How to Derive variance of Poisson distribution?

Answer 1

If X has the Poisson distribution with mean l

then Pr(X = k) = e-llk/k!

Mean of Poisson = Sum over all k of [k*P(X = k)] which happens to be l.

= Sum over all k of [k*e-llk/k!]

= Sum over all k of [e-llk/(k-1)!]

= Sum over all j of [le-llj/j!] where j has been substituted for k-1

= l*Sum over all j of [e-llj/j!]

But the quantity being summed is simply the pdf of the Poisson distribution and so its sum over all possible values is 1

So Mean = l

And then Variance of Poisson = Sum over all k of [k2*P(X = k)] - l2.

= Sum over all k of [k2*e-llk/k!] - l2

Then, since k2 = k*(k-1) + k

Variance = Sum over all k of [k*(k-1)e-llk/k!] +Sum over all k of [k*e-llk/k!] - l2

= Sum over all j of [l2e-llj/j!] where j has been substituted for k-2

+ Sum over all i of [le-lli/i!] where i has been substituted for k-1 - l2

= l2*Sum over all j of [e-llj/j!] + l*Sum over all i of [e-lli/i!] - l2

And since the sums are equal to 1,

Variance = l2 + l - l2 = l

Apologies: I did the answer using the symbol for lambda but this browser changed them all back to l. I cannot change them all t o something else but I hope it is clear. At least they are distinguishable from 1!