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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.

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1 Answer 1

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  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

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

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