more precisely, for all ϵ > 0, the probability that the difference is ≤ ϵ goes to 1 as n → ∞). Then

lim_{n→∞} \frac{2\overline{Y} − 1}{1 − \overline{Y}} = \frac{lim_{n→∞} 2\overline{Y} − 1}{lim_{n→∞} 1 − \overline{Y}}

= \frac{2\frac{θ+1}{θ+2} − 1}{1 − \frac{θ+1}{θ+2}}

= θ

Finally, the given sufficient statistic U = − ∑_{i=1}^{n} ln(Y_{i}) satisfies exp(−U) = Y_{1} ⋯ Y_{n} (the product of the sample values. The sample mean is not a function of U or exp(−U). (If it were, then we would have that the sum of the sample values was determined by the product, but that is clearly false. For instance, if n = 2,

Y_{1} = 1, Y_{2} = 1 ⇒ Y_{1}Y_{2} = 1, Y_{1} + Y_{2} = 2

but

Y_{1} = 2, Y_{2} = .5 ⇒ Y_{1}Y_{2} = 1, Y_{1} + Y_{2} = 2.5)

By the consequence of the Rao-Blackwell theorem we mentioned in class, this says that θ̂ cannot be the most efficient (minimum variance) unbiased estimator for this problem.

9.67/The given pdf has the form of the beta-distribution pdf with α = β = θ. This is a slightly tricky example for the method of moments, though. Setting μ′_{1} = E(Y) = \overline{Y} = m′_{1}, we get

\frac{θ}{θ + θ} = \overline{Y} ⇒ \frac{1}{2} = \overline{Y}

But there is no θ to solve for in this equation! This means that we can go on and set μ′_{2} = m′_{2} and try to use that to solve for θ, though. Using the formula for the variance of a beta-distributed random variable we get:

μ′_{2} = E(Y^{2}) = V (Y) + (E(Y))^{2} = \frac{θ^{2}}{2θ^{2}(2θ + 1)} + \frac{1}{4} = m′_{2}

After simplifying

m′_{2} = \frac{2θ + 2}{8θ + 4} ⇒ θ̂ = \frac{1 − 2m′_{2}}{4m′_{2} − 1}

9.75/a) The likelihood function is

L(y_{1}, …, y_{n}|θ) = (\frac{1}{2θ + 1})^{n}

(if 0 < y_{i} < 2θ + 1 for all i and zero otherwise). This function is decreasing for all θ > −1/2 (the vertical asymptote for L). Hence to maximize the likelihood, subject to the given sample values, we want to choose θ so that the right hand endpoint of the interval [0, 2θ] occurs as far to the left as possible, consistent with the sample values. This occurs at the sample maximum: Y_{(n)} = 2θ + 1. Hence

θ̂_{ML} = \frac{Y_{(n)} − 1}{2}

b) Use the invariance property of ML estimators (see page 452). The variance is

t(θ) = \frac{(2θ + 1)^{2}}{12}

which is a 1-1 function on θ > −1/2. Hence

t̂(θ)_{ML} = t(θ̂_{ML}) = \frac{(2 (\frac{Y_{(n)}−1}{2}) + 1)^{2}}{12} = \frac{(Y_{(n)})^{2}}{12}

9.80/The logarithm of the likelihood function is

ln(L) = ln((θ + 1)^{n}(y_{1} ⋯ y_{n})^{θ}) = n ln(θ + 1) + θ · ∑_{i=1}^{n} ln(y_{i})

To maximize ln(L), we differentiate with respect to θ and set equal to zero:

\frac{d ln(L)}{dθ} = \frac{n}{θ + 1} + ∑_{i=1}^{n} ln(y_{i}) = 0

Solving for θ, we get

θ̂_{ML} = −\frac{n + ∑_{i=1}^{n} ln(y_{i})}{∑_{i=1}^{n} ln(y_{i})} = \frac{−n}{∑_{i=1}^{n} ln(y_{i})} − 1

This is a maximum by the second derivative test: \frac{d^{2} ln(L)}{dθ^{2}} = \frac{−n}{(θ+1)^{2}} < 0 for all θ. This is quite different from the method of moments estimator from 9.61!