<s>[INST] Sum $\sum_{x}\sum_{y}xy\left(\frac{1}{2}\right)^{x-1}\cdot\left(\frac{1}{3}\right)^y$? $P(X=x,Y=y)=\left(\dfrac{1}{2}\right)^{x-1}\cdot\left(\dfrac{1}{3}\right)^y;x,y\ge1$. Find $E(XY)$ .
$$E(XY)=\sum_{x}\sum_{y}xy\left(\frac{1}{2}\right)^{x-1}\cdot\left(\frac{1}{3}\right)^y$$
$\sum_{y}y\left(\dfrac{1}{3}\right)^y\sum_{x}x\left(\dfrac{1}{2}\right)^{x-1}$
here $x,y\ge1$
I used this formula which is the sum of Arithmetic-Geometric progression.
$$a+(a+d)r+(a+2d)r^2\cdots = \frac{a}{1-r}+\frac{dr}{(1-r)^2}$$
$r$= common ratio(GP)
$d$= common difference(AP)
$a$= first term
$\bigg(1+2\cdot\dfrac{1}{2}+3\cdot\dfrac{1}{2^2}....\bigg)\cdot\bigg(1\cdot\dfrac{1}{3}+2\cdot\dfrac{1}{3^2}+3\cdot\dfrac{1}{3^3}....\bigg)=\bigg(\dfrac{1}{1-\frac{1}{2}}-\dfrac{1\cdot\frac{1}{2}}{(1-\frac{1}{2})^2}\bigg)\cdot\dfrac{1}{3}\bigg(\dfrac{1}{1-\frac{1}{3}}-\dfrac{1\cdot\frac{1}{3}}{(1-\frac{1}{3})^2}\bigg)=\bigg(4\cdot\dfrac{1}{3}\cdot\dfrac{3}{4}\bigg)=1$
But this is wrong.  Answer lies between $2.50$ to $3.50$
 [/INST] You have made a miscalculation.  The formula has a plus sign where you have a minus sign:  $$\frac{a}{1-r} \color{red}{+} \frac{dr}{(1-r)^2}.$$
The sum over $x$ is $$\frac{1}{1-1/2} + \frac{1/2}{(1-1/2)^2} = 4.$$  The sum over $y$ is $$\frac{1}{3}\left(\frac{1}{1-1/3} + \frac{1/3}{(1-1/3)^2}\right) = \frac{3}{4}.$$ Their product is $3$.  You have an extra $1/3$ in your calculation because when you calculated the sum over $y$, you already included the $1/3$, but forgot that you did that so you multiplied again by $1/3$.
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