# Partial fractions for geometric probability-generating function wrong

Let $X\sim \text{Geo}(1/4), Y\sim \text{Geo}(1/2)$ be given. First I have to compute $\mathbb{E}[z^{X+Y}]$: $$\mathbb{E}[z^{X+Y}]=\mathbb{E}[z^{X}]\cdot\mathbb{E}[z^{Y}]=\frac{\frac{1}{4}z}{1-\left(1-\frac{1}{4}\right)z}\cdot\frac{\frac{1}{2}z}{1-\left(1-\frac{1}{2}\right)z}=\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}.$$ The next task is to find $\alpha,\beta,\gamma\in\mathbb{R}$ such that $$\mathbb{E}[z^{X+Y}]=\alpha+\frac{\beta}{1-\frac{1}{2}z}+\frac{\gamma}{1-\frac{3}{4}z}.$$ Here is my procedure (which seems to be wrong...): $$\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=1+\frac{10z-8}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}$$ $$\frac{10z-8}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=2\cdot\frac{5z-4}{(z-2)(3z-4)}=\frac{A}{z-2}+\frac{B}{3z-4}$$ $$\frac{A(3z-4)+B(z-2)}{(z-2)(3z-4)}=\frac{(3A+B)z+(-4A-2B)}{(z-2)(3z-4)}$$ $$\left.\begin{cases} 3A+B&=5 \\ -4A-2B&=-4 \end{cases}\right\}\implies A=3,\;B=-4\implies 2A=6,\;2B=-8$$ $$\rightsquigarrow \mathbb{E}[z^{X+Y}]=1+\frac{6}{z-2}-\frac{8}{3z-4}=1-\frac{3}{1-\frac{1}{2}z}+\frac{2}{1-\frac{3}{4}z}\implies\alpha=1,\;\beta=-3,\;\gamma=2$$

However it seems that the values for $\alpha,\beta,\gamma$ should each be just a third of my result! Can anyone explain me, what I did wrong?

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The first line, $$\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=1+\frac{10‌​z-8}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)},$$ is off by a factor of $4$. – anon Jun 11 '12 at 23:02
@anon: Thanks. Answered my own question now. – Christian Ivicevic Jun 11 '12 at 23:17

$$\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=1+\frac{\frac{10}{3}z-\frac{8}{3}}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}$$
This leads to $\alpha=1/3,\;\beta=-1,\;\gamma=2/3$ which is correct.