which is the probability that $Y_1>Y_2$ with $Y_1, Y_2$ i.i.d [duplicate]

Question

I think that the answer is 0.5, independently of the common distribution of $Y_1$ and $Y_2$. Attempt to a proof, assuming continuous random variables:

$P(Y_1>Y_2)=\int_{\Omega}f_{Y_1,Y2}(y_1,y_2)dy_1dy_2=\int_{\mathbf {R}} \left( \int_{-\infty}^{y_1} f_Y(y_2)dy_2 \right) f_Y(y_1) dy_1 = \int_{\mathbf {R}} F_Y(y_1) f_Y(y_1) dy_1 $

Now, integrating by parts:

$ \int_{\mathbf {R}} F_Y(y) f_Y(y) dy = \left[(F_Y(y))^2\right]_{-\infty}^{+\infty} - \int_{\mathbf {R}} f_Y(y) F_Y(y) dy$

(EDITED because some users didn't understand the equation layout)

This implies that

$ 2\int_{\mathbf {R}} f_Y(y) F_Y(y) dy= \left[(F_Y(y))^2\right]_{-\infty}^{+\infty} = 1-0 \implies \int_{\mathbf {R}} f_Y(y) F_Y(y) dy =0.5 $

Is this correct? Does it hold for generic random variables too? With generic I mean $Y_1$ and $Y_2$ not necessarily continous, but still i.i.d.

Answer 1

There are three probabilities under consideration: $P\{Y_1 > Y_2\}$, $P\{Y_2 > Y_1\}$, and $P\{Y_1 = Y_2\}$. The sum of these three is $1$, and it must be that $$P\{Y_1 > Y_2\} = P\{Y_2 > Y_1\}. \tag{1}$$ One way to argue is to use symmetry, or fancier words such as exchangeable random variables. All of these essentially reduce to the argument that any alleged proof that anyone devises to show that $P\{Y_1 > Y_2\} > P\{Y_2 > Y_1\}$ can be changed to a proof that $P\{Y_1 > Y_2\} < P\{Y_2 > Y_1\}$ simply by interchanging $Y_1$ and $Y_2$ wherever they occur in the claimed proof. Since $P\{Y_1 > Y_2\}$ cannot be simultaneously be both larger than and also smaller than $P\{Y_2 > Y_1\}$ the alleged proof cannot be a valid proof at all: it must be the case that $(1)$ is true.

Now, as has been pointed out several times in the comments, $(1)$ does not imply that $$P\{Y_1 > Y_2\} = P\{Y_2 > Y_1\} = \frac 12 \tag{2}$$ unless we can also show that $P\{Y_1 = Y_2\} = 0$. A standard counterexample to $(2)$ is the case of discrete random variables for which $$P\{Y_1 = Y_2\} = \sum_{y} P\{Y_1 = y\}P\{Y_2 = y\} = \sum_y \left(P\{Y_1 = y\}\right)^2 > 0.$$ On the other hand, for continuous (iid) random variables, $P\{Y_1 = Y_2\} = 0$ as you already say you know (cf. your comment addressed to whuber), and so you can assert $(2)$ without needing to go through any integrations whatsoever.

Answer 2

There is a much simpler way to answer it: $Y_2$ and $Y_1$ are numbered arbitrarily (since they come from the same distribution). Thus it would make no sense to argue that, say the probability that $Y_1>Y_2$ is $0.6$, because the same argument would imply that the probability that $Y_2>Y_1$ is $0.6$. But plainly that's a contradiction.