# Sufficient statistics and UMVUE for joint poisson, bernoulli

Given a pair $(X,Y)$ of r.v.s such that: $$X \sim \text{Poisson}(\lambda)\quad \text{and}\quad Y \sim B(\frac{\lambda}{1+\lambda})$$ with $X,Y$ independent, determine a one-dimensional sufficient statistic for $\lambda$ and the UMVUE (uniformly minimum variance unbiased estimator) for $\lambda$.

Attempt:

$$P(X,Y) = P(X)P(Y) = \Pi_{i=1}^n\frac{\lambda^{x_i}e^{-\lambda}}{x_i!} \times \frac{\lambda^{y_i}}{1+\lambda} = \frac{\lambda^{\sum (x_i+y_i)} e^{-n\lambda}}{(1+\lambda)^n \Pi x_i!}$$

Using the factorisation theorem, $\sum(x_i+y_i)$ is a sufficient statistic. It is however not an unbiased estimator of $\lambda$ because:

$E(\sum(x_i+y_i)) = n(\lambda + \frac{\lambda}{1+\lambda})$

A trivial unbiased estimator would be $\frac{\sum x_i}{n}$, but then it is not sufficient. Clues?

• This sounds like self-study: if so, could you add the tag and read the wiki? Thank you. May 8 '15 at 20:51
• Hint: you are almost there: (i) you have a sufficient statistics and (ii) you have an unbiased estimator. All you need is the appropriate theorem linking the two towards the UMVUE. May 8 '15 at 20:55
• Not a self-study. Was an exam problem May 8 '15 at 23:01
• That leads to Rao-Blackwellising it by conditioning $\sum x_i$ on $\sum(x_i+y_i)$ which I am not able to comprehend May 9 '15 at 16:02
• Use the facts that $\sum X_i\sim \text{P}(n\lambda)$ and $\sum Y_i\sim\text{B}(n,\lambda/(1+\lambda))$ to derive the joint density of $(\sum X_i,\sum Y_i)$, then of $(\sum X_i,\sum X_i+\sum Y_i)$ then of $\sum X_i$ conditional on $\sum X_i+\sum Y_i$. From there you can get $\mathbb{E}[\sum X_i|\sum X_i+\sum Y_i]$. May 9 '15 at 16:06