Inequalities of Two i.i.d. Random Variables The Problem
Let $X_1$ and $X_2$ be a continuous i.i.d. random variables.  By definition, we know that...

*

*Independence: $\mathbb P(X_1 \in A \ \cap \ X_2 \in B)=\mathbb P(X_1 \in A) \ \mathbb P(X_2 \in B)$

*Identically Distributed: $\mathbb P(X_1 \le x) = \mathbb P(X_2 \le x) \ \ \forall \ \ x \in \mathbb R^1$
My hypothesis is that:
$$\mathbb P(X_1 > X_2) = 0.5$$
I have attempted to prove this by using the above definitions to form the expression:
$$\mathbb P(X_1 \ge x \ \cap \ X_2 < x)=\mathbb P(X_2 \ge x \ \cap \ X_1 < x) \ \ \forall \ \ x \in \mathbb R^1$$
My Questions

*

*Is this a valid proof, and am I missing anything?

*This seems like a fundamental property of i.i.d. random variables.  Is this already a named theorem?

 A: There is a standard trick involved here. Instead of giving you the full solution, let me point you in the right direction instead. Note that the probability you are interested in is equivalent to $\mathbb{P}(X_1 - X_2 > 0).$ Hence it would be enough to determine the cumulative distribution function of the variable $X_1 - X_2.$ For this it would be enough to first determine the CDF of $-X_2$ and then determine the CDF of the difference by explicitly computing the density of $X_1 - X_2$ in terms of the densities of $X_1$ and $-X_2$ (which you would have determined in the previous step). Is there any important Theorem that comes to mind when looking at it this way?
Please do let me know if you manage to fill in the blanks by yourself and if not, I shall do that.
A: Hint
(From Andrew Yung)

There is a standard trick involved here. Instead of giving you the full solution, let me point you in the right direction instead. Note that the probability you are interested in is equivalent to $\mathbb{P}(X_1 - X_2 > 0).$ Hence it would be enough to determine the cumulative distribution function of the variable $X_1 - X_2.$ For this it would be enough to first determine the CDF of $-X_2$ and then determine the CDF of the difference by explicitly computing the density of $X_1 - X_2$ in terms of the densities of $X_1$ and $-X_2$ (which you would have determined in the previous step). Is there any important Theorem that comes to mind when looking at it this way?
Please do let me know if you manage to fill in the blanks by yourself and if not, I shall do that.

Solution
Thank you for the hint, Andrew.  With your help I was able to find a solution.  Please let me know if you were thinking of an easier path to the answer.

*

*Because $X_1$ and $X_2$ are Identically Distributed, the CDF and PDF of each random variable can be represented as:
$$F_X(x)=\mathbb P(X_1 \le x)=\mathbb P(X_2 \le x)$$
$$f_X(x)=\frac d {dx} F_X(x)$$

*Note that:
$$F_{-X_2}(x)=\mathbb P(-X_2 \le x)=\mathbb P(X_2 > -x)=1-F_X(-x)$$

*Also, the CDF of the sum of two Independent Continuous Random Variables ($X$ and $Y$) is:
$$F_{X+Y}(a) = \mathbb P(X+Y \le a) = \int_{-\infty}^{+\infty} f_X(x)F_Y(a-x)dx$$

*Therefore:
$$\mathbb P(X_1 \le X_2)=\mathbb P(X_1 -X_2 \le 0)=$$
$$F_{X_1-X_2}(0) = F_{X_1+(-X_2)}(0) =$$
$$\int_{-\infty}^{+\infty} f_X(x)\Bigl(1-F_X\bigl(0-(-x)\bigr)\Bigr)dx =$$
$$\int_{-\infty}^{+\infty} f_X(x)dx+\int_{-\infty}^{+\infty} f_X(x) \Bigl(-F_X(x)\Bigr)dx=$$
$$1-\int_{-\infty}^{+\infty} f_X(x) \ F_X(x)dx = \ ...$$

*Then you can use Integration by Substution where $u=F_X(x)$:
$$1-\int_{x=-\infty}^{x=+\infty} \biggl(\frac {du} {dx}\biggr)(u) dx=1-\int_{u=0}^{u=1} (u) du=1-\biggl[ \frac {u^2} {2} \biggr]_{0}^{1}=$$
$$1-(0.5-0)=0.5$$
