Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

F y(w) >= F x(w), for all w. Prove P{y<x} >= 0.5

I had this question on the exam and boy did it stump me. It was simple to understand why it was true, but I tried several different paths and got nowhere.

Any ideas?

share|improve this question
    
What distribution are we talking about here? –  J. M. Oct 3 '10 at 10:56
    
Are $X$ and $Y$ independent? –  Jyotirmoy Bhattacharya Oct 3 '10 at 11:36
    
Any distribution, and I believe they were independent if it matters. –  Eruditass Oct 3 '10 at 17:28

3 Answers 3

For definiteness, let us assume that $X$ and $Y$ are positive random variables taking values in $[0, \infty)$,and $F_Y(w)\le F_X(w)$, where $F_X(x), F_Y(y)$ are the CDFs and the equalities take place only at zero and in the limit $w \rightarrow \infty$. For simplicity, let us assume further that both CDFs are strictly increasing. The latter condition implies that we can invert the x-CDF and substitute in the y-CDF to obtain the function $F_Y(F_X)$. In this function, both independent ($F_X$) and dependent ($F_Y$) variables range in $[0, 1]$ and the value of F_Y is always less than F_X (except at 0 and 1). (It is easy to see that, if one plots the x-CDF and y-CDF on the same axes, and see that at every value of the x-CDF, the y-CDF lies to the left of the x-CDF.

Now, $P(Y\lt X)$ is equal to the area under the line of $F_Y(F_X)$ , which is less than the area of a right isosceles rectangle of unit sides which is equal to 0.5.

Edit: Here is the proof (as asked by Eruditass) that the required probability can be computed from the area of under the $F_Y$ graph as a function of $F_X$.

Using the convolution formula to compute the probability density function of $Y-X$:

$f_{Y-X}(w) = \int_{0}^{\infty}f_X(x) f_Y(w+x) dx$

In terms of which, the required probability is given by:

$P(Y < X) = P(Y -X<0) = \int_{-\infty}^{0} f_{Y-X}(w) dw$

Substituting the convolution formula:

$P(Y < X) = \int_{-\infty}^{0} dw \int_{0}^{\infty} dx f_X(x) f_Y(w+x) $

Changing the order of integration (the integration is over bounded functions) and performing a change of variables $ u = w +x$.

$P(Y < X) = \int_{0}^{\infty} dx f_X(x) \int_{-\infty}^{x} du f_Y(u) $

Using the definition of the cumulative distribution function, the second integral is just the Y-CDF

$P(Y < X) = \int_{0}^{\infty} dx f_X(x) F_Y(x) $

Again using the definition of the CDF:

$f_X(x) dx = dF_X(x) $

Changing the integration variable from $x$ to $F_X$, we obtain the final result:

$P(Y < X) = \int_{0}^{1} dF_X F_Y$

share|improve this answer
    
Your conclusion is the opposite of the desired one. I think you mean that the value of $F_Y$ is always greater than $F_X$. Then the graph of $F_Y$ as a function of $F_X$ will lie above the line $F_Y=F_X$, and the area (which is $P(Y<X)$) will be greater than 0.5. –  Hans Lundmark Oct 3 '10 at 12:17
    
No, for a given $w$, $F_Y$ is greater than $F_X$ as given in the question, but, for a given $F_X$, $F_Y$ is smaller than $F_X$, this can be seen easily when you plot both CDFs on the same axes. –  David Bar Moshe Oct 3 '10 at 13:03
    
It's for all w and yes we wanted P(y<x) > 0.5 but Fy(w) > Fx(w), so it should logically work itself out. Why is P(y < x) equal to the line under Fy(Fx) ? Why not Fy(fx)? And what do the functions mean without something feeding in like (w)? –  Eruditass Oct 3 '10 at 17:44

Here's a rough sketch of one approach:

\begin{align} \mathbb{P}[Y\lt X] &= \int_{-\infty}^{\infty} \mathbb{P}[Y\lt X|X=x] \cdot f_X(x) \,dx \\ &= \int_{-\infty}^{\infty} F_Y(x) \cdot f_X(x) \,dx \\ &\geq \int_{-\infty}^{\infty} F_X(x) \cdot f_X(x) \,dx \\ &= \left[ \frac{1}{2} \left(F_X(x)\right)^2 \right]_{-\infty}^{\infty} \\ &= \frac{1}{2}. \end{align}

share|improve this answer

The assertion is false, even for independent random variables $X$ and $Y$. Fix $u$ in $(0,\frac12)$ and consider i.i.d. Bernoulli random variables $X$ and $Y$ with $$P(X=0)=P(Y=0)=1-u,\quad P(X=1)=P(Y=1)=u. $$ Then $[Y<X]=[Y=0,X=1]$, hence $P(Y<X)=u(1-u)$, which is at most $\frac14$ and can be as close to $0$ as desired. In particular $P(Y<X)<\frac12$.

Let us now consider the probability that $[Y\le X]$. Then the independence hypothesis is crucial. To see this, consider $u$ in $(0,1)$, $X$ uniformly distributed on the interval $(u,1+u)$ and $Y$ uniformly distributed on the interval $(0,1)$. Then $F_Y(x)\ge F_X(x)$ for every $x$. One can realize $(X,Y)$ in at least three ways.

(1) If $Y=X-u$, then $P(Y\le X)=1$ hence $P(Y\le X)\ge\frac12$.

(2) If $(X,Y)$ are independent, $P(Y\le X)=\frac12+u-\frac12u^2$ hence $P(Y\le X)\ge\frac12$.

(3) If $Y=\varphi(X)$ with $\varphi(x)=x+u$ if $x\le1-u$ and $\varphi(x)=x+u-1$ otherwise, one can check that $Y$ is uniformly distributed on $(0,1)$ and that $[Y\le X]=[X>1-u]$ hence $P(Y\le X)=2u$ and $P(Y\le X)$ is as close to $0$ as desired.

This shows that the result $P(Y\le X)\ge\frac12$ cannot hold in full generality.

Finally, for independent $X$ and $Y$, indeed $P(Y\le X)\ge\frac12$.

A proof which does not assume the existence of densities is as follows. Note that $$P(Y\le X)=E(P(Y\le X|X))=E(F_Y(X))\ge E(F_X(X))=P(X'\le X), $$ where $X'$ and $X$ are i.i.d. Now, by symmetry $[X'\le X]$ and $[X\le X']$ have the same probability and $P(X'\le X)+P(X\le X')=1+P(X=X')\ge1$ hence $P(X'\le X)\ge\frac12$ and we are done.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.