For all w, CDF of y at w is greater than the CDF of x at w. Prove that the probability ( y < x ) >= 0.5

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?

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 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

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}

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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:

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

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 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 $P(Y 0.5 but Fy(w) > Fx(w), so it should logically work itself out. Why is P(y < x) equal to the line under Fy(Fx) ? Why not Fy(fx)? And what do the functions mean without something feeding in like (w)? – Eruditass Oct 3 '10 at 17:44 The assertion is false, even for independent random variables$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.

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