# Find an unbiased estimate for λs (Poisson distribution)

So I have this problem to solve...

Let X denote the number of paint defects found in a square yard section of a car body painted by a robot.

These data are obtained: 8, 5, 0, 10, 0, 3, 1, 12, 2, 7, 9, 6

Assume that X has a Poisson distribution with parameter λs.

a) Find an unbiased estimate for λs.

b) Find an unbiased estimate for the average number of flaws per square yard.

c) Find an unbiased estimate for the average number of flaws per square foot.

I have no idea where to begin. I mean, how do I even find ANY unbiased estimate? The textbook is worthless imo and I can't find any good readings on the web either... please help.

-
The Wikipedia page en.wikipedia.org/wiki/Poisson_distribution#Parameter_estimation has a simple formula for an unbiased estimation of $\lambda$. It is even the one you might guess. – Ross Millikan Dec 7 '10 at 17:17
@Ross no, that's not it. I need an unbiased estimate, not maximum likelihood. 2 separate things (according to my textbook at least). – Mickel Dec 7 '10 at 17:48
It claims this is an unbiased estimator, with a short justification. – Ross Millikan Dec 7 '10 at 18:02
For any distribution with mean $\theta$ ($\theta = \lambda s$ in your example), the sample average is an unbiased estimator for $\theta$: $E(\bar X_n) = \theta$. – Shai Covo Dec 7 '10 at 19:30
Also, what's the difference between a) and b)? Maybe "Assume that X has a Poisson distribution with parameter $\lambda$" (rather than $\lambda s$)? – Shai Covo Dec 7 '10 at 19:43

In a somewhat more general setting, let $A(R)$ denote the area of region $R$. If the number of flaws found on region $R$ follows a Poisson distribution, then the mean is proportional to $A(R)$. That is, if $X(R)$ denotes the number of flaws found on region $R$, then $X(R)$ is Poisson distributed with mean $\lambda A(R)$, for some fixed $\lambda > 0$. Now, if $X_1,\ldots,X_n$ are i.i.d. Poisson$(\lambda)$ rv's (corresponding to the number of flaws found on a unit-area region, say square yard), then $\frac{1}{n}\sum\nolimits_{i = 1}^n {X_i }$ is an unbiased estimator for $\lambda$, and in turn, $\frac{1}{n}\sum\nolimits_{i = 1}^n {A(R) X_i }$ is an unbiased estimator for $\lambda A(R) = {\rm E}[X(R)]$. So, in order to estimate ${\rm E}[X(R)]$, it suffices to estimate $\lambda$ (and then just multiply by $A(R)$).