1
$\begingroup$

Let $X$ be a random variable with density function $f(x)$, so $$ \int_{-\infty}^{+\infty}f(x)dx=1 $$ and $$ E(X)=\int_{-\infty}^{+\infty}{xf(x)dx}. $$ Then, my question is if $$ \int_{-\infty}^{E(X)}{f(x)dx}=\frac{1}{2} ? $$ and if it is true, how can be proved? Otherwise, which would be a contra-example of this?

Thanks!

$\endgroup$
  • 3
    $\begingroup$ So, the question is whether the mean is equal to the median and this is always true if the density is symmetric about a middle value and sometimes otherwise. For example in the not very magnificent seven 0, 0, 0, 1, 1, 1, 4 the mean is equal to the median but symmetry is lacking. I don't know how much more notation and machinery you want to use. $\endgroup$ – Nick Cox May 3 at 8:18
  • $\begingroup$ Ok! I hadn't see that my question could be restated as asking if the mean is equal to the median. In this case, I see the answer it is clearly negative and I assume it should not be difficult to get a contra-example. Thanks! $\endgroup$ – iago May 3 at 8:48
  • $\begingroup$ ... take some skew distribution; not every asymmetric distribution has mean different from median but you'd find a counterexample in no time. $\endgroup$ – Glen_b May 4 at 0:10
4
$\begingroup$

Since you're writing what look to be Riemann integrals, I presume we're keeping to the case where $X$ is a continuous random variable.

If $m$ is a value for which $\int_{-\infty}^m f_X(x)\, dx =\frac12$ then $m$ is a median of $X$ (if the density is $>0$ in a neighborhood of $m$, then $m$ will be unique; the median).

Since here the upper limit of the integral is given as $m=E(X)$, this amounts - as Nick Cox pointed out in comments - to saying "is there a distribution for which the mean differs from the median", and the obvious thing to do when searching for a counterexample is to try some distributions that are skew*.

Here's an obvious example: the exponential distribution, which has its median at $\ln 2$ times the mean (i.e. about 70% of the mean).

You might like to also consider the density

$$f(x) = \cases{ \begin{array}{lr} 0, & \text{for } x\leq 0\\ 2x, & \text{for } 0< x\leq 1\\ 0, & \text{for } x> 1 \end{array}} $$

as a simple one to try for yourself (since the integrations are particularly simple).


*(not every asymmetric distribution has mean different from median, however)

$\endgroup$

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.