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Suppose $X,Y$ are iid random variables from a distribution function $F$, which is a continuous function. Then is it always true that $P(X=Y)=0$?

For me, the answer is trivially YES. We have $\int_y P(X=y)dF(y)=0$, and as $P(X=y)=0$, hence $P(X=Y)=0$.

It has however been claimed that it is false in general and counterexample exists, and a total of 10 points have been allotted to this problem. This makes me doubt: does there really exist such $F$?

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  • $\begingroup$ I know that. $F$ being a continuous function may not imply pdf exists, so I have been cautious not to use $f$ as pdf, but still, $P(X=Y)=\int_y P(X=y)dF(y)=0$ holds. Right? $\endgroup$
    – Landon Carter
    Apr 29, 2016 at 18:32
  • $\begingroup$ Please check edited question. Now the argument: $F$ is continuous, and we know a cdf is right continuous, so $F$ is lef continuous as well, so $F(x)=F(x-)$ implying $P(X=x)=0$ for any $x\in\mathbb R$. $\endgroup$
    – Landon Carter
    Apr 29, 2016 at 18:37
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    $\begingroup$ @Integral, No, because I would then have $P(X=X)=\int_x P(X=x|X=x)dF(x)=\int_xdF(x)=1$, the conditioning step is important. In this case, $X,Y$ are independent, that is why I could write $P(X=x|Y=x)=P(X=x)$. And this is standard argument. $\endgroup$
    – Landon Carter
    Apr 30, 2016 at 1:15
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    $\begingroup$ This doesn't address the actual question, but I'll add it to the discussion. The concept of coupling is interesting here. Since $X$ and $Y$ are i.i.d., then they can be coupled (constructed on a common probability space) so that $X=Y$ always (on the new probability space) which is even stronger than $P(X=Y)=1$. $\endgroup$
    – jdods
    Apr 30, 2016 at 2:35
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    $\begingroup$ @jdods Sorry but coupling, while an interesting concept in general, is offtopic here. Note that if $X$ and $Y$ are independent then they CANNOT be coupled so that $P(X=Y)=1$ (well, except if they are both almost surely constant and equal to the same value). Never. In particular, they cannot be coupled so that $[X=Y]=\Omega$. $\endgroup$
    – Did
    Apr 30, 2016 at 13:29

2 Answers 2

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More generally, if $X$ and $Y$ are independent, with cdfs $F_X$ and $F_Y$, then $$ \eqalign{\Bbb P[X=Y] &=\sum_{x\in\Bbb R}\Bbb P[X=x]\cdot\Bbb P[Y=x]\cr &=\sum_{x\in\Bbb R}[F_X(x)-F_X(x-)]\cdot[F_Y(x)-F_Y(x-)],\cr } $$ so $\Bbb P[X=Y]=0$ provided one or the other has a continuous cdf. (Or simply if their cdfs have no common discontinuities.) For a proof see my answer to this question: Calculating $P(X+Y=0)$ for independent random variables (Problem in Durrett)

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    $\begingroup$ Well, you cannot sum over reals like that; it's not defined because you can sum only over a countable set as probability of countable union of disjoint events can be written as countable sum of the probabilities of the disjoint events. So your argument is not correct, at least. $\endgroup$
    – Landon Carter
    Apr 30, 2016 at 1:18
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    $\begingroup$ Only countably many terms of the indicated sum are non-zero. See the referred-to post for more details. $\endgroup$
    – John Dawkins
    Apr 30, 2016 at 16:33
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Assuming they are real valued: $P(X=Y)=\int_{-\infty}^\infty P(Y=x|X=x) \ P(dx)=\int_{-\infty}^\infty P(Y=x) \ P(dx)=0$, because $P(Y=x)=0$ for any $x\in\mathbb{R}$ since $Y$ has a continuous distribution (pdf). If you meant for the cdf to be continuous, then the answer changes.

In general, though $P(X=Y)>0$ is possible, but you have to relax some of your conditions: independence, identically distributed, and/or continuous distribution.

$\textbf{Example $1$:}$ Take $X$ and $Y$ to be i.i.d. fair coin flips, then $P(X=Y)=0.5$ (discrete distributions).

$\textbf{Example $2$:}$ Let $\Omega=\Omega_1\cup\Omega_2$ (disjoint), and let $Y(\omega)=X(\omega)$ for $\omega\in\Omega_1$ and $Y(\omega)=X(\omega)+1$ for $\omega\in\Omega_2$. Then $P(X=Y)=P(\Omega_1)>0$ is possible.

If $X$ is uniform on $[0,2]$, and $\Omega_1=\{\omega \mid X(\omega)\leq 1\}$, then $Y$ is uniform on $[0,1]\cup (2,3]$. Then $Y$'s pdf is discontinuous, but it's cdf is continuous (both the pdf and cdf of $X$ are continuous). Of course, they are neither independent, nor identically distributed.

$\textbf{Example $3$:}$ Keep everything the same as Example $2$ above, except let $Y=\frac{1}{2}(X^2+1)$ on $\Omega_2$. Note that $Y=g(X)$ with both $g$ and $g^{-1}$ are continuous and differentiable (a fun calculus exercise). Then $Y$'s pdf and cdf are both continuous. Of course, $X$ and $Y$ are still neither independent, nor identically distributed.

Side Note: The last 2 examples could be described as couplings of $X$ and $Y$, since they are constructed on the same probability space. We could think of them as being "in principle independent" if we were to "simulate them separately". E.g. if I sample an $\omega$ and evaluate $X$, and you sample another $\omega$ and evaluate $Y$, then our experimental outcomes can be independent. But, this is equivalent to there being two distinct $\Omega$ spaces with identical structure -- i.e. one copy of $\Omega$ per experiment.

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