Your Answer : (Not Answered)

Correct Answer : C

\(\bar{X}\) is both Bayes and minimax.

Under the weighted loss \(L = (\theta - a)^{2}/[\theta(1 - \theta)]\), the Bayes estimator is \(E[\theta w(\theta) \mid x]/E[w(\theta) \mid x]\) with \(w = 1/[\theta(1 - \theta)]\). With the uniform prior the posterior is Beta(s + 1, n - s + 1), where \(s = \sum x_{i}\); multiplying by \(w\) gives a Beta(s, n - s) kernel, whose mean is \(s/n = \bar{X}\). So \(\bar{X}\) is Bayes.

Its risk is constant:

\[ \begin{aligned} \frac{E(\bar{X} - \theta)^{2}}{\theta(1 - \theta)} &= \frac{\theta(1 - \theta)/n}{\theta(1 - \theta)} \\ &= \frac{1}{n} \end{aligned} \]

A Bayes estimator with constant risk is minimax, and a unique Bayes estimator is admissible, so A, B and D are wrong.

Your Answer : (Not Answered)

Correct Answer : A

\(L_{x}\) is the number of years lived by the persons in the age interval \((x, x + 1)\).

In a life table, \(L_{x}\) is the total person-years lived between exact ages \(x\) and \(x + 1\) by the cohort, approximately \((l_{x} + l_{x+1})/2\). Summing \(L_{x}\) from age \(x\) onwards gives \(T_{x}\), and \(e_{x}^{0} = T_{x}/l_{x}\).

The number surviving to age \(x\) is \(l_{x}\) (B), to age \(x + 1\) is \(l_{x+1}\) (C), and the average remaining lifetime is the expectation of life \(e_{x}^{0}\) (D).

Your Answer : (Not Answered)

Correct Answer : A

The answer is both \(\{X_{n}\}\) and \(\{W_{n}\}\) are \(\mathbf{F}_{n}\)-martingales.

\(X_{n+1} = X_{n} + Y_{n+1}\), and \(Y_{n+1}\) is independent of \(\mathbf{F}_{n}\) with mean 0, so \(E[X_{n+1} \mid \mathbf{F}_{n}] = X_{n}\).

\(W_{n}\) is \(\mathbf{F}_{n}\)-measurable, so

\[ \begin{aligned} E[W_{n+1} \mid \mathbf{F}_{n}] &= W_{n} + W_{n}^{2}E[Y_{n+1}] \\ &= W_{n} \end{aligned} \]

(assuming the variables are integrable). But \(T_{n} = X_{n} + n\), so \(E[T_{n+1} \mid \mathbf{F}_{n}] = T_{n} + 1\): it is a submartingale, not a martingale. That rules out the options including \(T_{n}\), and 'only \(X_{n}\)' misses \(W_{n}\).

Your Answer : (Not Answered)

Correct Answer : B

The BLUEs exist and are uncorrelated with each other.

The regressor values are \(x_{i} - 2 = -1, 0, 1\), which sum to 0. The design matrix has full column rank, so the least squares estimators are the BLUEs (Gauss-Markov), and with a centred regressor

\[ \hat{B}_0 = \bar{Y},\ \ \hat{B}_1 = \frac{Y_3 - Y_1}{2} \]

Their covariance is \(-\sigma^{2}\bar{c}/S_{cc}\), where \(\bar{c} = 0\) is the mean of the centred regressor, so it is 0. Directly: \(\mathrm{Cov}(\bar{Y}, (Y_{3} - Y_{1})/2)\) is \(\tfrac{1}{6}(\sigma^{2} - \sigma^{2}) = 0\).

Independence would need normal errors, which are not assumed, so C goes too far; and they are not positively correlated.

Your Answer : (Not Answered)

Correct Answer : B

The incorrect statement is it requires exactly \(V^{3}\) experimental units.

A Latin square for V treatments is a \(V \times V\) arrangement, one plot per cell, so it uses \(V^{2}\) units, each treatment once in every row and every column.

