Joint density function where X=Y?

Question

Today I tried solving the following question as a preparation for my Intro. to Probability 1 exam.

Suppose X, Y have joint density f(x, y). what is the probability that X = Y ?

My first attempt was: let $Z = X - Y$, then $P(X=Y)=P(Z=0)=0$ because Z is a continuous random variable.

But I then thought about a a counter example, where $X\sim exp(1)$ and $Y=X$ then $P(X=Y)=1$

How would I go about finding the probability here, if I have no details regarding X,Y ?

Any help would be appreciated!

Answer 1

If $ X \sim exp(1), Y \sim exp(1) $, then $ P(X=Y) \neq $ 1.

Answer 2

Let $\Delta:=\{(x,x)|x\in\Bbb R\}$ and let $\lambda$ denote the Lebesgue measure on $\Bbb R^2$.

Then $\lambda(\Delta)=0$ and consequently:$$P(X=Y)=\int\int f(x,y)1_{\Delta}(x,y)dxdy=0$$

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In general for Borel-measurable $$A\subseteq\mathbb R^2$$ we have:$$P((X,Y)\in A)=\int\int 1_A(x,y)f(x,y)dxdy$$Applying this on $A=\Delta$ we find:$$P(X=Y)=\int\int1_{\Delta}(x,y)f(x,y)dxdy=$$$$\int\int1_{\{y\}}(x)f(x,y)dxdy=\int0dy=0$$