# Finding $\Pr(X_1 = X_2)$ for joint PDF

I am working on the following problem and I am having a hard time understanding how to put in the proper bounds for the following PDF. I have tried integrating from:

$$\int_0^{1}\int_0^{1} 4x_1x_2dx_1dx_2$$ under the idea that they should have the same bounds since, $$X_1 = X_2$$. But that didn't work. I know (from the book) that the result should be $$0$$. But don't know how to arrive there. Hope you can help.

The problem:

Let $$f(x_1,x_2) = 4x_1x_2$$ , $$0 < x_1 < 1$$ and $$0 < x_2 < 1$$, zero elsewhere be the PDF of $$X_1$$ and $$X_2$$.

Find $$\Pr(X_1 = X_2)$$.

• Write that $P(X_1=X_2)=E[1_{\{X_1=X_2\}}]$ and use Fubini's theorem to show that it is 0.
– Will
Nov 11, 2023 at 19:31
• The integral you show results in $1$, just confirming that $4x_1x_2$ is a pdf on the unit square. Nov 12, 2023 at 14:42

Note that, with $$A \mathrel{:=} \{(x_1,x_2) \in \mathbb R^2: x_1 = x_2\}$$, we have $$P(X_1 = X_2) = \int_A f(x_1, x_2) \,\mathrm d (x_1, x_2).$$ Since $$A$$ has two-dimensional Lebesgue measure zero (which is straightforward to show using, e.g., Tonelli's theorem), the integral is zero.
$$\mathbb P(X = Y) = \mathbb P(X - Y = 0) = \mathbb P(X - Y = 0 | Y = k) \mathbb P (Y = k)$$