"Generalized" monotonicity of the expected value

Question

Let $X$ and $Y$ be random variables with cumulative distribution functions $F_X(t)$ and $F_Y(t)$ respectively.

Suppose that $\forall t \in \mathbb{R} \ F_X(t) \geq F_Y(t)$. Does it imply that $E(X) \leq E(Y)$?

Note that $X \leq Y$ implies $F_X(t) \geq F_Y(t)$ but not vise versa:

$X \leq Y$ $\ \Rightarrow \ $ $F_Y(t) = P(Y \leq t) \leq P(X \leq t) = F_X(t)$.
If $X$ has the standard normal distribution and $Y=-X$, then $F_X(t) = F_Y(t)$ but $X$ can be greater than $Y$.

Yes. For example, you can couple $X$ and $Y$ so that $Y-X\geq 0$ always - e.g. let $U$ be uniform on $[0,1]$ and set $X=F_X^{-1}(U)$, $Y=F_Y^{-1}(U)$. Then $E(Y-X)$ is certainly non-negative so $EX\leq EY$. — James Martin, Commented May 15, 2012 at 22:43
Note that $X$ and $Y$ can be independent as well, and $F_X(t) \geq F_Y(t)$ does not imply that $X \leq Y$. I added an example to the question. — Stanislav, Commented May 16, 2012 at 10:23
That's not James Martin's point (I believe). You are basically asking a question about distribution functions (the distribution function determines a measure on $\mathbb{R}$, and from this you get the expectation). So if you can find any random variables X, Y with the distributions $F_X$ and $F_Y$ such that $E(X) \le E(Y)$, then the statement must be true for all random variables with these distributions. And James Martin's comment shows how to construct such $X, Y$. — Tom Bachmann, Commented May 16, 2012 at 10:56
By the way, is there any textbook or article, in which this property is listed? I have looked in several textbooks but have not found it. — Stanislav, Commented May 17, 2012 at 10:00

Cloudscape · Accepted Answer · 2015-02-10 20:53:00Z

2

Note that $E(X) = \int_{-\infty}^\infty x f(x) d x = -x (1-F(x))\vert_{-\infty}^\infty +\int_{-\infty}^\infty (1-F(x))d x.$

From this you should be able to deduce conditions under which the answer to your question is in the affirmative.

Using monotone convergence for the integrals on $(-\infty, 0)$ and $(0, \infty)$, we obtain:

$$ \int_{-\infty}^\infty x f(x) dx = \lim_{R \to \infty} \left( -x (1-F(x))\vert_{-R}^R +\int_{-R}^R (1-F(x))d x \right) $$

By monotony of limits, we obtain the result if a probability density function and the first moment exist.

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answered May 15, 2012 at 23:45

Igor Rivin

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$\begingroup$ Since $F(-\infty) = 0$, both summands diverge, so I do not see any advantage of this representation. $\endgroup$
– Stanislav
Commented May 16, 2012 at 9:39
$\begingroup$ They only diverge is the support is unbounded. $\endgroup$
– Igor Rivin
Commented May 16, 2012 at 12:27
1

$\begingroup$ Yes, if the support is bounded, this formula answers the question. Additionally, if the second moment exists then $1-F_X(t) \leq E(X^2)/t^2$ from Chebyshev's inequality, so that the limit and the integral exist even if the upper bound is infinite. And since integration by parts can be applied to the Riemann–Stieltjes integral, this formula holds for an arbitrary random variable (not only continuous) $\endgroup$
– Stanislav
Commented May 16, 2012 at 13:58

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