Let $(X,Y)$ be an absolute continuous R.V. Find $XY \mid Y = y$ and show $XY$ and $Y$ are independent. Also $XY \sim e(1)$.

Question

Let $(X,Y)$ be an absolute continuous R.V with density $f_{X,Y}(x,y) = ye^{-y(x+1)}, \ x,y >0$.

I've shown that $Y \sim e(1)$ and $X \mid Y = y$ density $x \mapsto ye^{-yx}$.

However I must find $XY \mid Y = y$ and show $XY$ and $Y$ are independent. Also $XY \sim e(1)$.

I've a theorem saying: $X=h(Y,Z)$ then $X \mid Y=y \sim h(y,Z) \mid Y=y$, but I can't figure out how to apply it or whether it should be applied.

Also I've tried to calculate $$P(XY \le k)=(X \le \frac k Y) = \int^{\infty}_0 \int^{\frac k y}_0 ye^{-y(x+1)} \ dx \ dy = 1-e^{-k}$$

but this is not $e(1)$, so I've done an error ? How can I proceed ?

Answer 1

To use the theorem you gave, let $XY = h(Y, Z)$ where $h(y,z) = yz$ and $Z = X$. Then $XY | Y = y \sim yX | Y = y$. Thus for $t > 0$

\begin{align} P(XY < t | Y = y) &= P(Xy <t | Y = y ) \\ &= P(X < \frac{t}{y} | Y = y) \\ & = \int_{0}^{\frac{t}{y}} y e^{-yx}dx \\ & = 1 - e^{-t} \end{align}

The cdf of $XY$ is not affected by the value of $Y$, thus they are independent. And $1-e^{-t}$ is exactly the cdf of $e(1)$