Starting from the log-likelihood function for a Gaussian,

$$ \mathcal{L}\mathcal{L} = - \frac{N}{2} \log{\left(2 \pi\sigma^2\right)} - \frac{1}{2 \sigma^2} \sum_{i=1}^{N}{{\left(x_i - \mu\right)}^2} \tag{1} $$

if we assume the population to be well-represented by the sample distribution such that $ \sigma^2 = \frac{\sum_{i=1}^{N}{{\left(x_i - \mu\right)}^2}}{N} $ , can we simplify the term on the right to $\frac{N}{2}$, resulting in:

$$ \mathcal{L}\mathcal{L} = -\frac{N}{2} \left(\log{\left(2 \pi\sigma^2\right)} + 1 \right) \tag{2} $$

  • 1
    $\begingroup$ What exactly are you trying to do here? $\endgroup$
    – Tim
    Dec 12, 2021 at 13:49

1 Answer 1


Not quite. However, you can set $\hat\mu=\bar x$, and $$\hat \sigma^2 = \frac{1}{N}\sum_i (x_i-\hat\mu)^2$$ to get the profile likelihood

$$\ell = -\frac{N}{2}\left(\log(2\pi\hat\sigma^2)+1\right)$$

This is useful for comparing different models. A similar approach is useful in fitting mixed models, where the profile likelihood with $\mu$ and $\sigma$ profiled out still depends on the other variance parameters.

You can't write $$\ell(\sigma^2) = -\frac{N}{2}\left(\log(2\pi\sigma^2)+1\right)$$ because that's only a good approximation if $\sigma^2\approx\hat\sigma^2$; it's not a good approximation as $\sigma^2$ varies.


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