# finding probability generating function and the sum of two independent random variables

Let $$X$$ be a discrete random variable with probability mass function

$$P_X(x) = p(1-p)^x,\qquad x=0,1,2,3,\ldots$$

(a) Find the probability generating function for $$X$$ and hence find its variance.

(b) $$X_1$$ and $$X_2$$ are independent random variables with probability generating functions $$e^{λ_1(t-1)}$$ and $$e^{λ_2(t-1)}$$ respectively. show that the probability generating function for $$X_1-X_2$$ is $$e^{(λ_1t+λ_2t^{-1})-(λ_1+λ_2)}$$ and hence find its expected value.

For part (a) I think you let $$1-p=q$$ so the p.g.f. will be the sum of $$pq^xt^x$$ which will be $$(1-p)/(1-pt)$$. I am unsure of how to find the mean/variance from this. Do you just substitute $$t$$ with 1 and 2?

For part (b) I have found good proofs for $$X_1 +X_2$$ and have tried to use it but I get down to $$e^{(λ_1(t-1))/(λ_2(t-1))}$$ and am unsure if I have just gone wrong or if I can simplify to the answer because it looks close.

• Dear Mathematics student at the University of Kent, The answer you get in part (b) simplifies to exp(λ1/λ2), which is not close to the answer. Try multiplying/dividing other parts, not just the exponents. – user105991 Nov 6 '13 at 22:58

If your generating function is $G_X(t)$, you need to take derivatives with respect to $t$ and evaluate them when $t=1$. If there is a problem with the radius of convergence being $1$ (not here) you may need to take the limit as $t$ approaches $1$ from below.
So if $G(t)= \sum_x p_x t^x$
then $G'(t)= \sum_x x p_n t^{x-1}$
and $G''(t)= \sum_x x(x-1) p_n t^{x-2}$
so $E[X] = G'(1)$ and $E[X(X-1)] = G''(1)$ meaning $Var(X)= G''(1)+ G'(1) - \left( G'(1)\right)^2$. Just apply these to your probability generating functions.