The other statements are the usual properties. The design classifies the variation three ways, by rows, columns and treatments; row, column and treatment contrasts are all estimated with the same variance \(2\sigma^{2}/V\) because the design is balanced; and the error degrees of freedom are \((V^{2} - 1) - 3(V - 1) = (V - 1)(V - 2)\).

Your Answer : (Not Answered)

Correct Answer : D

\(n\) is 19.

In a completely randomised design with \(t = 10\) treatments and \(n\) plots, the F-ratio is

\[ \begin{aligned} F &= \frac{SS_{T}/(t - 1)}{SS_{E}/(n - t)} \\ &= \frac{n - t}{t - 1}\cdot\frac{SS_{T}}{SS_{E}} \end{aligned} \]

Forgetting both divisions gives \(SS_{T}/SS_{E}\), which equals the correct F exactly when \(n - t = t - 1\), that is

\[ n - 10 = 9 \Rightarrow n = 19 \]

With 10, 20 or 30 plots the error degrees of freedom would be 0, 10 or 20, not 9.

Your Answer : (Not Answered)

Correct Answer : A

\(\mathrm{Var}(p) = \) \(\dfrac{PQ}{n}\dfrac{N - n}{N - 1}\).

A sample proportion is the mean of 0-1 values whose population variance (with divisor N) is PQ. Under SRSWOR the variance of a sample mean is \(\frac{\sigma^{2}}{n}\frac{N - n}{N - 1}\), so

\[ \mathrm{Var}(p) = \frac{PQ}{n} \cdot \frac{N - n}{N - 1} \]

The factor \((N - n)/(N - 1)\) is the finite population correction. \(PQ/n\) is the with-replacement variance, and the other two options use the wrong factor or denominator.

Your Answer : (Not Answered)

Correct Answer : D

The regression estimator is better than the sample mean if \(n\) is sufficiently large.

In large samples its variance is approximately \(V(\bar{y})(1 - \rho^{2})\), never larger than that of \(\bar{y}\), and smaller whenever the study and auxiliary variables are correlated. In small samples the regression estimator is biased and its gain is not guaranteed.

So it is not 'always' better (A), it does not have the same efficiency except when \(\rho = 0\) (B), and it is not worse (C) in large samples.

Your Answer : (Not Answered)

Correct Answer : B

The key's answer is that there exists an unbiased estimator of the population mean and it is different from the sample mean.

This is the Midzuno-Sen scheme. Because the first unit is drawn with unequal probabilities \(P_{i}\), the sample mean \(\bar{y}\) gives units different weights from their population shares and is biased (A fails). But the scheme has known inclusion probabilities, so the Horvitz-Thompson estimator, and the ratio-type estimator \(\bar{y}/\big(\sum_{i \in s} P_{i} \cdot \frac{N}{n}\big)\) used with this scheme, are unbiased, which also rules out D.

The procedure draws from the whole population at each stage; it does not divide the population into strata, so C is wrong.

Your Answer : (Not Answered)

Correct Answer : A

The answer is post-stratification is better than SRSWOR if the stratum sizes are large and errors in \(W_{h}\) can be ignored.

In post-stratification the sample is drawn by SRSWOR and then classified into strata with known weights \(W_{h}\). When the expected sample size in each stratum is large, its variance is approximately that of proportional stratified sampling, plus a small term of order \(1/n^{2}\) from the random stratum sample sizes. Stratification with proportional allocation is never worse than SRS, so under these conditions post-stratification is at least as good as SRSWOR.

Neither method is better always: with small strata the random sample sizes can make post-stratification worse. A large \(n\) alone is not the stated condition; it is the stratum sizes and the accuracy of the weights that matter.

Your Answer : (Not Answered)

Correct Answer : A

The \(100(2\alpha - 1)\%\) confidence interval is \([T_{1}, T_{2})\).

