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Let $X_1, ..., X_n$ be random variables with pdf $$\frac 1 \lambda e^{-x / \lambda} I(x > 0).$$

The goal is to find the best unbiased estimator of $h(\lambda) = e^{-\lambda}$ (incidentally, this is equal to $P(X_1 > \lambda^2)$). I'm currently tutoring some graduate students for a first year qualifying examination and in an embarrassing turn of events I've been unable to figure out the solution to this problem on a past exam. I'm sure I'm missing something obvious.

Some thoughts: the MLE of $h(\lambda)$ is $e^{-\bar X}$. The expectation can be calculated easily, but it is hopelessly biased:

$$E[e^{-\bar X}] = M_{\bar X} (-1) = M_{X_1} \left(-\frac 1 n\right)^n = \left(1 + \frac \lambda n\right)^{-n}$$

where $M_Y$ is the moment generating function of $Y$. There are only two ways of going about this that I can think of that are available to the students. One is to "guess" a $g(\bar X)$ such that $E[g(\bar X)] = e^{-\lambda}$. The other is to find an unbiased estimate of $e^{-\lambda}$ and Rao-Blackwell to finish it off (a prior part to the same question essentially sets up the ability to carry out such a calculation, so it is somewhat hinted at that this might work).

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I wonder why you restrict yourself to estimators of the form $g(\bar X)$. If not, consider $(g(X_1)+\cdots+g(X_n))/n$ where you choose $g$ such that $E(g(X_1))=h(\lambda)$ for every $\lambda$. A suitable function $g$ might be $g(x)=\sum(-1)^kx^k/(k!)^2$. –  Did Jul 24 '11 at 17:11
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@Didier Piau: $\bar X$ is complete sufficient, so the best unbiased estimator must be a function of $\bar X$. –  guy Jul 24 '11 at 17:22
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2 Answers

up vote 4 down vote accepted

The distribution of $S=n\bar X_n$ has density $$ P_S(\mathrm{d}s)=\frac{s^{n-1}}{(n-1)!}\mathrm{e}^{-s/\lambda}\frac{\mathrm{d}s}{\lambda^n}. $$ Hence, for every $k\ge0$, $$ E(S^k)=\frac{(n+k-1)!}{(n-1)!}\lambda^k. $$ This yields $$ \mathrm{e}^{-\lambda}=\sum_{k\ge0}(-1)^k\frac{\lambda^k}{k!}=\sum_{k\ge0}(-1)^k\frac{(n-1)!}{k!(n+k-1)!}E(S^k), $$ which proves that an unbiased estimator of $h(\lambda)=\mathrm{e}^{-\lambda}$ is $H_n(\bar X_n)$ with $$ H_n(x)=(n-1)!\sum_{k\ge0}(-1)^k\frac{n^k}{k!(n+k-1)!}x^k={}_0F_1(-nx;n). $$

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Thanks! It seems obvious now. One based on just $S$ is fine as well. So $\sum \frac{(-1)^k \Gamma(n) S^k}{k! \Gamma(n + k)}$ works fine as the final answer. –  guy Jul 24 '11 at 19:20
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$e^{-\lambda} = \Pr(X_1 > 1)$. So let $$ Y = \begin{cases}1 & \text{if }X_1> 1 \\ 0 & \text{otherwise}\end{cases} $$ Then $Y$ is an unbiased estimator of $e^{-\lambda}$. By Rao-Blackwell, you just need $E(Y\mid X_1+\cdots+X_n) = \Pr(X_1>1 \mid X_1+\cdots+X_n)$.

I think that's not hard to find.

And since you have completeness, you're done!

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This would be correct, but unfortunately you are using the rate parametrization whereas we are given the scale parametrization. $e^{-\lambda} = P(X_1 > \lambda^2) \ne P(X_1 > 1)$. –  guy Jul 25 '11 at 15:32
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