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We need to establish the given inequality : $$F_X(x) + F_Y(y) - 1 \leq F_{X,Y}(x,y) \leq \sqrt{F_X(x) F_Y(y)}$$ where $X$ and $Y$ are any random variables.

First I tried the R.H.S :

I started working on p.d.f's ( probability density functions first ).

$ f_{X|Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)}\implies \dfrac{f_{X|Y}(x|y)}{f_X(x)} = \dfrac{f_{X,Y}(x,y)}{f_Y(y)f_X(x)}$

But I don't think that's gonna work . Can anyone suggest some other way ?

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  • $\begingroup$ Since $X$ and $Y$ are said to be any random variables, using density functions will at nest lead to a proof restricted to jointly continuous $X$ and $Y$. @StubbornAtom's answer is very much to the point (+1 to him) and deserves acceptance. $\endgroup$ Mar 19, 2020 at 14:57

2 Answers 2

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Hint for the right hand side:

$F_X(x) = P(X < x) \ge P(X < x \wedge Y < y) = F_{X, Y}(x, y)$.


Hint for the left hand side:

The following figure shows the $XY$ plane, with the horizontal and vertical lines intersecting at $(x, y)$.

enter image description here

The region $C$ is the area where $X < x \wedge Y < y$ - you need to integrate over it to obtain $F_{X, Y}(x, y)$. Over which regions do you need to integrate to obtain $F_X(x)$, $F_Y(y)$, and $1$?

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  • $\begingroup$ How did you conclude the first one , i.e , $P(X<x) \geq P(X<x , Y<y)$ ? $\endgroup$
    – User9523
    Oct 16, 2016 at 18:44
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    $\begingroup$ $P(X < x \wedge Y < y) = P(X < x) P(Y < y | X < x)$. A conditional probability is nevertheless a probability. What is the range of a probability? $\endgroup$
    – Ami Tavory
    Oct 16, 2016 at 18:45
  • $\begingroup$ You're welcome. All the best. $\endgroup$
    – Ami Tavory
    Oct 16, 2016 at 18:53
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It is just the elementary inequality $$P(A)+P(B)-1\le P(A\cap B)\le \sqrt{P(A)P(B)}$$ for events $A=\{X\le x\}$ and $B=\{Y\le y\}$. There is no need to go into distributions.

Let $I_A$ be the indicator of $A$, i.e. $I_A=1$ if $A$ occurs and $I_A=0$ if $A$ does not occur.

Then by Cauchy-Schwarz inequality,

$$\left(E\left[I_AI_B\right]\right)^2 \le E\left[I_A^2\right]E\left[I_B^2\right] \implies (P(A\cap B))^2\le P(A)P(B)$$

And $P(A^c\cup B^c)\le P(A^c)+P(B^c)=1-P(A)+1-P(B)$ implies $$P(A\cap B)=1-P(A^c\cup B^c)\ge P(A)+P(B)-1$$

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