By definition \(P(T_{1} \le \theta) = \alpha\) and \(P(\theta \lt T_{2}) = \alpha\), so the two ways the interval can miss \(\theta\) each have probability \(1 - \alpha\). They cannot happen together, since \(T_{1} \lt T_{2}\) almost surely. Hence

\[ \begin{aligned} P(T_{1} \le \theta \lt T_{2}) &= 1 - 2(1 - \alpha) \\ &= 2\alpha - 1 \end{aligned} \]

The condition \(\alpha \gt 1/2\) makes this positive. Dividing the limits by \(\alpha\) or scaling them changes the coverage in an uncontrolled way, so options B, C and D have no stated coverage.

Your Answer : (Not Answered)

Correct Answer : A

The correct statements are I, II and III.

The two equations give \(P(E)P(F \cap G) = P(E)P(F)P(G)\). Assuming \(P(E) \gt 0\), this yields \(P(F \cap G) = P(F)P(G)\), so F and G are independent (I).

The second equation says E and \(F \cap G\) are independent. Independence is preserved under complements, so \(E^{c}\) and \(F \cap G\) are independent (III), and E and \((F \cap G)^{c} = F^{c} \cup G^{c}\) are independent (II).

IV does not follow: mutual independence also needs \(P(E \cap F) = P(E)P(F)\) and \(P(E \cap G) = P(E)P(G)\), which are not given. So options including IV, or omitting I, are wrong.

Your Answer : (Not Answered)

Correct Answer : C

The probability is 0.2.

Given \(X_{2} = 1\), the event \(X_{1} + X_{2} = 0\) is the event \(X_{1} = -1\). The variables are independent, so conditioning on \(X_{2}\) and \(X_{3}\) does not change the distribution of \(X_{1}\):

\[ \begin{aligned} &P[X_{1} = -1 \mid X_{2} = 1, X_{3} = 2] \\ &= P[X_{1} = -1] = 0.2 \end{aligned} \]

The condition on \(X_{3}\) is irrelevant. 0.3 is \(P[X = 0]\), the value for \(X_{1} + X_{2} = 0\) only if \(X_{2}\) were 0, and 0.11 and 0.13 come from multiplying probabilities that conditioning removes.

Your Answer : (Not Answered)

Correct Answer : A

The keyed answer is the non-central F-distribution.

For \(T^{2} = n(\bar{x} - \mu_{0})'S^{-1}(\bar{x} - \mu_{0})\) from a sample of size \(n\) from \(N_{p}(\mu, \Sigma)\), Anderson's theorem states that \(\dfrac{T^{2}}{n - 1}\dfrac{n - p}{p}\) has a non-central F-distribution with \(p\) and \(n - p\) degrees of freedom and non-centrality \(n(\mu - \mu_{0})'\Sigma^{-1}(\mu - \mu_{0})\).

When the null hypothesis \(\mu = \mu_{0}\) holds the non-centrality is zero and the distribution is the central F, which is the special case used to set critical values; the general statement, with no hypothesis assumed, is the non-central one. A chi-square or t-distribution is not the distribution of this ratio.

Your Answer : (Not Answered)

Correct Answer : D

The statement that is not true is \(\hat{Y} \sim N_{n}(\beta, \sigma^{2}H)\).

With \(H = X(X'X)^{-1}X'\) the hat matrix, \(\hat{Y} = HY\). Since \(HX = X\), \(E(\hat{Y}) = HX\beta = X\beta\) (A). Since H is symmetric and idempotent, \(Cov(\hat{Y}) = H\sigma^{2}IH' = \sigma^{2}H\) (B). And \(Cov(HY, (I - H)Y) = \sigma^{2}H(I - H) = 0\) (C).

Under normality \(\hat{Y}\) is normal with mean \(X\beta\), not \(\beta\). The vector \(\beta\) has \(p\) components while \(\hat{Y}\) has \(n\), so the stated mean is not even of the right size.

Your Answer : (Not Answered)

Correct Answer : B

The odds against are 39 : 1.

