## Given the pdf of independent RVs I and R, how to find cdf of W=I2R?

The simplest and surest way to compute the distribution density or probability of a random variable is often to compute the means of functions of this random variable. In the case at hand, one wants to write E(g(W)) as
E(g(W))=g(w)f(w)dw,
for every bounded measurable function g, then one can be sure that f is the density of the distribution of W. So, in a way, the functions g play the role of a dummy variable and one wants the equality above to hold for every g. Naturally W=I2R hence E(g(W)) is a priori a double integral, but one can be sure that a change of variable will save the day. So, applying the definitions, E(g(W))=E(g(I2R)) and
E(g(I2R))=g(x2y)[0x1]6x(1x)[0y1]2ydxdy,
where, for every property A, Iverson bracket [A] denotes 1 if A holds and 0 otherwise. (Begin of rant: no, I do not like to put the limits of the domain of integration on the integral signs, and yes, I prefer to use the notation [A] or its cousin 1A because they are more systematic and, at least to me, less error prone. End of rant.)
Now, what change of variable? For one of the two new variables, we want w=x2y, of course. For the other, a sensible choice (but not the only one) is z=x. The new domain is 0wz21 and the Jacobian is given by dxdy=z2dwdz, hence
E(g(W))=g(w)[0w1]([wz21]6z(1z)(2wz2)z2dz)dw.
By identification, the density f(w) is the quantity enclosed by the parenthesis, that is, for every 0w1,
f(w)=[wz21]6z(1z)(2wz2)z2dz=12w1wz3(1z)dz,
Finally,
f(w)=6(1w)2[0w1].