How to prove that if two independent variables have different densities then they are different?

Question

Suppose that we have two continuous independent random variable $X$ and $Y$ with respective densities $f_x$ and $f_y$, how can we prove that $X$ is different than $Y$.

I thought about proving it by contradiction using the definition of the probability of a random variable ( using integral equalities to prove that densities are equals therefore the absurdity ) but I didn't utilize independency there so I felt like it's wrong ( also I'm not sure if two lebesgue integrals are equals in the same interval then the functions integrated are also equals ).

Answer 1

If we have that $X \perp \!\!\! \perp Y$ then we know that $\displaystyle f_{X,Y}(x,y) = f_X(x)\cdot f_Y(y)$.

We can either have $P(X=Y) = 0$ or $f_X \neq f_Y$ as our " measure" of them being different.

For a given $z$ we have

$P(X =z, Y =z) = P(X=z)P(Y=z) = 0 \cdot 0$.

More work is required to get that $P(X=Y)=0$. See @angryavian's answer.

Answer 2

In general, if $(X,Y)$ has joint density $f_{X,Y}$, then the probability that $(X,Y)$ lies in some subset $A \subseteq \mathbb{R}^2$ is the two-dimensional integral $$P((X,Y) \in A) = \iint_A f_{X,Y}(x,y) \, dx \, dy.$$ In this case, you do have a joint density: $f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y)$. And in your case, $A = \{(x,y) : x=y\}$ is a diagonal line in the $(x,y)$-plane. The two-dimensional integral of a function over a region with no area (like a line) is zero.

[Update: I noticed you mentioned Lebesgue integrals in your question, so I am guessing you would understand that the integral over a set of measure zero is zero.]