Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.