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Problem: Let $X_i \sim Pois(\lambda)$. Find the UMVUE of the probability that $X_1$ is odd.

My attempt:

I don't think there's any obvious unbiased estimator to use conditioning. So instead I write

$$P(X_1 \text{ is odd}) = \sum_{k=0}^\infty \frac{\lambda^{2k+1}}{(2k+1)!}e^{-\lambda}$$

The complete sufficient statistic is $T(X)= \sum_{i=1}^n X_i \sim Pois(n\lambda)$, so an unbiased estimator must satisfy

$$\sum_{k=0}^\infty \delta(k)\frac{(n\lambda)^k}{k!}e^{-n\lambda} =\sum_{k=0}^\infty \frac{\lambda^{2k+1}}{(2k+1)!}e^{-\lambda}$$

But here I am stuck. I know I need to write out the power series of each exponential, but then I get left with the product of two sums which does not seem very helpful.

$$\sum_{k=0}^\infty \delta(k)\frac{(n\lambda)^k}{k!}\sum_{j=0}^\infty \frac{\lambda^j}{j!}=\sum_{j=0}^\infty \frac{(n\lambda)^j}{j!}\sum_{k=0}^\infty \frac{\lambda^{2k+1}}{(2k+1)!}$$

Using Fubinis I thiink I can write this as

$$\sum_{k=0}^\infty \sum_{j=0}^\infty \delta(k)\frac{(n\lambda)^k}{k!} \frac{\lambda^j}{j!}=\sum_{k=0}^\infty\sum_{j=0}^\infty \frac{(n\lambda)^k}{k!} \frac{\lambda^{2j+1}}{(2j+1)!}$$

This seems to suggest $\delta(t)$ is of the form $\delta(t) = \frac{t!}{(2t+1)!}$ when $t$ is odd, and 0 otherwise?

I am not confident with my attempt at all.

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We have the exact expression, verifiable using the power series expansion of $e^{\lambda}$: $$\sum_{k=0}^\infty \frac{\lambda^{2k+1}}{(2k+1)!}=\frac{1}{2}(e^{\lambda}-e^{-\lambda})$$

So that reduces the probability to $$P(X_1=\text{odd})=\frac{1}{2}(1-e^{-2\lambda})=g(\lambda)\,,\text{ say }$$

Since $T\sim \mathsf{Poi}(n\lambda)$ we have $E(a^T)=e^{n\lambda(a-1)}$. This equals $e^{-2\lambda}$ for $a=1-\frac{2}{n}$.

So UMVUE of $e^{-2\lambda}$ based on a sample of $n$ observations is $$h(T)=\left(1-\frac{2}{n}\right)^T$$

This means UMVUE of $g(\lambda)$ is $$\frac{1}{2}\left[1-h(T)\right]=\frac{1}{2}\left[1-\left(1-\frac{2}{n}\right)^T\right]$$

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  • $\begingroup$ $T=\sum X_i$ to clarify. $\endgroup$ – StubbornAtom May 24 at 9:53
  • $\begingroup$ Oh that makes so much more sense. Thank you! $\endgroup$ – Xiaomi May 24 at 10:05

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