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proof_id |
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true |
Joram Soch |
BCCN Berlin |
joram.soch@bccn-berlin.de |
2020-08-25 14:11:00 -0700 |
Extreme points of the probability density function of the normal distribution |
Probability Distributions |
Univariate continuous distributions |
Normal distribution |
Extreme points |
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Wikipedia |
2021 |
Normal distribution |
Wikipedia, the free encyclopedia |
retrieved on 2021-08-25 |
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P251 |
norm-extr |
JoramSoch |
Theorem: The probability density function of the normal distribution with mean $\mu$ and variance $\sigma^2$ has a maximum at $x = \mu$ and no other extrema. Consequently, the normal distribution is a unimodal probability distribution.
Proof: The probability density function of the normal distribution is:
$$ \label{eq:norm-pdf}
f_X(x) = \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] ; .
$$
The first two deriatives of this function are:
$$ \label{eq:norm-pdf-der1}
f'_X(x) = \frac{\mathrm{d}f_X(x)}{\mathrm{d}x} = \frac{1}{\sqrt{2 \pi} \sigma^3} \cdot (-x + \mu) \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right]
$$
$$ \label{eq:norm-pdf-der2}
f''_X(x) = \frac{\mathrm{d}^2f_X(x)}{\mathrm{d}x^2} = -\frac{1}{\sqrt{2 \pi} \sigma^3} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] + \frac{1}{\sqrt{2 \pi} \sigma^5} \cdot (-x + \mu)^2 \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] ; .
$$
The first derivative is zero, if and only if
$$ \label{eq:norm-pdf-der1-zero}
-x + \mu = 0 \quad \Leftrightarrow \quad x = \mu ; .
$$
Since the second derivative is negative at this value
$$ \label{eq:norm-pdf-der2-extr}
f''_X(\mu) = -\frac{1}{\sqrt{2 \pi} \sigma^3} < 0 ; ,
$$
there is a maximum at $x = \mu$. From \eqref{eq:norm-pdf-der1}, it can be seen that $f'_X(x)$ is positive for $x < \mu$ and negative for $x > \mu$. Thus, there are no further extrema and $\mathcal{N}(\mu, \sigma^2)$ is unimodal.