Point and Interval Estimation Questions and Answers

Answer: B

The true statement is \(\sum_{i=1}^{n} X_{i}^{2}\) is a sufficient statistic for \(\sigma^{2}\).

The joint density factorises as

\[ \begin{aligned} f(x) &= (2\pi\sigma^{2})^{-\frac{n}{2}} \\ &\quad \times \exp\left(-\frac{\sum x_{i}^{2}}{2\sigma^{2}}\right) \end{aligned} \]

which depends on the data only through \(\sum x_{i}^{2}\), so by the factorisation theorem that statistic is sufficient.

\(\sum X_{i}\) is not sufficient: with the mean fixed at 0, it carries little information about the spread. Nor is it complete, since \(\sum X_{i} \sim N(0, n\sigma^{2})\) has \(E(\sum X_{i}) = 0\) for every \(\sigma\) although \(\sum X_{i}\) is not zero. Finally \(E(X_{i}^{2}) = \sigma^{2}\), so \(\frac{1}{n}\sum X_{i}^{2}\) is an unbiased estimator of \(\sigma^{2}\), not a biased one.

Answer: D

Neither P nor Q is correct.

For \(Exp(\beta)\) with mean \(\beta\), the maximum likelihood estimator of \(\beta\) is \(T = \bar{X}\). By invariance, the MLE of any function of \(\beta\) is that function evaluated at \(T\).

P: the minimum of 10 independent \(Exp(\beta)\) variables is \(Exp\left(\frac{\beta}{10}\right)\), whose median \(m\) solves \(1 - e^{-10m/\beta} = \frac{1}{2}\):

\[ \begin{aligned} m &= \frac{\beta\log_{e}2}{10} \\ \hat{m} &= \frac{(\log_{e}2)T}{10} \end{aligned} \]

\((\log_{e}2)T\) is the MLE of the median of \(X_{1}\) itself, ten times too large for \(X_{(1)}\).

Q: \(P(X_{1} \lt 2) = 1 - e^{-2/\beta}\), so its MLE is \(1 - e^{-2/T}\). \(e^{-2/T}\) estimates the complement \(P(X_{1} \gt 2)\).

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 \(\sum e^{x_{i}}\).

Answer: C

The conditional density is \(\dfrac{(n - 1)(t - x_{1})^{n - 2}}{t^{n - 1}}\), \(0 \lt x_{1} \lt t\).

\(X_{1}\) is exponential with mean \(\theta\), and \(S = X_{2} + \ldots + X_{n}\), independent of it, is Gamma with shape \(n - 1\). T = \(X_{1} + S\) is Gamma with shape n. Then

\[ \begin{aligned} f(x_{1} \mid t) &= f_{X_{1}}(x_{1}) \\ &\quad \times \frac{f_{S}(t - x_{1})}{f_{T}(t)} \end{aligned} \]

In the numerator the exponentials combine to \(e^{-t/\theta}\), which cancels with the denominator, along with every power of \(\theta\). What remains is

\[ \begin{aligned} &\frac{\Gamma(n)}{\Gamma(n - 1)} \cdot \frac{(t - x_{1})^{n - 2}}{t^{n - 1}} \\ &= \frac{(n - 1)(t - x_{1})^{n - 2}}{t^{n - 1}} \end{aligned} \]

It is free of \(\theta\), as it must be because T is sufficient. In other words \(X_{1}/T\) is Beta(1, n - 1). The first two options are this same formula with n replaced by \(n - 1\) and \(n + 1\), that is, the answers for the wrong sample size, and the last still contains \(\theta\), which sufficiency rules out.

Answer: C

The log likelihood is a concave function of \(\lambda\).

For a Poisson sample the log likelihood is

\[ \begin{aligned} L(\lambda) &= -n\lambda + \Big(\sum x_{i}\Big)\log\lambda \\ &\quad - \sum \log(x_{i}!) \end{aligned} \]

Its second derivative is \(L''(\lambda) = -\dfrac{\sum x_{i}}{\lambda^{2}} \le 0\), so L is concave on \(\lambda \gt 0\) (strictly so whenever some \(x_{i} \gt 0\)). This is why the maximum likelihood estimate \(\hat{\lambda} = \bar{x}\), where \(L'(\lambda) = 0\), is the unique maximum.

L is not constant, since it depends on \(\lambda\). It is not monotonic in general: it rises up to \(\bar{x}\) and falls after it. And it is not convex, because the second derivative is negative, not positive.

Answer: D

\(\left(\frac{n+1}{n}\right)T_{1} - 2T_{2}\) and \(\left(\frac{n+1}{n}\right)T_{1}\) are uncorrelated.

