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Let $Pr_{(X,Y)}$ be a probability distribution of a random vector $(X,Y)$. Let $F$ be the cumulative distribution function of $(X,Y)$. Define $$ \mathcal{A}\equiv \{(x,y): x\leq 2 \text{ and }x-y\leq 3\} $$ Is there a way to express $Pr_{(X,Y)}(\mathcal{A})$ as $$ (*)\quad \sum_{k=1}^K F(a_k, b_k)\times c_k $$ for some finite $K$ and $\{a_k, b_k, c_k\}_{k=1}^K$?


Note: if $(X,Y)$ had a uniform distribution, then we could write $$ Pr_{(X,Y)}(\mathcal{A})=Pr_{(X,Y)}((-\infty,2]\times [-1,\infty))+\frac{1}{2}Pr_{(X,Y)}((-\infty,2]\times (-\infty,-1]) $$ which can be rewritten as $(*)$. However, I wonder whether we could do something similar for the general case without uniformity.

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    $\begingroup$ There is no such thing as "a probability measure of a random vector". There is the notion, though, of the probability distribution of a random vector (which is a measure). $\endgroup$ Commented Nov 22, 2021 at 14:47
  • $\begingroup$ Thank you for pointing this out. $\endgroup$
    – Star
    Commented Nov 22, 2021 at 14:51
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    $\begingroup$ In your formula $$ Pr_{(X,Y)}(\mathcal{A})=F((-\infty,2]\times [-1,\infty))+\frac{1}{2}F((-\infty,2]\times (-\infty,-1]), $$ you treat $F$ as a function of a set, whereas in fact $F$ is a function of pair of real numbers. $\endgroup$ Commented Nov 22, 2021 at 14:54
  • $\begingroup$ Yes, I have edited that. Thanks $\endgroup$
    – Star
    Commented Nov 22, 2021 at 14:56

1 Answer 1

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No, this cannot be done in general. Indeed, let $A:=\mathcal A$. You want to express $$P((X,Y)\in A)=Ef(X,Y)$$ as $$\sum_{k=1}^K c_k F(a_k,b_k)=Eg(X,Y),$$ where $$f(x,y):=1(x\le2,x-y\le3)$$ and $$g(x,y):=\sum_{k=1}^K c_k 1(x\le a_k,y\le b_k).$$ However, for any choice of the numbers $a_k,b_k,c_k$, there will be some $(x_*,y_*)\in\mathbb R^2$ such that $f(x_*,y_*)\ne g(x_*,y_*)$. Letting the random pair $(X,Y)$ take value $(x_*,y_*)$ with probability $1$, we get $$P((X,Y)\in A)=Ef(X,Y)=f(x_*,y_*)\ne g(x_*,y_*)=Eg(X,Y)=\sum_{k=1}^K c_k F(a_k,b_k),$$ so that $P(A)\ne\sum_{k=1}^K c_k F(a_k,b_k)$.

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  • $\begingroup$ Thanks. What is $E$ in your notation (in front of $f$ and $g$)? $\endgroup$
    – Star
    Commented Nov 22, 2021 at 14:38
  • $\begingroup$ @TEX : As usual, $E$ denotes the expectation. $\endgroup$ Commented Nov 22, 2021 at 14:40

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