Q) Let $(x_1, x_2, \cdots, x_n)$ be an observed sample from an exponential($\lambda$) distribution. Note: Exponential($\lambda$) distribution has probability density function $f(x) = \lambda e^{-\lambda x}$ for $x \geq 0$ and zero otherwise. It can be shown that $T_n = \bar{X}$ is an maximum likelihood of $\psi(\lambda) = \frac{1}{\lambda}$

(a) Is $T_n$ an unbiased estimator of $\psi_1(\lambda)$? Prove your answer.

$$E(T_n) = E(\bar{X}) = E(X_1) = \frac{1}{\lambda} = \psi_1(\lambda)$$

(b) Is $U_n = \bar{X}^2$ an unbiased estimator of $\psi_2(\lambda)$? Prove your answer.

$$E(U_n) = E(\bar{X}^2) = V(X) + E(X)^2 = \frac{1}{n}V(X_1) + E(X_1)^2 = \frac{1}{n\lambda^2} + \frac{1}{\lambda^2} \neq \frac{1}{\lambda^2}$$ so no

(c) Is $U_n = \bar{X}^2$ an asymptotically unbiased estimator of $\psi(\lambda) = \frac{1}{\lambda^2}$. Prove your answer.

$$\lim_{n\to\infty} E(U_n) = \lim_{n\to\infty} \left(\frac{1}{n\lambda^2} + \frac{1}{\lambda^2} \right) = \frac{1}{\lambda^2} = \psi_2(\lambda)$$

(d) An observed sample of ten observations from this distribution had mean $\bar{x} = 0.6$. Use this information to calculate the standard error of $T_n = \bar{X}$.

$Var(T_n) = V(\bar{X}) = \frac{1}{n}V(X_1) = \frac{1}{n\lambda^2}$

MLE of $\lambda = \frac{1}{\bar{x}} = \frac{1}{0.52}$ so

$var(T_n) = \frac{1}{n\lambda^2} = \frac{1}{4 (0.52)^2} = 0.0676$

$SE(X) = \sqrt{V(T_n)} = 0.26$

Is this correct?

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    $\begingroup$ If you could pinpoint your doubt, that would be better than a yes/no question. $\endgroup$ – StubbornAtom Mar 15 '19 at 15:43

Your reasoning, while largely correct, lacks rigor.

In (a), you state without proof $$\operatorname{E}[\bar X] = \operatorname{E}[X_1].$$ While true, you do not explain why it is true. The reason is due to linearity of expectation and the fact that the $X_i$ are identically distributed: $$\operatorname{E}[\bar X] = \operatorname{E}\left[\frac{1}{n} \sum_{i=1}^n X_i\right] = \frac{1}{n} \sum_{i=1}^n \operatorname{E}[X_i] = \frac{1}{n} \sum_{i=1}^n \operatorname{E}[X_1] = \frac{1}{n} \cdot n \operatorname{E}[X_1] = \operatorname{E}[X_1].$$ We used linearity in the second equality from the left, and we used the identically distributed property of the $X_i$ in the third equality from the left. Technically, one does not need that that $X_i$ are identically distributed, since all that is required is that $\operatorname{E}[X_i] = 1/\lambda$ for each $i = 1, 2, \ldots, n$ for the claim $\operatorname{E}[\bar X] = 1/\lambda$ to be true.

In (b), you make the assertion $$\operatorname{E}[\bar X^2] = \operatorname{Var}[X] + \operatorname{E}[X]^2$$ without proof. This is not correct, unless you meant to write $$\operatorname{E}[\bar X^2] = \operatorname{Var}[\bar X] + \operatorname{E}[\bar X]^2.$$ If you meant the latter, then the following step requires substantiation along the same lines as in (a); namely $$\operatorname{Var}\left[\frac{1}{n} \sum_{i=1}^n X_i\right] = \frac{1}{n^2} \sum_{i=1}^n \operatorname{Var}[X_i] = \frac{1}{n^2} \sum_{i=1}^n \operatorname{Var}[X_1] = \frac{1}{n} \operatorname{Var}[X_1],$$ where the first equality holds due to independence of the $X_i$, and the second due to the $X_i$ being identically distributed.

(c) is correct.

(d) seems to have some inconsistencies. Is $\bar x = 0.6$, or is $\bar x = 0.52$? Which one? Also, is $n = 10$, or is $n = 4$?

In any case, the estimate of the standard error of the sample mean is given by $$\operatorname{SE}[\bar X] = \sqrt{\widehat{\operatorname{Var}}[\bar X]}.$$ From part (b) above, we know that $$\operatorname{Var}[T_n] = \frac{1}{n}\operatorname{Var}[X_1] = \frac{1}{n \lambda^2} = \frac{1}{n} \cdot \operatorname{E}[\bar X]^2.$$ This suggests that an appropriate choice would be $$\widehat{\operatorname{Var}}[\bar X] = \frac{\bar X^2}{n} = \frac{U_n}{n}.$$ This is by no means the only possible choice for the estimated variance of $\bar X$, since $U_n$ is biased. To compensate, you could note $$\operatorname{E}\left[\frac{n}{n+1} U_n\right] = \frac{n}{n+1} \left( \frac{1}{n\lambda^2} + \frac{1}{\lambda^2} \right) = \frac{n}{n+1} \left( \frac{1}{n} + 1 \right) \frac{1}{\lambda^2} = \frac{1}{\lambda^2}$$ is unbiased, so an unbiased estimator for the variance of the sample mean would be $U_n/(n+1)$.


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