Let $X_1....X_n$ be a random sample from the Poisson distribution with parameter $\lambda$. Find the uniformly minimum-variance unbiased estimator of

a) $\lambda^3$

b) $E(e^X)=e^{\lambda (e-1)}$

I'm having some difficulty with this. So far, I've found the function at the same time as the estimator using $a(\theta)[W(x)-\tau(\theta)]=\frac{\delta}{\delta \theta} Log(L(\theta|x)$ where W(x) is the estimator and $\tau(\theta)$ is the function.

Any assistance with a theorem or idea to begin here? Or can that be the right track?

  • $\begingroup$ I'm no expert in this, but what you did is proving that the score function can be written in efficient form. This means that $W(x)$ is an efficient estimator for $\tau(\theta)$, i.e. its variance is the smallest possible and reaches the Cramer-Rao lower bound. Now all you have to do is show that $W(x)$ is unbiased: $\mathrm{E}[W(x)|\theta]=\tau(\theta)$ and you're done,... I think. $\endgroup$ – StijnDeVuyst Mar 23 '15 at 23:49

There is one main strategy that I use for this type of problem. The key theorem is the Lehmann-Scheffe theorem which tells us that if we can get an unbiased estimator for $\tau(\theta)$ that is purely a function of a complete and sufficient statistic for $\theta$ [being a bit casual with definitions] then we have the unique BUE of our target.

The Poisson distribution is an exponential family so we know that $S := \sum\limits_{i=1}^n X_i$ is complete and sufficient for $\lambda$. We also know that $S/n$ is the MLE of $\lambda$.

For (1) we need to find an unbiased estimator of $\lambda^3$ that is only a function of $S$. We know by the invariance principle of the MLE that $S^3 / n^3$ is the MLE for $\lambda^3$ so we'll start there.

The $X_i$ are iid so we know that $S \sim Poisson(n\lambda)$ so its MGF is $M_S(t) = e^{n\lambda(e^t-1)}$. Computing the first three derivatives of this gives us that

$$ E(S^3/n^3) = \lambda^3 + 3\lambda^2 / n + \lambda / n^2. $$

We now need to eliminate the lower order powers of $\lambda$ and can do so by a linear combination of powers of $S$, giving us that

$$ E(\bar{X}^3 - \frac{3}{n}\bar{X}^2 + \frac{2}{n^2}\bar{X}) = \lambda^3. $$

Since the statistic $T(\vec{X}) = \bar{X}^3 - \frac{3}{n}\bar{X}^2 + \frac{2}{n^2}\bar{X}$ is purely a function of the complete and sufficient statistic $S$ it is the unique UMVUE for $\lambda^3$. Note that this requires $n$ to be non-random.

For (2) we will proceed by Rao-Blackwellization with $S$. This really is just Lehmann-Scheffe again but it looks a little different. The idea is that if $W$ is an unbiased estimator for $\tau(\theta)$ and $T$ is a sufficient statistic for $\theta$ then $\Phi(T) = \mathbb{E}(W|T)$ is often a better and never a worse estimator of $\tau(\theta)$ than $W$ alone. Note that $\Phi$ is only a function of $T$ so if $T$ is also complete then by Lehmann-Scheffe we have the unique UMVUE. So this is really just a different algebraic route to doing the same thing.

We've been handed our unbiased estimator by being told that $E(e^X) = e^{\lambda(e-1)}$ [note that this is $M_X(1)$].

This tells us that $\mathbb{E}(e^{X_1}|S)$ is our UMVUE [note that by the tower property we haven't changed its mean]. Let's work out what this is.

$$ \mathbb{E}(e^{X_1}|S = s) = \sum\limits_{x_1=0}^s e^{x_1} P(X_1 = x_1 | S = s) $$

$$ = \sum e^{x_1} \frac{P(X_1 = x_1 , S = s)}{P(S = s)} = \sum e^{x_1} \frac{P(X_1 = x_1) P(\sum\limits_{i=2}^n X_i = s - x_1) }{P(\sum\limits_{i=1}^n X_i = s)}. $$

Using the fact that the $X_i$ are iid and Poisson you should be able to work this out.


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