I have a gaussian distribution such as $$P(x)=\frac {1}{\sqrt {2\pi}\sigma}e^{-\frac {(x-\mu)^2}{2\sigma^2}} $$ As my knowledge, $P(x)$ is non convex function interm of $x$. However, if I map it to $log$ space, Does it become convex function? If it is convex, please prove help me. Thanks in advance

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    $\begingroup$ Just use $\log(ab)=\log a+\log b$ and $\log(a^b)=b\log a$...... $\endgroup$ Commented Jun 1, 2015 at 3:45

2 Answers 2


Well, note that $$ \log P(x)=\log\left[\frac{1}{\sqrt{2\pi}\sigma}\right]-\frac{(x-\mu)^2}{2\sigma^2}. $$

This is a downward-facing parabola; its second derivative is $$ \frac{d^2}{dx^2}\left[\log P(x)\right]=-\frac{1}{\sigma^2}<0. $$

  • $\begingroup$ So, It will becomes convex function in log space, Right? $\endgroup$
    – Jame
    Commented Jun 1, 2015 at 3:51
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    $\begingroup$ Yes. It has a negative second derivative. $\endgroup$ Commented Jun 1, 2015 at 3:52
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    $\begingroup$ By standard nomenclature, its log is concave, not convex. $\endgroup$
    – p.s.
    Commented Jun 2, 2015 at 0:01
  • $\begingroup$ @p.s. Very true. $\endgroup$ Commented Jun 2, 2015 at 0:49

The Gaussian density function is quasiconcave but not concave. Moreover, it is log-concave because log P(x) is essentially a negative quadratic function.


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