Gamma Distribution Mean And Variance

If the probability function of Geometric Distribution is \(P(X=x)=\theta(1-\theta)^{x-1}\), where \(x = 1, 2, 3,...\), then without using the generating function prove that \(\mu=\frac{1}{\theta}\) and \(\sigma^{2}=\frac{(1-\theta)}{\theta^2}\).

 

hi asmkdas,
the probability function which you have stated is of GEOMETRIC distribution and not GAMMA..
your question is a simple bookwork from ct3 notes..you will find it in the notes..
all the best!

 

Geometric Distribution's mean and variance

Thank for the reply and rectifying me Jyoti, but my problem is not solved yet. My solution is not given on the book at all. However, I am jotting down as far as I have reached to the solution.

We got \(P(X=x)=\theta(1-\theta)^{x-1}\)
\(E(X)=\sum_{x=1}^{\propto}x{\theta}(1-\theta)^{x-1}\)
\(=\theta(1+2(1-\theta)+3(1-\theta)^2+4(1-\theta)^3+...)\)
\(=\theta{.}\frac{1}{(1-\theta)^2}\)

This long I have reached. Now please help me how to get the exact figure of \(\mu=\frac{1}{\theta}\).

 

Somehow you got the denominator in the last term wrong. It should come out to be \(\theta^2\)

Ok let's denote the series in the bracket by S.

Now,

\(S = 1+2(1-\theta)+3(1-\theta)^2+4(1-\theta)^3+.....\)

and

\(S(1-\theta) = (1-\theta)+2(1-\theta)^2+3(1-\theta)^3+.....\)

Therefore

\(S - S(1-\theta) = 1+(1-\theta)+(1-\theta)^2+(1-\theta)^3+.....\)

which is an infinite GP.

So,

\(S\cdot\theta=\frac{1}{1-(1-\theta)}\) or

\(S=\frac{1}{\theta^2}\)

P.S. If you're interested in all these proofs send me a PM.

 

Thanks Suraj.

 

I got your message. Really a good one. Thanks again.