show that if a random variable (x) has a poisson distribution with parameter (lander),i.e. x~p(0) then e(x)=var(x)

Shaun, I think you mean parameter "lambda" (). :)

For a Poisson distribution, we have the following:



To find the mean (a.k.a. expected value) we compute E(x):



Notice that when x=0, the argument of the sum is 0, so we can start our summation at 1 instead. Also notice that the x in front of the fraction cancels one of the x's in the factorial in the denominator, so we can rewrite the above expression as follows:



Now if we substitute y=x-1, we can rewrite the equation again:



Since is a constant, we can factor a lambda out of the summation.



Now notice that the summation term is exactly the Poisson distribution! Like all probability density functions, the sum (or integral for PDF's that work for continuous values) of all possible values of x from 0 to infinity is always, always equal to 1. This is a fundamental property of PDFs. The same is true of a Gaussian or binomial or any other distribution - the total area under the curve is always 1!

Hence, the summation in the above equation is identically equal to 1 so we have



Now let's move on to the variance. Recall the definition of variance:



We already know , but we need to compute . We can do that by first recognizing that



Again, we already know what E(X) is, so we simply need to compute E(X(X-1)).



Employing a similar strategy to what we did before, now notice the argument of the summation goes to zero for both x=0 and x=1. Also, the x(x-1) terms cancel out part of the factorial in the denominator, so we can rewrite the expression as:



Now substituting in y=x-2 we have



Factoring out a from the summation, we now have



Once again, this summation is identically equal to the Poisson distribution function, so it's value is equal to 1. Hence



Finally, we have



Which means

Sweet. I would give you a greenie, but it won't let me.