Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mu,\sigma>0$ and define the function $f$ as follows: $$ f(x) = \frac{1}{\sigma\sqrt{2\pi}}\mathrm \exp\left(-\frac{(x-\mu)^2}{2\sigma ^2}\right) $$ How can I show that $$ \int\limits_{-\infty}^\infty x\log|x|f(x)\mathrm dx\geq \underbrace{\left(\int\limits_{-\infty}^\infty x f(x)\mathrm dx\right)}_\mu\cdot\left(\int\limits_{-\infty}^\infty \log|x| f(x)\mathrm dx\right) $$ which is also equivalent to $\mathsf E[ X\log|X|]\geq \underbrace{\mathsf EX}_\mu\cdot\mathsf E\log|X|$ for a random variable $X\sim\mathscr N(\mu,\sigma^2).$

share|improve this question
Is this inequality actually true? –  Did May 30 '12 at 8:43
@did: the motivation is to show that $\mathsf E\log|\mu+\sigma Y|$ for a standard normal $Y$, for any fixed $\sigma>0$ is increasing in $\mu$ along positive values of $\mu$. This is equivalent to the monotonicity of $\mathsf E\log|X|$ where $X\sim\mathscr N(\mu,\sigma^2)$. I've found the derivative of the latter with respect to $\mu$ and that is $$ (\mathsf E\log|X|)^\prime_\mu = \frac{1}{\sigma^2} (\mathsf E[X\log|X|] - \mu \mathsf E\log|X|) $$ which should be hence non-negative. –  Ilya May 30 '12 at 8:53
@did: but why did you ask - do you have a particular counterexample in mind? –  Ilya May 30 '12 at 9:59

3 Answers 3

up vote 1 down vote accepted

$\int x log \vert x \vert f(x) dx - \mu \int log \vert x \vert f(x) dx = \int( x - \mu) log \vert x \vert f(x) dx = \int x log \vert x + \mu \vert \phi(x) dx = \int_0^{\infty} x log(\vert \frac {\mu +x}{\mu - x}\vert )\phi(x) dx$ and the integrand is positive. $\phi$ is symmetric is all that gets used.

share|improve this answer

Below is a probabilistic and somewhat noncomputational proof.

We ignore the restriction to the normal distribution in what follows below. Instead, we consider a mean-zero random variable $Z$ with a distribution symmetric about zero and set $X = \mu + Z$ for $\mu \in \mathbb R$.

Claim: Let $X$ be described as above such that $\mathbb E X\log|X|$ is finite for every $\mu$. Then, for $\mu \geq 0$, $$ \mathbb E X \log |X| \geq \mu \mathbb E \log |X| \> $$ and for $\mu < 0$, $$\mathbb E X \log |X| \leq \mu \mathbb E \log |X| \>.$$

Proof. Since $X = \mu + Z$, we observe that $$ \mathbb E X \log |X| = \mu \mathbb E \log |X| + \mathbb E Z \log |\mu + Z| \>, $$ and so it suffices to analyze the second term on the right-hand side.

Define $$ f(\mu) := \mathbb E Z \log|\mu+Z| \>. $$

Then, by symmetry of $Z$, we have $$ f(-\mu) = \mathbb E Z \log|{-\mu}+Z| = \mathbb E Z \log|\mu-Z| = - \mathbb E \tilde Z \log|\mu + \tilde Z| = - f(\mu) \>, $$ where $\tilde Z = - Z$ has the same distribution as $Z$ and the last equality follows from this fact. This shows the $f$ is odd as a function of $\mu$.

Now, for $\mu \neq 0$, $$ \frac{f(\mu) - f(-\mu)}{\mu} = \mathbb E \frac{Z}{\mu} \log \left|\frac{1+ Z/\mu}{1- Z/\mu}\right| \geq 0\>, $$ since $x \log\left|\frac{1+x}{1-x}\right| \geq 0$, from which we conclude that $f(\mu) \geq 0$ for all $\mu > 0$.

Thus, for $\mu > 0$, $\mu \mathbb E \log |X|$ is a lower bound on the quantity of interest and for $\mu < 0$, it is an upper bound.

