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If $X$ and $Y$ are identically distributed, independent, continuous random variables, then $P(X=Y) = 0$.

I know that for any particular value $x$, $P(X=x)=0$, but how to show the above rigorously?

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    $\begingroup$ $P (X=Y) = P (X-Y = 0) $ $\endgroup$ – G. Snapsmath Dec 5 '16 at 14:03
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    $\begingroup$ A limiting argument shows the situation pretty clearly. Suppose that the variables are not continuous but finite with $n$ equally likely possibilities: then $P(x = y) = n/n^2 = 1/n$. Now let $n \to \infty$. $\endgroup$ – Tom Collinge Dec 5 '16 at 14:10
  • $\begingroup$ Actually, "identical" may be a wrong choice of terminology. "Identically distributed" seems better. Because, to me, "identical" means $X\equiv Y$, i.e. $X(\omega)=Y(\omega)$ for any $\omega\in\Omega$ where $(\Omega,\Sigma,P)$ is our probability space. And this makes the statement false. $\endgroup$ – MoebiusCorzer Dec 5 '16 at 14:10
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    $\begingroup$ @G.Snapsmath So you know how to prove that if $X$ and $Y$ are continuous and independent then $P(X-Y=0)=0$? Is the proof any easier than solving the original question? $\endgroup$ – Did Dec 5 '16 at 14:18
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    $\begingroup$ @fizis OK, let us assume this, if you wish, and then... what? What does this imply regarding the continuity of the distribution of $X-Y$? $\endgroup$ – Did Dec 5 '16 at 14:32
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Hint: $$P(X = Y) = E[I\{X = Y\}] = \int \int I\{x = y\}\, dP^X(x)\, dP^Y(y)$$

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Let $\epsilon\to0$, $F,f$ be the CDF and PDF of r.v. $X$. From the total probability theorem: $$ P(X-Y\le \epsilon)=\int_{-\infty}^\infty P(X-t\le\epsilon)P(Y=t)dt=\int_{-\infty}^\infty F(t+\epsilon)f(t)dt $$

$$ P(X-Y\le -\epsilon)=\int_{-\infty}^\infty P(X-t\le-\epsilon)P(Y=t)dt=\int_{-\infty}^\infty F(t-\epsilon)f(t)dt $$

$$ P(X-Y\le \epsilon)-P(X-Y\le -\epsilon)\approx2\epsilon\int_{-\infty}^\infty f(t)f(t)dt\to 0 $$

Therefore, the CDF of $X-Y$ is continuous at $0$.

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  • $\begingroup$ Do you mean $P(X-Y \geq -\varepsilon)$? And then calculate $P(X-Y \in [-\varepsilon,\varepsilon])$ in the third line? $\endgroup$ – Therkel Dec 5 '16 at 15:06
  • $\begingroup$ @Therkel The first line is $F_{X-Y}(\epsilon)$, the second line is $F_{X-Y}(-\epsilon)$, the third line is showing that they are equal in the limit of $\epsilon\to0$ $\endgroup$ – fizis Dec 5 '16 at 15:10
  • $\begingroup$ Then you may have shown that $F_{X-Y}(0) = 0$. What does that have to do with $P(X-Y ~{\color{red}=}~ 0)$? Also, where does your approximation in line 3 come from? $\endgroup$ – Therkel Dec 5 '16 at 15:13
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    $\begingroup$ There are a lot of mistakes in your calculation, starting with the fact that a merely continuous random variable doesn't need to have a density. $\endgroup$ – Dominik Dec 5 '16 at 15:13
  • $\begingroup$ @Therkel $F(t+\epsilon)-F(t-\epsilon)\approx f(t)\cdot 2\epsilon$ $\endgroup$ – fizis Dec 5 '16 at 15:15

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