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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2$\begingroup$ Just use $\log(ab)=\log a+\log b$ and $\log(a^b)=b\log a$...... $\endgroup$– symplectomorphicJun 1, 2015 at 3:45
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2 Answers
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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 $$ \frac{d^2}{dx^2}\left[\log P(x)\right]=-\frac{1}{\sigma^2}<0. $$
answered Jun 1, 2015 at 3:45
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$\begingroup$ So, It will becomes convex function in log space, Right? $\endgroup$– JameJun 1, 2015 at 3:51
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1$\begingroup$ Yes. It has a negative second derivative. $\endgroup$ Jun 1, 2015 at 3:52
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3$\begingroup$ By standard nomenclature, its log is concave, not convex. $\endgroup$– p.s.Jun 2, 2015 at 0:01
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The Gaussian density function is quasiconcave but not concave. Moreover, it is log-concave because log P(x) is essentially a negative quadratic function.
