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The number of breakdowns Y per day for a certain machine is a Poisson random variable with mean $\lambda$. The daily cost of repairing these break downs is given by $C=3Y^2$ If $Y_1, Y_2, ..., Y_n$ denote the observed number of breakdowns for $n$ independently selected days find an MVUE for $E(C)$.

We can use the Rao-Blackwell Theorem.

We know that $E(C) = E(3Y^2)=3[V(Y) + (E(Y))^2]$ and $E(Y)=\lambda=V(Y)$. With some calculations we see that $E(Y^2)= \lambda + \lambda^2$

$\sum_{i=1}^n Y_i=\bar Y$ is a sufficient statistics for $ \lambda$ So I am assuming we can replace $\lambda$ with $\bar {Y}$

I am just unsure how to go from here. Can you guys help me pull strings together?

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    $\begingroup$ Emulate the Wikipedia example. $\endgroup$ – whuber Feb 12 '13 at 21:40
  • $\begingroup$ Thanks for the link :) I prefer if someone gave me some guidance though $\endgroup$ – math101 Feb 12 '13 at 21:53
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The Rao-Blackwell Theorem says that if you have a sufficient statistic $T$ and an unbiased estimator $U$ (i.e., $E\{U(X)\} = \theta$), then $E\{U(X) | T\}$ has variance no larger than $V\{U\}$.

You have shown that $3Y_1^2$ is an unbiased estimator of $E\{C\}$.

You can now improve that estimator by finding its expectation with respect to $T = \sum Y_j$. (Point of pedantry: $\bar{Y}$ is usually taken to be $\frac{1}{n}\sum Y_j$. In dealing with the Poisson distribution it's usually easier to use $T$ rather than $\bar{Y}$ as the sufficient statistic. The sample mean has a lattice distribution, which $T$ is Poisson.)

What you will find is that the result is a function of $T$, and because the Poisson distributions form an exponential family, $T$ is a complete, sufficient statistic. Consequently, your estimator derived by Rao-Blackwellizing $3Y_1^2$ is the (unique) MVUE.

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