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.

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.

  • $\begingroup$ No, $E[X] = \int_{-\infty}^\infty x\; f(x)\; dx$. $\int_{-\infty}^\infty f(x)\; dx = 1$. $\endgroup$ – Robert Israel Mar 24 '17 at 19:57
  • $\begingroup$ @RobertIsrael Sorry about the typo, still familiarizing with MathJax. $\endgroup$ – lydias Mar 24 '17 at 20:00
  • $\begingroup$ 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. $\endgroup$ – 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 $a=0$.

  • $\begingroup$ Hi, do we have to add the difference between G(x) and f(x) that is the area under the intersecting point? $\endgroup$ – lydias Mar 24 '17 at 23:31
  • $\begingroup$ @lydias : You shouldn't write $G(x)$ if you mean $g(x).$ The area under just one point is $0. \qquad$ $\endgroup$ – Michael Hardy Mar 24 '17 at 23:58
  • $\begingroup$ @MichaelHardy thanks Michael. For some reason my cellphone keyboard just wanted to capitalize that G lol okay it did it again. $\endgroup$ – lydias Mar 24 '17 at 23:59
  • $\begingroup$ Since there seems to be some general interest about the questen, I've formulated the shifting argument. Thanks @MichaelHardy for the formatting improvement. ;) $\endgroup$ – tofurind Mar 25 '17 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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