3
$\begingroup$

I am having lots of trouble proving the following statement:

Let $X,Y$ be two real valued random variables on a probability space $(\Omega,\mathcal{F},P)$. These two variables are independent and identically distributed, iid, in the sense that they share the same distribution $P_X=P_Y$ and the cumulative distribution function $F_X(x)=P_X[X\leq x]$ is continuous. The probability measure $P$ might not necessarily be absolutely continuous wrt a sigma-finite measure. Therefore there might not be a Radon-Nikodym derivative with respect to a sigma-finite measure. If there was a Radon-Nikodym derivative with respect to a sigma-finite measure, it is easier to show what I want.

I need to prove that, under the above conditions $P[X=Y]=0$

Any insights?

Best regards,

Juan Manuel

$\endgroup$
6
$\begingroup$

For any increasing sequence $-\infty=x_0<x_1<\ldots < x_n=\infty$, we have $$\tag0P[X=Y]\le \sum_{i=1}^n P[x_{i-1}<X\le x_i,x_{i-1}<Y\le x_i]=\sum_{i=1}^n (F_X(x_i)-F_X(x_{i-1}))^2.$$ Since $F_X\colon\mathbb R\to[0,1]$ is continuous and at least covers $(0,1)$, we can pick $x_i$ so that $F_X(x_i)=\frac in$. Then $(0)$ gives us $P[X=Y]\le \frac 1n$.

$\endgroup$
  • $\begingroup$ Thank you Hagen, this one does the trick very intuitively. $\endgroup$ – Julio May 28 '13 at 19:19
6
$\begingroup$

Here is a big machinery argument. Since the cdf is continuous, the probability of each point is zero. let $D$ be the diagonal on $\mathbb{R}^2$. By Fubini's theorem, you get

$$P(X=Y)=\int_\mathbb{R^2}1_DdP_{(X,Y)}=\int_\mathbb{R}\int_\mathbb{R}1_D(x,y)~dP_X(x)~ dP_Y(y) =\int_\mathbb{R}0~dP_Y(y)=0.$$

$\endgroup$
  • $\begingroup$ Thank you Michael, it was very helpfull. $\endgroup$ – Julio May 28 '13 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.