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I have a weird question about probability and density functions : Let's take a random variable X whose p.d.f exists and let's denote it $f_{X}\left(x\right)$. Does the definition of the joint probability $f_{X,X}\left(x,x\right)$ exist ? clearly it's not continuous but i wanted to "check" that the marginal of X ($f_{X}\left(x\right)$) would be the integral of this joint distribution...

Can you give me more insight about it?

thanks,

Romain

Romain227
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1 Answers1

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You can define the probability distribution of $(X,X)$ but there is no density with respect to the Lebesgue measure of $\mathbb{R}^2$, because $(X,X)$ is supported by $\Delta=\{(x,x),\, x\in\mathbb{R}^2\}$, whose Lebesgue measure in $\mathbb{R}^2$ is $0$. So the pdf of $(X,X)$ does not exist, it's a degenerate distribution.

However, you can calculate it's cdf, which always exists. $$P(X\leq x,X\leq y)=P(X\leq\min(x,y))=F_X(\min(x,y))$$ where $F_X$ denotes the cdf of $X$.

Augustin
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    thanks for the reply, but can we not see things like this ? that is, the joint distribution exists, but the probability mass on the straight with slope 1 in 2D space (that is, for (x0, x0) 2D vectors) are "infinity", and 0 elsewhere (that is, for (x0, x1) 2D vectors). then $f_{X,X}\left(x,x\right)$ could be defined as equal to $f_{X}\left(x\right)$ * the dirac function, and when integrating the integral of a dirac is 1...so it remains $f_{X}\left(x\right)$..and we have gotten our marginal – Romain227 Apr 19 '16 at 09:34
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    Yes, in a way we could write $dP_{(X,X)}(x,y)=f_X(x)dx\delta_x(dy)$ but be careful, this is not a function, there is no density. – Augustin Apr 19 '16 at 09:50
  • Augustin and @Romain227 , what about saying $P(B \le A | B= t) = P(t \le A | B= t)$ ? Please see here: When continuous random variable is in both the conditional and the conditioned – BCLC Mar 26 '21 at 11:13