Is this a composition of two functions of random variables, or two independent functions?

Question

I'm reviewing some topics that were tricky for me in the past, and this problem from Grimmett and Stirzaker (4.7.1) came up:

Let $X,Y,Z$ be independent and uniformly distributed on $[0,1]$. Find the joint density function of $XY$ and $Z^2$, and show that $P(XY < Z^2) = \frac{5}{9}$.

I'm not clever with these types of problems, so I just stick with the change of variables, but I don't know if I'm preparing the right statements to evaluate $P(XY < Z^2)$ (as the second part of the problem is clearly just a self-test):

1. Let $u = xy, v = x, w = z^2 \implies t_1(v) = v, t_2(u,v) = \frac{u}{v}, t_3(w) = \sqrt{w}$.

2. Compute Jacobian:

   \begin{equation} \left| \begin{matrix} 0 & \frac{1}{v} & 0 \\ 1 & -\frac{u}{v^2} & 0 \\ 0 & 0 & \frac{1}{2\sqrt{w}} \end{matrix} \right| = -\frac{1}{2v\sqrt{w}} \end{equation}

3. Stating the joint density function:

   \begin{equation} f_{U, V, W} (u, v, w) = f(t_1(v), t_2(u,v), t_3(w))|J(v,w)| \end{equation}

   (in the domain of the variables)

   \begin{equation} f_{U, W} (u, w) = f(x, \frac{u}{x}, \sqrt{w}) \left| \frac{1}{2x\sqrt{w}}\right| \end{equation}

   (for $w \geq 0$)

4. Assuming I have that right, I think $P(XY < Z^2)$ can be phrased as

   \begin{equation} P(XY < Z^2) = F_{U} (u=w) = \int_{-\infty}^w \int_{-\infty}^\infty f(x, \frac{u}{x}) \left| \frac{1}{x}\right| dxdu \end{equation}

   (for $w \geq 0$), since $Z^2$ is independent of $XY$.

I don't have any confidence in what I did in (3) and (4), however (despite the $XY$ function being derived as an example in the same text), and I'm not sure how to come up with a PDF to substitute in the last integrand to compute with. Am I on the right track?

Answer 1

Let me offer an alternate derivation of the density. First, let us call $U=XY,V=Z^2$ and note that by independence $$f_{U,V}(u,v)=f_U(u)f_V(v).$$

Now, let's derive the density of $U$ by a conditioning argument. First, we note that $$\begin{align}P(U \leq u) &= P(XY \leq u) = \int_0^1P(Xy\leq u|Y=y)f_Y(y)dy \\ &=\int_0^1 P(X \leq u/y)f_Y(y)dy\end{align},$$ where the second equality comes from the law of total probability.

Now, we know by the distribution of $X$ that $$P(X\leq u/y) = \begin{cases} 0, & u/y < 0 \\ u/y, & 0 \leq u/y < 1 \\ 1, & u/y \geq 1\end{cases}.$$ Then, noting that $u/y <1 \iff u<y,$ we see that the above integral can be split into $$\begin{align}P(U \leq u) &=\int_0^1 P(X \leq u/y)f_Y(y)dy \\ &=\int_0^uP(X\leq u/y)f_y(y)dy + \int_u^1P(X\leq u/y)f_y(y)dy \\ &=\int_0^u1 \cdot f_y(y)dy + \int_u^1\frac{u}{y}\cdot f_y(y)dy \\ &=\int_0^udy + \int_u^1\frac{u}{y}\cdot dy \\ &= u(1-u\ln(u)), \qquad u\in [0,1]\end{align},$$ where the second to last equality comes from noting that $f_Y(y)=1$ for $y \in [0,1]$. Hence, the density of $U$ is given by $$f_U(u) = \frac{d}{du}u(1-u\ln(u)) = -\ln(u), \qquad u \in [0,1].$$ A quick gut check on this is that we know (by independence) that $E(U) = E(X)E(Y) = 1/4, $ and can verify easily that $$E(U) = \int_0^1 -u\cdot \ln(u)du = 1/4.$$ So, then we move on to the density of $V = Z^2$. This one is quite a bit easier to derive, we simply consider that $$P(V\leq v) = P(Z \leq \sqrt{v}) = \sqrt{v}.$$ Hence, $$f_V(v) = \frac{d}{dv}\sqrt{v} = \frac{1}{2\sqrt{v}}.$$ Therefore, the joint density, by independence, is given by $$f_{U,V}(u,v) = f_U(u)f_V(v) = \frac{-\ln(u)}{2\sqrt{v}}.$$ Now, the probability that $U<V$ is then given by $$\begin{align}P(U<V) &= \int_0^1 \int_u^1 \frac{-\ln(u)}{2\sqrt{v}}dvdu \\ &= \int_0^1 \ln(u)\left(\sqrt{u} - 1\right)du = \frac{5}{9},\end{align}$$ as desired. If you need clarification on any of the steps, let me know.