For \(U(0, \theta)\), \(E(X_{(n)}) = \dfrac{n\theta}{n + 1}\) and \(E(\bar{X}) = \dfrac{\theta}{2}\). So \(\frac{n+1}{n}T_{1}\) and \(2T_{2}\) are both unbiased for \(\theta\), and their difference \(D = \frac{n+1}{n}T_{1} - 2T_{2}\) has expectation 0: it is an unbiased estimator of zero.

\(X_{(n)}\) is complete and sufficient, so by Lehmann-Scheffe \(\frac{n+1}{n}T_{1}\) is the UMVUE of \(\theta\). A standard characterisation says an unbiased estimator is UMVUE if and only if it is uncorrelated with every unbiased estimator of zero. Hence D and \(\frac{n+1}{n}T_{1}\) are uncorrelated.

The other options fail: \(T_{1}\) itself is biased, so it is not UMVUE (and \(T_{2}\) is unbiased for \(\theta/2\), not for \(\theta\)); \(T_{2}\) is unbiased for \(\theta/2\) but is not a function of the sufficient statistic, so it is not UMVUE; and 'correlated' contradicts the characterisation.

Answer: B

The true statement is Risk of \(T_{2} = (1 - \mu + \mu^{2})/2\).

Risk under squared error is variance plus squared bias. \(T_{1} = X\) is unbiased, so its risk is \(\mu^{2} + 1\), not \(\mu^{2}\). For \(T_{2} = X/2 + 1/2\): bias \((\mu + 1)/2 - \mu = (1 - \mu)/2\), variance \((\mu^{2} + 1)/4\), so

\[ \begin{aligned} R(T_{2}) &= \frac{(1 - \mu)^{2} + \mu^{2} + 1}{4} \\ &= \frac{1 - \mu + \mu^{2}}{2} \end{aligned} \]

Then \(R(T_{1}) - R(T_{2}) = (\mu^{2} + \mu + 1)/2 \gt 0\) for every \(\mu\), so \(T_{2}\) is uniformly better. That makes C false and shows \(T_{1}\) is inadmissible, so D is false too.

Answer: A

The MLE is 2.

With \(x = 2\), \(y = 3\), the log-likelihood (dropping constants) is \(\ell(\lambda) = -\lambda + 2\log\lambda - (\lambda + 1) + 3\log(\lambda + 1)\). Setting the derivative to zero:

\[ \begin{aligned} -2 + \frac{2}{\lambda} + \frac{3}{\lambda + 1} &= 0 \\ 2\lambda^{2} - 3\lambda - 2 &= 0 \end{aligned} \]

after multiplying by \(\lambda(\lambda + 1)\). The roots are \(\lambda = 2\) and \(\lambda = -1/2\), and only the positive one is admissible, so \(\hat{\lambda} = 2\). The value 2.5 is the average of 2 and 3, which ignores the different means, and 3 is the observed \(y\).

Answer: B

The MLE is \((X_{(3)})^{1/2}\).

The likelihood is \((\theta^{2} - \theta)^{-3}\) when every observation lies in \([\theta, \theta^{2}]\), and 0 otherwise. That requires \(\theta \le X_{(1)}\) and \(\theta^{2} \ge X_{(3)}\), so

\[ \sqrt{X_{(3)}} \le \theta \le X_{(1)} \]

For \(\theta \gt 1\), \(\theta^{2} - \theta\) is increasing, so the likelihood decreases in \(\theta\) and is largest at the smallest admissible value, \(\sqrt{X_{(3)}}\).

\(X_{(1)}^{2}\) and \(X_{(1)}^{2} + X_{(3)}\) are not even in the admissible interval in general, and the weighted average in D is a different estimator, not the maximiser.

Answer: C

The statement that is not correct is \(\bar{X}\) is sufficient for \(\mu\) and \(s^{2}\) is sufficient for \(\sigma^{2}\).

With both parameters unknown, the normal likelihood depends on the data through \((\sum X_{i}, \sum X_{i}^{2})\), so that pair, and its one-to-one function \((\bar{X}, s^{2})\), is jointly sufficient (A, with the paper's subscript 1 read as \(i\), and B). If \(\sigma^{2}\) is known, \(\bar{X}\) alone is sufficient for \(\mu\) (D).

Sufficiency does not split this way when both are unknown: \(\bar{X}\) alone is not sufficient for \(\mu\), because the likelihood also involves \(\sigma^{2}\) through \(\sum(X_{i} - \mu)^{2}\), and \(s^{2}\) alone is not sufficient for \(\sigma^{2}\). Only the pair is.