mydaruma25

2022-09-27

A certain discrete random variable has probability generating function: ${\pi }_{x}\left(q\right)=\frac{1}{3}\frac{2+q}{2-q}$
Compute . (Hint: the formula for summing a geometric series will help you expand the denominator)."

altaryjny94

Step 1
$\begin{array}{rl}{\pi }_{x}\left(q\right)& =\frac{1}{3}\left(2+q\right)\left(\frac{1}{2}\frac{1}{1-\left(\frac{q}{2}\right)}\right)\\ & =\frac{2+q}{6}\left(\sum _{k=0}^{\mathrm{\infty }}{\left(\frac{q}{2}\right)}^{k}\right)\\ & =\sum _{k=0}^{\mathrm{\infty }}\frac{1}{3\cdot {2}^{k}}{q}^{k}+\sum _{k=0}^{\mathrm{\infty }}\frac{1}{3\cdot {2}^{k+1}}{q}^{k+1}\\ & =\frac{1}{3}+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{3\cdot {2}^{k}}{q}^{k}+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{3\cdot {2}^{k}}{q}^{k}\\ & =\frac{1}{3}+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{3\cdot {2}^{k-1}}{q}^{k}.\end{array}$
Step 2
Now recall the definition of the generating function:
${\pi }_{x}\left(q\right)=\sum _{k=0}^{\mathrm{\infty }}\mathbb{P}\left(X=k\right){q}^{k}.$
Step 3
By unicity of generating functions you gain $\mathbb{P}\left(X=k\right)=\frac{1}{3\cdot {2}^{k-1}}$ if $k\ge 1$, and $\mathbb{P}\left(X=0\right)=\frac{1}{3}$.

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