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Let $X_1, X_2, ...$ be i.i.d. (real) random variables with continuous distribution function (in particular, we're not assuming absolute continuity).

How can I show that $P[X_n = X_m] = 0$ for $n \ne m$?

It's pretty clear for an absolutely continuous distribution function, where one can even drop the assumption that the $X_i$ are identically distributed. I'm not sure how to handle the only continuous case though.

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2 Answers 2

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It suffices to consider $m=1,n=2$. Given $\epsilon > 0$, take $N$ so that $P[|X_2| \le N] > 1 - \epsilon$. Now the event $(X_1 = X_2 \in [-N,N])$ is contained in the union of rectangles $(x_i \le X_1, X_2 \le x_{i+1})$ where $x_{i+1} - x_i = \delta$. Take $\delta$ small enough that ...

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If $n\neq m$ then $P\left\{ X_{n}=X_{m}\right\} =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x,y\right)dF_{X_{n}}\left(x\right)dF_{X_{m}}\left(y\right)$ where $f\left(x,y\right)=1$ if $x=y$ and $f\left(x,y\right)=0$ otherwise. If $F_{X_{n}}$ or $F_{X_{m}}$ is continuous then $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x,y\right)dF_{X_{n}}\left(x\right)dF_{X_{m}}\left(y\right)=0$. The independency is used, but you don't need $F_{X_{n}}=F_{X_{m}}$

Edit:

Generally $P\left\{ X_{n}=X_{m}\right\} =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x,y\right)dF_{X_{n},X_{m}}\left(x,y\right)$ and the independency gives $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x,y\right)dF_{X_{n},X_{m}}\left(x,y\right)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x,y\right)dF_{X_{n}}\left(x\right)dF_{X_{m}}\left(y\right)$

Here:

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x,y\right)dF_{X_{n}}\left(x\right)dF_{X_{m}}\left(y\right)=\int_{-\infty}^{\infty}P\left\{ X_{n}=y\right\} dF_{m}\left(y\right)$

and a continuous distributed $X_{n}$ means exactly that $P\left\{ X_{n}=y\right\} =0$ for each $y$.

Also

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x,y\right)dF_{X_{n}}\left(x\right)dF_{X_{m}}\left(y\right)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x,y\right)dF_{X_{m}}\left(y\right)dF_{X_{n}}\left(x\right)=\int_{-\infty}^{\infty}P\left\{ X_{m}=x\right\} dF_{m}\left(x\right)$

and a continuous distributed $X_{m}$ means exactly that $P\left\{ X_{m}=x\right\} =0$ for each $x$.

So in both cases the integrand is the zero-function, leading to $0$ as outcome.

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absolutely continuous means that there is a density. I do not demand the existence of a density. Function $f$ here is no density, but a characteristic function. I only use that the cumulative distribution functions are continuous. –  drhab Nov 25 '13 at 18:59
    
@Balerion_the_black. I do not assume $F_{X}$ to be absolutely continuous. You should explain your downvote or withdraw it. –  drhab Nov 25 '13 at 19:12
    
Sorry I overlooked your definition of the characteristic function. However, could you explain where independence is used, and why continuity implies that the integral is 0? (I don't know so much about Lebesgue integration.) –  Balerion_the_black Nov 25 '13 at 19:48
    
Independence takes care of $\int\int f\left(x,y\right)dF_{X_{n},X_{m}}\left(x,y\right)=\int\int f\left(x,y\right)dF_{X_{n}}\left(x\right)dF_{X_{m}}\left(y\right)$. If $F_{X_{n}}$ is continuous then $\int f\left(x,y\right)dF_{X_{n}}\left(x\right)=P\left\{ X_{n}=y\right\} =0$ for every $y$ and consequently $\int\int f\left(x,y\right)dF_{X_{n}}\left(x\right)dF_{X_{m}}\left(y\right)=0$. The order of integration can be changed, so this will also work if $F_{X_{m}}$ is continuous. –  drhab Nov 25 '13 at 20:08
    
I have made an edit to my answer, trying to explain things. –  drhab Nov 25 '13 at 20:29

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