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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 unsure where to go from here. Can someone help me pull the strings together?

math101
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2 Answers2

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Basically you're trying to estimate $\lambda + \lambda^2$. In a Poisson, the sample mean (or just the sum) is a complete and sufficient stat, so if you can write an estimator as a function of it which is unbiased for $E(C)$, you win. We have

$ \sum_i X_i $ is distributed as Poi$(n\lambda)$. Thus,

$E\left(\sum_i X_i \right)^2 = n\lambda + n^2 \lambda^2 $ and $E(\left(\sum_i X_i \right) = n\lambda$

Then you can see that $$\frac{\left(\sum_i X_i \right)^2}{n^2} + \frac{(n-1)\left(\sum_i X_i \right)}{n^2} $$

should work

(Edit- multiply everything by 3 since we're really estimating $3(\lambda + \lambda^2)$)

blitzer
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$$E(Y^2)=V(Y) + E(Y)^2$$

It is from the definition of variance. Just moved around.

Stefan Hansen
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grad_student
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