so if the PDF attains its maximum at 0 it means that $f'(0) = 0$:

$$f'(x) = -\frac{x}{\sigma^2 \sqrt{\pi}}\exp({-\frac{(x - \mu)^2}{2\sigma^2}}) $$

$$f'(0) = 0 \iff 0 =0$$

yep, no valuable information whatsoever,

now onto the second derivative criterion : $f''(0) < 0$

$$f''(0) = 0 -\frac{1}{\sigma^2 \sqrt{\pi}}\exp({-\frac{\mu^2}{2\sigma^2}}) < 0$$

again, no valuable information whatsoever.

maybe if we try using the fact that $f(0) \geq f(y), \, \forall y \in \mathbb{R}$ for $y = \mu$

then :

$$\frac{1}{\sigma\sqrt{2 \pi}}\exp(-\frac{\mu^2}{2\sigma^2}) \geq\frac{1}{\sigma\sqrt{2 \pi}}\iff \exp(-\frac{\mu^2}{2\sigma^2}) \geq1 \iff -\frac{\mu^2}{2\sigma^2} \geq 0 \iff \mu^2 \leq 0$$

therefore $\mu = 0$ since $0$ is the only 'negative' positive number.

I'm pretty confident that my work is correct but I'd like a confirmation, thanks !

  • 6
    $\begingroup$ For the first derivative, try a more careful use of the chain rule. $\endgroup$
    – BruceET
    Commented Oct 21, 2019 at 22:35

3 Answers 3


We have

$$\begin{align*} f(x\mid\mu,\sigma^2) &=\frac{1}{\sqrt{2\pi\sigma^2}}\text{exp}\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\\\\ \end{align*}$$

Taking the log we get


Setting the derivative equal to zero we get

$$\begin{align*} \frac{d}{dx}\text{log }f(x\mid\mu,\sigma^2) &=-\frac{(x-\mu)}{\sigma^2}\\\\ &=0\\\\ &\Rightarrow x=\mu\\\\ \end{align*}$$

so the mean and mode are the same for a normal distribution. Hence $\mu=0$.

For completeness, since

$$\frac{d^2}{dx^2}\text{log }f(x\mid\mu,\sigma^2)=-\frac{1}{\sigma^2}<0$$

for all $x\in\mathbb{R}$ then we can conclude that this function is concave with a global maximum at $x=\mu$.


Assuming $\sigma>0$, the derivative of $$f(x)=\frac{1}{\sqrt{2\pi}\sigma} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$$ is not quite what you wrote but

$$f'(x)=-(x-\mu)\frac{1}{\sqrt{2\pi}\sigma^3} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$$

so $f'(x)=0$ when and only when $x-\mu=0$, i.e. $x=\mu$, as the other terms are all positive.

Since you have a probability density function which is differentiable on the whole of $\mathbb R$, this unique zero means that the mode of the distribution must be at $x=\mu$. You could check the second derivative if you wish, but that is not necessary

  • 1
    $\begingroup$ Do you mean $\mu$ instead of $m$ in the exponent? $\endgroup$
    – mai
    Commented Oct 22, 2019 at 18:27
  • $\begingroup$ @MichaelChen yes - thank you $\endgroup$
    – Henry
    Commented Oct 22, 2019 at 19:57

The mode of any symmetrical, unimodal distribution is always equal to the mean, and also to the median of that distribution. A unimodal distribution is one with only one peak. All normal distributions fall in that category, so their mode and mean are identical.

  • $\begingroup$ Is there anything wrong in my statement? Isn't it obvious, not requiring any proof? $\endgroup$ Commented Oct 22, 2019 at 20:43
  • 4
    $\begingroup$ How do you know that a normal distribution is symmetrical and unimodal without checking? Even then, a Cauchy distribution is symmetrical and unimodal but you cannot say its mode is equal to its mean $\endgroup$
    – Henry
    Commented Oct 22, 2019 at 23:02
  • $\begingroup$ @Henry: For the same reason I know 2 + 3 = 5. If I had to prove why $2 + 3 = 5$, I would be unable to do it. These are things that are just so rudimentary that any proof will be less obvious than the fact itself. You could argue that the mean of a Cauchy distribution does not exist, but that's splitting hair and too advanced for a novice in statistics. It just depends what you mean by "mean", whether you use a Lebesgue measure to compute it, or more advanced stuff, which for the layman is of no use $\endgroup$ Commented Oct 23, 2019 at 0:53
  • 1
    $\begingroup$ Henry's objection is not an esoteric argument about angels dancing on pins, but a serious point with practical consequences; for example, as a consequence the Cauchy and other distributions lacking a finite mean (but some of which arise as models for real data, for example in physics) don't obey the law of large numbers -- their sample means don't converge, not least because they don't have a mean to converge to. They're just as "real" as the normal distribution, and they're symmetric unimodal distributions for which the mean is quite literally not equal to the mode. $\endgroup$
    – Glen_b
    Commented Oct 23, 2019 at 5:01
  • 1
    $\begingroup$ @Henry: I fail to see the argument. The question's title is "if the mode of a normal distribution is 0, then what's the value of the mean". If a question assumed knowledge of the normal distribution, but not of its symmetry (which is pretty much the first thing you learn about the normal, and is also immediately obvious from the density the OP starts off), then that's a very strange question indeed. Without further information from the OP, e.g., that this is self study and only certain things should be assumed known about the normal, Vincent's answer is absolutely the best one. +1. $\endgroup$ Commented Oct 23, 2019 at 6:11

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