P1

• Suppose that $X_1$ and $X_2$ are independent Poisson random variables each with parameter $\lambda$
• Let the parameter $\theta$ by
  $\theta=e^{-\lambda}$

1. Show that $X_1+X_2$ is a sufficient statistic for $\theta$
   (Assume $\lambda$ ranges of $(0,\infty))$
2. Show that $X_1+X_2$ is also complete
   (Note that sum of independent Poisson random variables is a Poisson random variable
   In addition, if the sum is a power series and is zero, then all the coefficients are zero.)
3. Define an estimate $\hat \theta$ by
   $\hat\theta(\text{x})=\frac{1}{2}(f(x_1)+f(x_2))$
   where $f$ is defined by
   $f(x) = \begin{cases} 1, & \text{if } x = 0 \\ 0, & \text{if } x \neq 0 \end{cases}$
   Show that $\hat\theta$ is an unbiased estimate of $\theta$
4. Find an MVUE of $\theta$

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Solution 1

• Let $Y=X_1+X_2$
  $p(X_1=x_1,X_2=x_2|Y=y;\theta)\\[0.3cm] =\frac{p(X_1=x_1,X_2=x_2,Y=y;\theta)}{p(Y=y;\theta)}\\[0.3cm] Y \text{ is Poisson Distribution paramter }2\lambda\\[0.3cm] =\frac{p(X_1=x_1,X_2=x_2;\theta)\delta(y-(x_1+x_2))}{p(Y=y;\theta)}\\[0.3cm] =\frac{\frac{(-\ln\theta)^{x_1}\theta}{x_1!}\cdot\frac{(-\ln\theta)^{x_2}\theta}{x_2!}}{\frac{(-2\ln\theta)^y\theta^2}{y!}},\;(x_1+x_2=y)\\[0.3cm] =\frac{(-1)^{x_1+x_2}}{(-2)^{x_1+x_2}}\cdot\frac{(x_1+x_2)!}{x_1!\cdot x_2!} \text{ not depend on }\theta\\[0.3cm] \rightarrow Y=X_1+X_2 \text{ sufficient statistics}$

Solution 2

Solution 3

Solution 4

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P2

• Assume that $x[n]$ is the result of a Bernoulli trial (a coin toss) with
  $\Pr\{x[n]=1\}=\theta\\[0.3cm] \Pr\{x[n]=0\}=1-\theta$
• And that $N$ IID observations have been made
• Assuming that Neyman-Fisher factorization theorem holds for discrete random variables
• Find a sufficient statistic for $\theta$
• Then, assuming completeness, find the MVUE for $\theta$

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Solution