1
$\begingroup$

Let's say that we want to find Unbiased Minimum Variance estimator of $\lambda^2$ for a sample $X=(X_1,\dots,X_n)$ from Poisson distribution. We can consider $\overline{X}^2$ and notice that

$$\mathbb{E}(\overline{X}^2) = Var(\overline{X}) + \left(\mathbb{E}(\overline{X})\right)^2 = \frac{1}{n^2} \cdot n \cdot Var(X_1) + \lambda^2 = \lambda^2 + \frac{\lambda}{n} = \lambda^2 + \mathbb{E}\left(\frac{1}{n}\overline{X} \right)$$

Which gives us that $\overline{X}^2 + \frac{1}{n}\overline{X}$ is what we are searching for: because it's a function of $\overline{X}$, which for exponential families is sufficient, and in this case also complete, statistic, then from Lehmann-Scheffe theorem we conclude that it's UMVU estimator.


Now, let's consider a function like: $$g(\lambda) = \frac{\lambda^3}{3!}e^{-\lambda}.$$

Which is actually a probability $P(X_i=3)$. What can we do with this, to find an UMVU estimator of this one? I was wondering to start, just like above, from considering

$$T(X)= \frac{\overline{X}^3}{3!}e^{-\overline{X}},$$

and eventually adjust it with some other components, but trying to find expected value of this almost made me cry. I was thinking about transforming it somehow and using Basu theorem make things easier, but not sure how.

Any advice how to do this? Or maybe I can find the UMVUE easier here?

$\endgroup$
1
$\begingroup$

Start from $$ g_n = \mathcal{I}\{X_1 = 3\}, $$ as an unbiased estimator. Then, using Rao-Blackwell, compute $$ g_n^{RB} = \mathbb{E}[g_n|\sum _{i=1}^n X_i =t]. $$ Note that $g_n^{RB}$ is an unbiased estimator and function of the complete minimal sufficient statistic $\sum_{i=1}^n X_i$. Thus, by Lehmann-Scheffe, it is a UMVUE.

Namely, \begin{align} g_n^{RB} &= \mathbb{E}[g_n|\sum _{i=1}^n X_i =t]\\ &= \frac{\mathbb{P}(X_1 = 3) \mathbb{P}( \sum _{i=2}^n X_i =t - 3)}{\mathbb{P}( \sum _{i=1}^n X_i =t)}\\ & =\frac{e^{-\lambda}\lambda^3/3! \times e^{-\lambda (n-1)}\lambda^{t-3} (n-1)^{t-3} / (t-3)!}{e^{-\lambda n}\lambda^t n^t/t!} \\ & = \left( \frac{n-1}{n} \right)^{t} \binom{t}{3}(n-1)^{-3} \\ & = \left( 1 - \frac{1}{n} \right)^{n\bar{x}_n} \binom{n\bar{x}_n}{3}(n-1)^{-3}. \end{align} Now, for validation, you can use the continuous mapping theorem and the WLLN. Note that $\bar{X}_n \xrightarrow{p} \lambda$, same is true for $\frac{1}{n-1}\sum X_i \xrightarrow{p} \lambda$, and note that $(1-1/n)^n \xrightarrow{} e^{-1}$. Hence, combining it all, the estimator converges to $g(\lambda)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What does $\mathcal{I}$ in $g_n$ stands for? Fisher Information? $\endgroup$ – Kusavil Feb 6 '18 at 21:13
  • $\begingroup$ $\mathcal{I}$ - indicator function, $g_n$ - an estimator of $g(\lambda)$ $\endgroup$ – V. Vancak Feb 6 '18 at 21:15
  • $\begingroup$ So we have: $\mathbb{P}(X_1 = 3)=\frac{\lambda^3}{3!}e^{-\lambda}$, and because for $X\sim Poiss(\lambda)$ we have $\sum_{i=1}^n \sim Poiss(n \lambda)$, thus we have $\mathbb{P}( \sum _{i=1}^n X_i =t) = \frac{(n \lambda)^t}{t!}e^{-n \lambda}$, then similarly $\mathbb{P}(\sum_{i=1}^n X_i =t-3)= \frac{(n \lambda)^{t-3} }{(t-3)!} e^{-n \lambda}$, which gives me $\frac{\mathbb{P}(X_1 = 3) \mathbb{P}( \sum _{i=1}^n X_i =t - 3)}{\mathbb{P}( \sum _{i=1}^n X_i =t)} = \binom{t}{3}e^{-\lambda}(1-\frac{1}{n})^3(n-1)^{-3}$. How to I transform this into your form? $\endgroup$ – Kusavil Feb 6 '18 at 21:46
  • $\begingroup$ Also, what does $t$ stands for? Where do we know $t$ from? $\endgroup$ – Kusavil Feb 6 '18 at 21:47
  • 1
    $\begingroup$ Please the edited answer $\endgroup$ – V. Vancak Feb 6 '18 at 23:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.