Index: The Book of Statistical ProofsStatistical Models ▷ Univariate normal data ▷ Univariate Gaussian with known variance ▷ Maximum likelihood estimation

Theorem: Let there be univariate Gaussian data with known variance $y = \left\lbrace y_1, \ldots, y_n \right\rbrace$:

$\label{eq:ugkv} y_i \sim \mathcal{N}(\mu, \sigma^2), \quad i = 1, \ldots, n \; .$

Then, the maximum likelihood estimate for the mean $\mu$ is given by

$\label{eq:ugkv-MLE} \hat{\mu} = \bar{y}$

where $\bar{y}$ is the sample mean

$\label{eq:y-mean} \bar{y} = \frac{1}{n} \sum_{i=1}^n y_i \; .$

Proof: The likelihood function for each observation is given by the probability density function of the normal distribution

$\label{eq:ugkv-yi} p(y_i|\mu) = \mathcal{N}(x; \mu, \sigma^2) = \frac{1}{\sqrt{2 \pi \sigma^2}} \cdot \exp \left[ -\frac{1}{2} \left( \frac{y_i-\mu}{\sigma} \right)^2 \right]$

and because observations are independent, the likelihood function for all observations is the product of the individual ones:

$\label{eq:ugkv-LF-s1} p(y|\mu) = \prod_{i=1}^n p(y_i|\mu) = \sqrt{ \frac{1}{(2 \pi \sigma^2)^n} } \cdot \exp \left[ -\frac{1}{2} \sum_{i=1}^{n} \left( \frac{y_i-\mu}{\sigma} \right)^2 \right] \; .$

This can be developed into

$\label{eq:ugkv-LF-s2} \begin{split} p(y|\mu) &= \left( \frac{1}{2 \pi \sigma^2} \right)^{n/2} \cdot \exp \left[ -\frac{1}{2} \sum_{i=1}^{n} \left( \frac{y_i^2 - 2 y_i \mu + \mu^2}{\sigma^2} \right) \right] \\ &= \left( \frac{1}{2 \pi \sigma^2} \right)^{n/2} \cdot \exp \left[ -\frac{1}{2 \sigma^2} \left( y^\mathrm{T} y - 2 n \bar{y} \mu + n \mu^2 \right) \right] \end{split}$

where $\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i$ is the mean of data points and $y^\mathrm{T} y = \sum_{i=1}^{n} y_i^2$ is the sum of squared data points.

Thus, the log-likelihood function is

$\label{eq:ugkv-LL} \mathrm{LL}(\mu) = \log p(y|\mu) = -\frac{n}{2} \log (2 \pi \sigma^2) - \frac{1}{2 \sigma^2} \left( y^\mathrm{T} y - 2 n \bar{y} \mu + n \mu^2 \right) \; .$

The derivatives of the log-likelihood with respect to $\mu$ are

$\label{eq:ugkv-dLLdl-d2LLdl2} \begin{split} \frac{\mathrm{d}\mathrm{LL}(\mu)}{\mathrm{d}\mu} &= \frac{n \bar{y}}{\sigma^2} - \frac{n \mu}{\sigma^2} = \frac{n}{\sigma^2} (\bar{y}-\mu) \\ \frac{\mathrm{d}^2\mathrm{LL}(\mu)}{\mathrm{d}\mu^2} &= - \frac{n}{\sigma^2} \; . \\ \end{split}$

Setting the first derivative to zero, we obtain:

$\label{eq:ugkv-dLLdl} \begin{split} \frac{\mathrm{d}\mathrm{LL}(\hat{\mu})}{\mathrm{d}\mu} &= 0 \\ 0 &= \frac{n}{\sigma^2} (\bar{y}-\hat{\mu}) \\ 0 &= \bar{y}-\hat{\mu} \\ \hat{\mu} &= \bar{y} \\ \end{split}$

Plugging this value into the second derivative, we confirm:

$\label{eq:ugkv-d2LLdl2} \frac{\mathrm{d}^2\mathrm{LL}(\hat{\mu})}{\mathrm{d}\mu^2} = -\frac{n}{\sigma^2} < 0 \; .$

This demonstrates that the estimate $\hat{\mu} = \bar{y}$ maximizes the likelihood $p(y \vert \mu)$.

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Metadata: ID: P207 | shortcut: ugkv-mle | author: JoramSoch | date: 2021-03-24, 03:48.