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I am looking at elementary properties of joint distribution functions and I want to show that

$$F_{X, Y}(\infty, y)=\lim _{x \rightarrow \infty} F_{X,Y}(x,y)= F_{Y}(y)$$

This is the definition for the marginal distribution function. Any clues on how to prove this without recourse to the density function ?

If not, my approach would be to show, by the definition of $F_{X,Y}$ ;

$$F_{x, y}(x, y)=\int_{-\infty}^{x} \int_{-\infty}^{y} f_{X,Y}(s,t) \,ds\, dt$$

that:

$$\lim_{x \rightarrow \infty} F_{x, y}(x, y)= \lim_{x \rightarrow \infty} \int_{-\infty}^{x} \int_{-\infty}^{y} f_{X,Y}(x,y) \,dx \,dy$$

... $$= \int_{-\infty}^{y} \lim_{x \rightarrow \infty} \int_{-\infty}^{x} f_{X,Y}(x,y) \,dx \,dy = \int_{-\infty}^{y} f_{Y}(y)\,dy = F_{Y}(y)$$

Unsure though, how to swap here the $x,y$ for the placeholder variables $s,t$ without altering the definitions.

StubbornAtom
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JKM
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1 Answers1

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No integrals required. Instead, go back to the definition $F_{X,Y}(x,y):=P(X\le x,Y\le y)$. It's enough to prove for each fixed $y$ that $$\lim_{n\to\infty}P(X\le x_n, Y\le y)=P(Y\le y) $$ whenever $x_n\uparrow\infty$. To prove this, write the LHS as $\lim P(A_n)$ where $A_n:=\left\{X\le x_n\right\}\cap \left\{Y\le y\right\}$. As $n\to\infty$, the events $A_n$ increase to the event $A:=\left\{Y\le y\right\}$, so by continuity from below $$\lim P(A_n)=P(A).$$

grand_chat
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