Comparing the mean of two continuous random variable when one's CDF is always greater than the other

Question

Suppose that for two continuous random variables $X_1$ and $X_2$, the below condition holds.

$$ F_{X_2}(x) \geq F_{X_1}(x) $$

for all $x \in R$ and $F$ stands for CDF.

Then can I say that the mean of $X_1$ is always greater than the mean of $X_2$? (assume that mean of both of them are finite)

If so, what would be the proof of that?

Also relevant: Monotone Likelihood Ratio. See bit about stochastic dominance later in the article. — Aaron Hendrickson, Mar 12 '23 at 18:09

score 3 · Answer 1 · answered Mar 12 '23 at 18:05

3

If $X$ is nonnegative, you can use the following relation to compute the expectation from the distribution. $$ E(X) = \int_0^\infty 1-F_X(x) \, dx. $$ The result you are looking for follows from that equality.

answered Mar 12 '23 at 18:05

coudy

5,843

Thanks for your comment, but what if the X can be negative?? – Neo Mar 12 '23 at 18:23

score 2 · Answer 2 · answered Mar 12 '23 at 21:57

2

If $X$ can be negative, use the identity $$E(X)=\int_0^\infty[1-F_X(t)]dt-\int_{-\infty}^0 F_X(t)dt,$$ to show that $E(X_2)\le E(X_1)$.

(The identity is valid whenever $E(X)$ exists, and is proved here.)

answered Mar 12 '23 at 21:57

grand_chat

38,951

Comparing the mean of two continuous random variable when one's CDF is always greater than the other

2 Answers2