The sample mean of 100 children has standard error \(2.24/\sqrt{100} = 0.224\). Standardise the value 41.7:

\[ \begin{aligned} z &= \frac{41.7 - 41.26}{0.224} \\ &= \frac{0.44}{0.224} \approx 1.96 \end{aligned} \]

So \(P(\bar{X} \gt 41.7) = P(Z \gt 1.96) \approx 0.025\), and the probability of the opposite event is 0.975. The odds against the event are

\[ 0.975 : 0.025 = 39 : 1 \]

1 : 39 gives the odds in favour, not against. 40 : 1 and 1 : 40 compare the event with the whole, 1 in 40, rather than with its complement.

Your Answer : (Not Answered)

Correct Answer : D

The statement that is not true is E(U|F) and E(V|F) are independent.

Given \(F = \sigma(X)\), X is known and Y is independent of it with mean 0, so \(E(Y|F) = 0\). Hence \(E(U|F) = X + 0 = X\) and \(E(V|F) = X - 0 = X\), and \(E(U + V|F) = E(2X|F) = 2X\). Options A, B and C hold almost surely.

Both conditional expectations equal the same random variable X, which has variance 1. A non-degenerate random variable is not independent of itself, since \(P(X \le 0, X \le 0) = 1/2 \ne 1/4\). (U and V themselves are independent, being uncorrelated jointly normal variables, which is the trap.)

Your Answer : (Not Answered)

Correct Answer : C

A test is unbiased if \(E_{H_{0}}\phi(X) \le \alpha\) and \(E_{H_{1}}\phi(X) \ge \alpha\).

Unbiasedness means the test rejects at least as often when the alternative is true as when the null is: its size is at most \(\alpha\) and its power is at least \(\alpha\) at every alternative. Otherwise there would be an alternative under which the test is less likely to reject than under the null, which is perverse.

Each of options A and B states only one half (A is just the size condition). Option D demands power equal to \(\alpha\), which describes a useless test that ignores the data, not an unbiased one.

Your Answer : (Not Answered)

Correct Answer : C

The maximum likelihood estimator is \(\frac{1}{3n}\sum_{i=1}^{n} e^{X_{i}}\).

The log-likelihood is

\[ \begin{aligned} \ell(\beta) &= -n\log 2 - 3n\log\beta \\ &\quad + 3\sum x_{i} - \frac{1}{\beta}\sum e^{x_{i}} \end{aligned} \]

\[ \begin{aligned} \ell'(\beta) &= -\frac{3n}{\beta} + \frac{\sum e^{x_{i}}}{\beta^{2}} = 0 \\ \Rightarrow \hat{\beta} &= \frac{\sum e^{x_{i}}}{3n} \end{aligned} \]

\(\ell'\) is positive below this value and negative above it, so it is the maximum. (Substituting \(U = e^{X}\) shows \(U\) is gamma with shape 3 and scale \(\beta\), mean \(3\beta\), which explains the 3.)

\(\frac{3}{n}\sum e^{X_{i}}\) multiplies by 3 instead of dividing. \(\frac{1}{n}\sum X_{i}\) and \(\frac{1}{n}\sum X_{i}^{3}\) ignore the fact that the data enter the likelihood only through \(\sum e^{x_{i}}\).

Your Answer : (Not Answered)

Correct Answer : D

The true statement is 3 is a recurrent state.

Read the matrix row by row. From state 1 the chain can move to 2 or 3, but the first column is zero below the top entry, so no state leads back to 1. From 2 the chain goes to 2 or 3, and from 3 it always goes to 2.

So \(\{2, 3\}\) is a closed communicating class: once there, the chain never leaves, and 2 and 3 reach each other. A finite closed communicating class is recurrent, so both 2 and 3 are recurrent. State 1 is left with probability \(\frac{1}{2}\) at each step and never re-entered, so it is transient.

That rules out the other options: the chain is not irreducible, since 1 cannot be reached from 2 or 3; 1 is transient, not recurrent; and 2 is recurrent, not transient.