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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.

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

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Thanks! That does help a lot. – Robert Feb 26 '12 at 5:00

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