Question 1 [16 marks] Let X1, …, Xn1 be a random sample from...

Question 1 [16 marks] Let $X_1, \dots, X_{n_1}$ be a random sample from Poisson($\mu$) distribution, and $Y_1, \dots, Y_{n_2}$ be another sample from Poisson($\mu$). We call them $X$ and $Y$ sample in this questions. Assume $X$ and $Y$ samples are independent. Let $X = \sum_{i=1}^{n_1} X_i$ and $Y = \sum_{j=1}^{n_2} Y_j$. Consider the following two estimators of $\mu$: $\hat{\mu}_1 = \frac{n_1X + n_2Y}{n_1 + n_2}$ and $\hat{\mu}_2 = \frac{X + Y}{2}$. Note that the mean and variance of a Poisson($\lambda$) distribution are $\lambda$ and $\lambda$, respectively. (a) Demonstrate that both $\hat{\mu}_1$ and $\hat{\mu}_2$ are unbiased estimators of $\mu$. (b) Show that the variances of $\hat{\mu}_1$ and $\hat{\mu}_2$ are given by, respectively, $\text{var}(\hat{\mu}_1) = \frac{\mu}{n_1 + n_2}$ and $\text{var}(\hat{\mu}_2) = \frac{n_1 + n_2}{4n_1n_2}\mu$. (c) Show that both $\hat{\mu}_1$ and $\hat{\mu}_2$ are consistent estimators of $\mu$. (d) Derive the relative efficiency of $\hat{\mu}_1$ to $\hat{\mu}_2$. (e) Which estimator do you prefer? Give reasons to your answer. (f) Demonstrate that $\hat{\mu}_1$ achieves the Cramer-Row lower bound among all unbiased esti- mators created from the combined $X$ and $Y$ samples. (g) Find normal approximations for $\hat{\mu}_1$ and $\hat{\mu}_2$ when both $n_1$ and $n_2$ are large.

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00:04 By definition, we know mx t is equal to e bracket e to the power px which is equal to mean upper limit n lower limit x is equal to 0 f multiplied by x element of to the power tx which is equal to mean upper limit n lower limit x is equal to 0 bracket n x bracket close theta to the power x one subtracted from theta to the power n subtracted from x e to the power tx which is equal to mean upper limit n lower limit x is equal to 0 nx one subtracted from theta to the power n subtracted from x so theta e to the power t whole x which is equal to mode setting bracket theta e to the power t added to one subtracted from theta to the power n hence mx t is equal to bracket theta e to the power t added to one subtract from theta bracket close n with respect to gives m dash x t is equal to n theta e to the power t added to one subtracted from theta to the power n subtracted from one theta e to the power t so t is equal to zero gives e x is equal to m dash x zero is equal to n theta again m double dash x t is equal to n theta bracket theta e to the power t added to one subtract from theta to the power n subtracted from e e to the power t added to n subtracted from one bracket close bracket opens theta e to the power t added to one subtracted from theta n theta to the power n subtracted from two theta e to the power t bracket close so e x square is equal to m double dash x zero is equal to n theta bracket open theta added to one subtracted from theta to the power n subtracted from one added to n subtracted from one theta added to one subtracted from theta to the power n subtracted from two theta bracket close which is equal to n theta one added to n subtracted from one theta so therefore radius x is equal to e x square subtracted from e x whole square which is equal to n theta bracket one added to n subtracted from one theta bracket close subtracted from n square theta square which is equal to n theta one subtracted from the question to that is b by definition.

04:07 We know mx t is equal to e bracket e to the power tx which is equal to mean upper limit infinity lower limit x is equal to 0 s x e to the power tx which is equal to mean upper limit infinity lower limit x is equal to 0 e to the power minus lambda lambda to the power x divided by x e to the power tx which is equal to e to the power minus lambda mean upper limit infinity lower limit x is equal to 0 bracket lambda e to the power t pulse x divided by x to e to the power lambda e to the power lambda e to the power t which is equal to e to the power lambda e to the power t subtracted from one hence mx t is equal to e lambda e to the power t subtracted from one so differentiate with respect to t gives m dash x t is equal to e to the power lambda bracket e to the power t subtracted from one bracket lambda e to the power t bracket closed.

05:48 P t is equal to 0 gives...

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