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Can someone please explain why is $P(X > Y) = \frac{1}{2}$, when $X, Y$ are i.i.d. continuous random variables? I have seen people use the symmetry argument to justify this answer. The argument goes as follows: there are two ways of arranging two numbers $x$ and $y$, and out of these arrangements only one has $x > y$. So, the probability using symmetry is $0.5$.

I don't understand this conclusion. Doesn't this argument make an assumption that the values of $X$ and $Y$ that are drawn are not equal? To take a more concrete example, if we assume $X, Y$ are standard normal, wouldn't the sample space be divided into three events: $X > Y, X < Y, X = Y$? Based on this, we can say that $X > Y$ and $X < Y$ must be equal using symmetry, and let's call this value $\alpha$. So,

$2\alpha + P(X=Y) = 1$.

Clearly, $\alpha < 0.5$, contrary to the first argument. So, which is the correct argument and why?

Edit: based on the comments I am adding the calculations for $X=Y$ in case of normal distributions. Can someone point out the mistake? Thanks!

$P(X=Y) = \int_{-\infty}^{\infty} P(X=Y|Y=y) P(Y=y) dy$

$P(X=Y) = \int_{-\infty}^{\infty} \frac{1}{2\pi} e^{-y^{2}} dy$

$P(X=Y) = \frac{1}{2\sqrt\pi}$.

Intuitively, the symmetry argument feels like an approximation (probably a very good one). Imagine we have a bivariate normal distribution, which is formed using $X$ and $Y$. $P(X>Y)$ represents the region below the line $X=Y$ (in the 1st quadrant). Similarly, we can argue about the values in the other 3 quadrants. By geometry, the area of line is zero (because the line has no width?), and hence you arrive at the 1D analogy that the probability it takes a specific value is zero. Still not sure though why it doesn't show up in the calculations above.

Edit: I realize the mistake I made in the above calculations. Going from step 1 to step 2, when I replaced $P(X=Y|Y=y)$ with $f_{X}(y)$, this is wrong. As pointed out in the comments and answer below, this must be equal to zero. I confused (/abused) the notation for the discrete and continuous cases. Thanks everyone for an interesting discussion.

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    $\begingroup$ Are we talking about continuous or discrete random variables? $\endgroup$ Commented Apr 17, 2020 at 17:23
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    $\begingroup$ The point is the probability that two continuous random variables assume the same value is $0$ which gives $\alpha = 1/2$. Had they been discrete, then yes one would very much need to account for that probability. $\endgroup$
    – sudeep5221
    Commented Apr 17, 2020 at 17:23
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    $\begingroup$ Let $X,Y$ two continuous random variables on $\mathbb R$. Then $X=Y$ implies $X-Y=0$. If $Z=X-Y$ then $Z$ is continuous as well. And therefore $P(Z=0)=0$. $\endgroup$ Commented Apr 17, 2020 at 17:30
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    $\begingroup$ Your $\frac1{2\sqrt{\pi}}$ is maybe the pdf, but the probability is still $P(X=Y)=0$ $\endgroup$ Commented Apr 17, 2020 at 17:34
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    $\begingroup$ Just to complete the previous coments: in general we have the equality of events $(X=x)=(X\leq x)\setminus (X<x)$, so $P(X=x)=F(x)-F(x-)$, $F$ being the cdf of $X$. If $F$ is a continuous function (this is the case for any continuous rv), then $F(x)=F(x-)$ and $P(X=x)=0$. Then if $Y,X$ are continuous iid $P(X=Y)=E[1_{X=Y}]=\int P(X=x,Y=x)dF(x,y)=0$. So as @callculus pointed out, for the gaussian rv $P(X=Y)=0$. $\endgroup$
    – RLC
    Commented Apr 17, 2020 at 17:47

2 Answers 2

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Another perspective (using measure-theoretic probability): $$P(X > Y) = \int_{-\infty}^\infty P(X > y)dF(y) = \int_{-\infty}^\infty (1 - F(y))dF(y) = 1 - \int_{-\infty}^\infty F(y)dF(y) = 1/2.$$

Literally, we can evaluate $\int_{-\infty}^\infty F(y)dF(y) = \int_0^1 zdz = 1/2$ given $F$ is continuous. A more rigorous proof for it rests on Fubini's theorem. Interested people may refer to Theorem $18.4$ of Probability and Measure by Patrick Billingsley.

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The distribution of $X-Y$ is symmetric around zero, that is $\mathsf{P}(X-Y\le v)=\mathsf{P}(Y-X\le v)$ for all $v\in \mathbb{R}$ (for example, you may show that the characteristic function of $X-Y$ and $Y-X$ are equal, i.e. $\varphi_{X-Y}(t)=\varphi_{Y-X}(t)=\varphi_X(t)\varphi_X(-t)$). Then as you noticed $$ \mathsf{P}(X-Y>0)+\mathsf{P}(Y-X>0)+\mathsf{P}(X-Y=0)=1, $$ and the result follows assuming that $\mathsf{P}(X-Y=0)=0$. (This is always true for continuous r.v.s.)

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