Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've come across a statistics problem that I can't seem to figure out how to solve:

"A certain discrete random variable has probability generating function: $$ \pi_x(q) = \dfrac{1}{3}\dfrac{2+q}{2-q} $$ Compute p(x) for x = 0,1,2,3,4,5. (Hint: the formula for summing a geometric series will help you expand the denominator)."

I'm not entirely sure what sort of answer this problem requires. q is not given, so is it only possible to solve this in terms of q? What would be a good way to start solving this problem (especially since I don't know of any way to solve for p(x) given a probability generating function)? Any help would be greatly appreciated.

share|improve this question
    
You're looking for the coefficients of $q^0,q^1,\cdots$ –  user21436 Feb 26 '12 at 3:10

1 Answer 1

up vote 1 down vote accepted

Use the hint.

\begin{align*} \pi_x(q) & = \frac 13 (2+q) \left( \frac 12 \frac 1{1-\left( \frac q2 \right)} \right) \\ & = \frac {2+q}6 \left( \sum_{k=0}^{\infty} \left( \frac q2 \right)^k \right) \\ & = \sum_{k=0}^{\infty} \frac 1{3 \cdot 2^k} q^k + \sum_{k=0}^{\infty} \frac 1{3 \cdot 2^{k+1}} q^{k+1} \\ & = \frac 13 + \sum_{k=1}^{\infty} \frac 1{3 \cdot 2^k} q^k + \sum_{k=1}^{\infty} \frac 1{3 \cdot 2^k} q^k \\ & = \frac 13 + \sum_{k=1}^{\infty} \frac 1{3 \cdot 2^{k-1}} q^k. \end{align*}

Now recall the definition of the generating function : $$ \pi_x(q) = \sum_{k=0}^{\infty} \mathbb P(X = k) q^k. $$ By unicity of generating functions you gain $\mathbb P(X = k) = \frac 1{3 \cdot 2^{k-1}}$ if $k \ge 1$, and $\mathbb P(X = 0) = \frac 13$.

Hope that helps,

share|improve this answer
    
Thanks! That does help a lot. –  Robert Feb 26 '12 at 5:00

Your Answer

 
discard

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.