Conditional Expectation with complete sufficient statistics

Question

Let $X_1,...,X_n$ be a sample from a poisson distribution with parameter $θ > 0$. I need to find the $E(X_1^2 | \sum_{i=1}^nX_i)$. The only thing i noticed is that $\sum_{i=1}^nX_i$ is a complete sufficient statistic. I'm confused about this problem and how to approach it, could anyone help? Thank you!

Answer 1

$$P(X_1 = p | \sum_i X_i = q) = \frac{P(\sum_i X_i = q,X_1 = p)}{P(\sum_i X_i = q)} = \frac{P(\sum_i X_i = q|X_1 = p)P(X_1 = p)}{P(\sum_i X_i = q)} = \frac{P(\sum_{i \geq 2} X_i = q-p)P(X_1 = p)}{P(\sum_i X_i = q)}$$

$$P(\sum_{i = 1}^n X_i = q) = \frac{e^{-(n\theta)} (n\theta)^q}{q!}$$

$$P(X_1 = p | \sum_i X_i = q) = \frac{P(\sum_{i \geq 2} X_i = q-p)P(X_1 = p)}{P(\sum_i X_i = q)} = \frac{q!}{(q-p)! p!} \left(\frac{n-1}{n}\right)^{q-p} \left(\frac{1}{n}\right)^{p}$$

Hence conditional mass function $P(X_1 = p | \sum_i X_i = q)$ follows binomial distribution.

Now $$E(X_1^2|\sum_i X_i = q) = q(q-1) \left(\frac{1}{n}\right)^2 + q \left(\frac{1}{n}\right)$$

Answer 2

I don't know if there is a more elegant/general solution (perhaps relating with the suficient statistics thing) but it's quite easy to show that $X_1$ conditioned on $\sum_{i=1}^n X_i$ follows a Binomial distribution. From that, we can calculate the (conditional) mean and variance, and from that the desired $E(X_1^2 | \sum_{i=1}^n X_i)$

Answer 3

Using a hint which @StubbornAtom wrote in the comments, i get the following. $E(X_{1}^{2} | \sum_{i=1}^nX_i)$ by the corollary on the optimal estimate for the Kolmogorov-Blackwell-Rao theorem, i need to find optimal estimate of the parameter $E(X_1^2) = \theta^2 + \theta$. Sufficient statistic of possion distribtuion is $\sum_{i=1}^nX_i$. Also this statistic is a complete. So considering $\phi(\sum_{i=1}^nX_i) =(\frac{1}{n}\sum_{i=1}^nX_i)^2 + \frac{1}{n}\sum_{i=1}^nX_i(1-\frac{1}{n})$, we get that $(\frac{1}{n}\sum_{i=1}^nX_i)^2 + \frac{1}{n}\sum_{i=1}^nX_i(1-\frac{1}{n})$ is a optimal estimator(best unbiased) of $E(X_1^2)$. This answer is the same as @Balaji sb's answer.