# Find $P(XY)$, where $X\sim U(0,1)$ and $Y\sim\exp(2)$

Let $X$ have a uniform distribution in unit interval $[0,1]$ and let $Y$ have an exponential distribution with parameter $=2$. Assume that $X$ and $Y$ are independent. Let $Z=X+Y$.

a) Find $P(YX)$.

b) Find the conditional PDF of $Z$ given $Y=y$.

c) Find the conditional PDF of $Y$ given that $Z=3$.

I am not sure how to express $P(YX)$ in general. Also, since it is stated that they are independent would this just amount to finding the $f_{XY}(x,y)$, which in this case would be equal to $f_X(x)\cdot{f}_Y(y)$? I know the function $f_{XY}(x,y)$ does not constitute a probability on its own, as it must be integrated over an interval but in this case what would that integral be?

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It isn't quite clear what your exact question(s) is but I will try and answer at a broad level.

If $X$ and $Y$ are independent, the joint probability distribution of $(X,Y)$ will indeed factor into a product of their individual CDFs like you mention. As for the range of values that $(x,y)$ can take, think of what independence means. Any value that the random variable $X$ takes is independent of any value that the random variable $Y$ takes and they individually take values in the intervals $(0,1)$ and $(0,\infty)$ respectively. Therefore, the range of values for the ordered pair $(X,Y)$ is just the Cartesian product of these two intervals.

Update: Think of $A = (X,Y)$ as a random point in a 2-D graph. Then this random point $A$ can lie anywhere in the (infinite) rectangle with base between $0 \leq x \leq 1$ and height $0 \leq y \leq \infty$. Every point in this rectangle occurs with a certain probability the density of which is given by the joint distribution of $X$ and $Y$. In this case, the coordinates are independent which implies that joint probability is the product of the individual PDFs.

The first part of the problem asks you to find the PDF of the product of $X$ and $Y$ which is different from the joint distribution of $(X,Y)$. One possible approach to do this is by conditioning. Let $M = XY$.

\begin{align} f_M(m) &=& \int_0^1 f_{Y|X} (m/x) f_X (x) dx \\ &=& \int_0^1 f_Y (m/x) f_X (x) dx \end{align} where the second step follows from the independence of $X$ and $Y$ (which implies that $f_{Y|X}(\cdot) = f_Y(\cdot)$). Since we know the PDFs of $X$ and $Y$, we can evaluate this integral to find the PDF of $M$.

Update: In terms of the rectangle picture above, you can think of finding the density of $M$ as follows: For every point in the rectangle, you can multiply its coordinates to come up with a value for $M$. Think of what it takes for the product to be equal to some positive number $m$. If $X$ takes the value $x$, $Y$ must take the value $m/x$. The probability of this event is $P((X,Y) = (x,m/x)) = P(X=x) \cdot P(Y = m/x)$ by independence. Now we integrate over all possible values that $X$ takes (i.e., integrate from 0 to 1) to arrive at the total probability that $M$ can equal $m$.

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So the second step is made because the interval of x [0,1] is contained within [0,$\inf$]? I got everything up til there. Also, so the question what is P(XY) could I just see it like XY=M from the beginning of the problem? – Olivia Irving Dec 14 '12 at 6:21
The fact that a Jacobian is $1$ is crucial to explain the first expression of $f_M(m)$, you might add something on this. – Did Dec 14 '12 at 6:41
I have added more details. I have been loose in my usage of some terms like PDF vs probability. If this example isn't clear, I would suggest that you work out a similar problem in a discrete setting and then come back to the continuous setting. – Dinesh Dec 14 '12 at 6:44