I'm trying to show that the mode (i.e. the maximum) of a Gaussian distribution is given by its mean.
Consider that the Gaussian distribution is given by Equation 1: $$N(x|\mu, \sigma^2)=\frac{1}{(2\pi\sigma^2)^{1/2}}exp(-{\frac{1}{2\sigma^2}}(x-\mu)^2)$$
where $\mu$ is the mean and $\sigma^2$ is the variance.
To solve the problem, I know that I should differentiate Equation 1 and set it to zero. Doing so, I can go as far as:
$$exp(-\frac{1}{2\sigma^2}(x-\mu)^2) \times (-\frac{1}{\sigma^2}(x-\mu))\frac{}{} \times \frac{1}{(2\pi\sigma^2)^{1/2}} = 0$$
My question is: how do I solve this equation?
Intuitively, I would say that we must have $x=\mu$ to make one of the terms zero and then, since all the terms are multiplying, we know that $x=\mu$ is a solution of the problem. However, I would like to know how to solve the equation analytically.
Best Answer
It's not always wise to jump straight into taking derivatives (indeed there are a number of widely used densities where this will be useless, or if you're sufficiently rash in going about it, may lead you to select values that are not the mode).
Let's think a little first. That often pays handsome dividends.
The location of its mode will be given by the location of the mode of its log -- the height is changed but not which arguments correspond to the largest value. (Strictly monotonic transformation of the density doesn't alter where the modes are)
note further that shifting the function up or down by additive constants and scaling by positive multiplicative constants don't alter the location of a mode
So in the case of the normal, after we take logs, then drop the additive and multiplicative constants, it suffices to find modes of the function $h(x) = -(x-\mu)^2$
Since this is the negative of a perfect square we can identify the (now plainly) unique mode by inspection, though if you feel compelled to use calculus at this stage it's quite straightforward.
Indeed, as whuber notes in comments, we can simply recognize that ($x-\mu$) is shifting the location of the function right by $μ$. That shifts its mode by $μ$, reducing the problem to finding the vertex of the parabola $h(x)=−x^2$, whose location is trivially implied by the fact that $x^2$ is non-negative (& only zero at $x=0$). [In more complicated problems, if all occurrences of $x$ are in the form $(x-\kappa)$ for some $\kappa$, consider $x^*=x-\kappa$, find the mode of the new function of $x^*$ ($x_0^*$, say) and then the original mode is at $x_0=x_0^*+\kappa$.]
Such simple notions often greatly simplify finding the optimum.