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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 $\Pr(X_1 = X_2)$.

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    $\begingroup$ Write that $P(X_1=X_2)=E[1_{\{X_1=X_2\}}]$ and use Fubini's theorem to show that it is 0. $\endgroup$
    – Will
    Nov 11, 2023 at 19:31
  • $\begingroup$ The integral you show results in $1$, just confirming that $4x_1x_2$ is a pdf on the unit square. $\endgroup$
    – AlgTop1854
    Nov 12, 2023 at 14:42

2 Answers 2

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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.

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If you can compute the marginals, you can also consider

$$\mathbb P(X = Y) = \mathbb P(X - Y = 0) = \mathbb P(X - Y = 0 | Y = k) \mathbb P (Y = k)$$

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