Logarithm of Gaussian function is whether convex or nonconvex?

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

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

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 $$\frac{d^2}{dx^2}\left[\log P(x)\right]=-\frac{1}{\sigma^2}<0.$$

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