If I have a cumulative distribution function of one discrete variable $\mathbb{P}(X\le a)$, then I can use differencing to find the probability mass function: $$\mathbb P(X=a)=\mathbb{P}(X\le a)-\mathbb P(X\le a-1).$$

If I am given $\mathbb P(X\le a, Y\le b)$, how can I find $P(X=a,Y=b)$?

I don't think it is $\mathbb {P}(X\le a,Y\le b)-\mathbb P(X\le a-1, Y\le b-1).$ I think more terms are needed.

  • $\begingroup$ Just a remark. It is more professional to write out acronyms when using them the first time in your text. $\endgroup$ – MrYouMath Sep 6 '17 at 8:24

If $F(a,b)=P(X\le a,Y\le b)$ is the DF of the random vector $(X,Y)$, then the amount of jump of $F$ at $(a,b)$ is given by $$P(X=a,Y=b)=F(a,b)-F(a-0,b)-F(a,b-0)+F(a-0,b-0)$$

where $F(a-0)=\lim_{h\to0+}F(a-h)$

This mainly follows from the property that for every $(x_i,y_i)_{i=1,2}$ with $x_1<x_2$ and $y_1<y_2$,

$P(x_1\le X\le x_2,y_1\le Y\le y_2)$

$=P(X\le x_2,Y\le y_2)-P(X\le x_1,Y\le y_2)-P(X\le x_2,Y\le y_1)+P(X\le x_1,Y\le y_1)$


  • $\begingroup$ Is there an analogue of this when we have $n$ variables instead of 2? $\endgroup$ – Iced Palmer Sep 6 '17 at 9:22
  • $\begingroup$ It could be extended to the n-dimensional case I guess, but I don't see how off the top of my head. $\endgroup$ – StubbornAtom Sep 6 '17 at 10:00

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