# Joint distribution of X and XY

Suppose X and Y, independent, are distributed as Uniform[0,1].

(Disclaimer: this relates to a homework problem, but is not itself a homework problem. The problem itself asks to find the conditional density of X given XY=t, i.e. $$f_{X|XY=t}(x)$$).

Now, I am proceeding by using the definition: $$f_{X|XY=t}(x)=\frac{f_{X,XY}(x,t)}{f_{XY}(t)}$$

I have already calculated the density $$f_{XY}(t)$$, but have gotten a little lost on the numerator. I am trying to calculate the numerator by first calculating the $$F_{X,XY}(x,t)$$, and then differentiating with respect to x and t. Now, I am wondering if, using the independence of X and Y, I can split this joint density up as: \begin{align} F_{X,XY}(x,t)=P(X \leq x, XY \leq t) \\ = P(X \leq x, Y \leq \frac{t}{X}) \\ = \int_{t}^{1}P(Y \leq \frac{t}{X}|X=s)P(X=s) ds \\ =\int_{t}^{1}P(Y \leq \frac{t}{s})P(X=s) ds \\ =\int_{t}^{1}\frac{t}{s}ds=-t log(t) \end{align}

But differentiating this with respect to x and t yields 0, which I know is again wrong.

The areas of concern that I have noted while working through the problem are as follows:

1. Confusion over the use of the conditional division formula when using CDFs as opposed to pdfs (can we just substitute the pdfs for CDFs?)
2. Confusion if the conditioning on X=s I use to split up the densities is valid

If someone can point out the error, I would be much obliged.

1. Using $$\mathbb{P}(X = x)$$ to denote the PDF of $$X$$ is a very bad convention. It overrides the meaning of $$\mathbb{P}(X = x)$$ as probability of the event $$\{X = x\}$$, and there is no clear advantage for doing so.

2. I am not sure how you came up with integral limits $$t$$ and $$1$$.

A systematic approach is as follows. First, using the fact that $$X$$ has continuous distribution with the PDF $$f_X$$, we can write

\begin{align*} \mathbb{P}(X \leq x, XY \leq t) &= \int_{-\infty}^{\infty} \mathbb{P}(X \leq x, XY \leq t \,|\, X = s) f_X(s) \, \mathrm{d}s, \\ &= \int_{0}^{1} \mathbb{P}(X \leq x, XY \leq t \,|\, X = s) \, \mathrm{d}s \end{align*}

where the second step follows from the $$X \sim \operatorname{Uniform}([0, 1])$$.

Next, let us simply the conditional probability. Before doing so, notice that the range of $$(X, XY)$$ lies in the triangle $$\mathcal{T} = \{ (x, t) : 0 \leq t \leq x \leq 1\}$$. So, for the purpose of computing the joint PDF of $$(X, XY)$$, it suffices to assume that $$0 < t < x < 1$$. Then for $$0 < s < 1$$,

\begin{align*} \mathbb{P}(X \leq x, XY \leq t \,|\, X = s) &= \mathbb{P}(s \leq x, sY \leq t \,|\, X = s) \\ &= \mathbb{P}(s \leq x, sY \leq t) \\ &= \mathbb{P}(s \leq x, Y \leq t/s) \\ &= \begin{cases} 0, & \text{if } s > x; \\ t/s, & \text{if } s \leq x \text{ and }t \leq s; \\ 1, & \text{if } s \leq x \text{ and }t > s. \\ \end{cases} \end{align*}

Plugging this back, we are led to integrate a piecewisely-defined function, thus we get

\begin{align*} \mathbb{P}(X \leq x, XY \leq t) = \int_{t}^{x} \frac{t}{s} \, \mathrm{d}s + \int_{0}^{t} \mathrm{d}s = t \log (x / t) + t. \end{align*}

Differentiating w.r.t. $$x$$ and $$t$$, we get

\begin{align*} f_{X,XY}(x, t) = \frac{\partial^2}{\partial t \partial x}\mathbb{P}(X \leq x, XY \leq t) = \frac{\partial^2}{\partial t \partial x}\left( t \log (x / t) + t \right) = \frac{1}{x}. \end{align*}

We recall that this result is derived on the assumption that $$0 < t < x < 1$$. In a full-form, $$f_{X,XY}$$ can be written as

$$f_{X,XY}(x, y) = \begin{cases} \frac{1}{x}, & \text{if } 0 < t < x < 1; \\ 0, & \text{otherwise}. \end{cases}$$