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I have two independent random variables X and Y with given pmf. I want to calculate joint probability of W=X+Y and Z=XY, is the below formula correct for calculating it?

$P(W=w,Z=z) = \sum_{k}P(X=k)P(Y=w-k \ \land Y = \frac{z}{k})$

and by this, I want to show whether these two RV are dependent or independent.

I have to mention that I already calculate pmf for both W and Z.

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The formula is "almost" correct. It is not unthinkable that $P(X=0)>0$ and in that case $k$ ranges over a set that contains $0$. That gives troubles because $\frac{z}{k}$ is part of the expression.

With more caution it can be deduced that:

$$P\left(W=w,Z=z\right)=P\left(X+Y=w,XY=z\right)=\sum_{k}P\left(X=k\right)P\left(X+Y=w,XY=z\mid X=k\right)=$$$$\sum_{k}P\left(X=k\right)P\left(k+Y=w,kY=z\mid X=k\right)=\sum_{k}P\left(X=k\right)P\left(Y=w-k,kY=z\right)$$where the last equality is based on independence. Observe that $Y=\frac{z}{k}$ is replaced by $kY=z$.

Further note that for almost every $k$ the event $\{Y=w-k,kY=z\}$ is empty, so that you can hardly speak of "summation" because almost all terms are $0$.

A route that avoids the summation is:$$P\left(W=w,Z=z\right)=P\left(X+Y=w,XY=z\right)=P\left(X\left(w-X\right)=z\right)$$ You know the PMF of $X$ and based on that knowledge you can find the PMF of $X(w-X)$ for every $w$.

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  • $\begingroup$ Thanks for your explanations. $\endgroup$ – Peyman Tahghighi Feb 22 at 9:09
  • $\begingroup$ You are welcome. $\endgroup$ – drhab Feb 22 at 9:10

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