NB. In the particular case of a normal distribution, $X \sim \mathcal N(\mu,\sigma^2)$ and $Z \sim N(0,\sigma^2)$. The moment condition stated in the claim is satisfied.

share|improve this answer
I accepted the answer by mike (which follows similar lines) since it answers the case of OP. But thank you very much for the formal solution for the general case –  Ilya Jul 6 '12 at 8:19
@Ilya: Sure. I actually originally obtained my answer independently of the other two. I see some similarities with mike's (and robjohn's, too, actually), but I have a bit of a hard time parsing it. There appears to also be an error or two. –  cardinal Jul 6 '12 at 12:16

Change the dummy variable $x\mapsto x+\mu$: $$ \begin{align} \int_{-\infty}^\infty x\log|x|f(x)\,\mathrm{d}x &=\int_{-\infty}^\infty(x+\mu)\log|x+\mu|f(x+\mu)\,\mathrm{d}x\\ &=\mu\int_{-\infty}^\infty\log|x+\mu|f(x+\mu)\,\mathrm{d}x\\ &\phantom{=}+\int_{-\infty}^\infty x\log|x+\mu|f(x+\mu)\,\mathrm{d}x\\ &=\mu\int_{-\infty}^\infty\log|x|f(x)\,\mathrm{d}x\\ &\phantom{=}+\int_{-\infty}^\infty x\log|x+\mu|f(x+\mu)\,\mathrm{d}x\tag{1} \end{align} $$ Next $$ \begin{align} \varphi(x) &=\int_{-\infty}^\infty x\log|x|f(x)\,\mathrm{d}x-\mu\int_{-\infty}^\infty \log|x|f(x)\,\mathrm{d}x\\ &=\int_{-\infty}^\infty x\log|x+\mu|f(x+\mu)\,\mathrm{d}x\\ &=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^\infty x\log|x+\mu|\;e^{-\frac12(x/\sigma)^2}\,\mathrm{d}x\\ &=\frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^\infty x\log|x+\mu/\sigma|\;e^{-x^2/2}\,\mathrm{d}x\\ &=\frac{\sigma}{\sqrt{2\pi}}=\hspace{-11.5pt}\int_{-\infty}^\infty\frac{1}{x+\mu/\sigma}\;e^{-x^2/2}\,\mathrm{d}x\tag{2} \end{align} $$ $\varphi(x)$ is an odd function of $\mu$, which is positive when $\mu>0$.

First note that $$ \begin{align} \varphi(\mu)+\varphi(-\mu) &=\frac{\sigma}{\sqrt{2\pi}} =\hspace{-11.5pt}\int_{-\infty}^\infty\frac{2x}{x^2-(\mu/\sigma)^2}\;e^{-x^2/2}\,\mathrm{d}x\\ &=0\tag{3} \end{align} $$ since $(3)$ is the integral of an odd function times an even function. Therefore, $\varphi$ is an odd function.

Furthermore, since $\varphi$ is odd, for $\mu>0$, $$ \begin{align} 2\varphi(\mu) &=\varphi(\mu)-\varphi(-\mu)\\ &=\frac{\sigma}{\sqrt{2\pi}}=\hspace{-11.5pt}\int_{-\infty}^\infty\frac{1}{x}\;\left(e^{-(x-\mu/\sigma)^2/2}-e^{-(x+\mu/\sigma)^2/2}\right)\,\mathrm{d}x\\ &=\frac{\sigma}{\sqrt{2\pi}}=\hspace{-11.5pt}\int_{-\infty}^\infty\frac{2}{x}\sinh\left(\frac{x\mu}{\sigma}\right)e^{-\frac12\left(x^2+(\mu/\sigma)^2\right)}\,\mathrm{d}x \tag{4} \end{align} $$ Thus, the integrand in $(4)$ is positive for $\mu>0$. Therefore, $\varphi(\mu)>0$ for $\mu>0$.

Thus, your inequality is true for $\mu>0$.

share|improve this answer
Thanks a lot for the answer! –  Ilya Jul 6 '12 at 8:12
Sorry for the long answer, but as I was modifying my previous, erroneous answer, and there were already answers using the symmetry, I simply stayed with this. –  robjohn Jul 6 '12 at 8:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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