# Show $\mathbb{E}[X] \le \mathbb{E}[Y]$

I have a problem that I'm having trouble to identify the correct integral intervals for proof. I would like to have some help on the thought process.

Problem:
Let $X$ and $Y$ have densities $f$ and $g$, respectively, and $f(x) \begin{cases} \ge g(x), & \text{if$x$}\le a;\\ \le g(x), & \text{if$x$}\ge a \end{cases}$

Show that $\mathbb{E}[X] \le \mathbb{E}[Y]$

Since $\mathbb{E}[X]= \displaystyle \int_{-\infty}^{\infty} xf(x)\, dx$

I will have to identify intervals (one positive and one negative) that are able to prove $\mathbb{E}[X] \le \mathbb{E}[Y]$ for all cases. However, I feel there may be an intersection, which leads to 3 intervals total, but I'm not sure how to divide up the segments. I would like some help with laying out the proof.

• No, $E[X] = \int_{-\infty}^\infty x\; f(x)\; dx$. $\int_{-\infty}^\infty f(x)\; dx = 1$. – Robert Israel Mar 24 '17 at 19:57
• @RobertIsrael Sorry about the typo, still familiarizing with MathJax. – lydias Mar 24 '17 at 20:00
• A quick proof is to note that the hypothesis implies that the CDFs of $X$ and $Y$ are such that $$F_X\geqslant F_Y$$ everywhere and to recall that expectations have simple expressions in terms of CDFs. – Did Mar 25 '17 at 10:33

W.l.o.g. let be $a=0$ (otherwise shift the random variables, see Addendum). Then \begin{align} \int_{-\infty}^\infty xf(x)\,dx & = \int_{-\infty}^0 xf(x)\,dx + \int_0^\infty xf(x)\,dx \\[12pt] & \leq \int_{-\infty}^0 xg(x)\,dx + \int_0^\infty xg(x)\,dx = \int_{-\infty}^\infty xg(x)\,dx \end{align} holds.
Addendum (maybe some extra notes about the "$a=0$" assumption above do not harm): With the linearity of the expectation value the following equivalence holds $$\Bbb E(X-a)\leq \Bbb E(Y-a) \Leftrightarrow \Bbb E(X)-a\leq \Bbb E(Y)-a \Leftrightarrow \Bbb E(X)\leq \Bbb E(Y),$$ which justifies just to consider the translated variables. For the CDF of $X-a$ (and for $Y-a$ analogously) we have $$P_X(X-a\leq x) = P_X(X\leq x+a) = \int_{-\infty}^{x+a}f(x)\,dx = \int_{-\infty}^x f(x+a)\,dx.$$ For the translated desnsity function $f(x+a)$ (this is the density function of $X-a$) we obtain now $$f(x+a) \begin{cases} \ge g(x+a), & \text{if x }\le 0\\ \le g(x+a), & \text{if x }\ge 0 \end{cases}.$$ This allows us just to study the case $a=0$.
• @lydias : You shouldn't write $G(x)$ if you mean $g(x).$ The area under just one point is $0. \qquad$ – Michael Hardy Mar 24 '17 at 23:58