For a sample of size $m$ from a Normally distributed random variable with paramters $\mu$ and $\sigma$, the sample variance, which is the sum of the squared deviations from the mean over the number of samples, the estimator $\mathbb{E}[\hat{\sigma'}^2_m] = \mathbb{E}\left[\frac{1}{m}\sum_{i=1}^m \left(x^{(i)} - \hat{\mu}_m\right)^2\right]$ Is a biased estimator for the variance $\sigma^2$.

To see why, consider that the sample variance, $\mathbb{E}[\hat{\sigma}^2_m] =\mathbb{E}\left[\frac{1}{m - 1}\sum_{i=1}^m \left(x^{(i)} - \hat{\mu}_m\right)^2\right]$ is an unbiased estimator for the variance $\sigma^2$. Then $\textrm{bias}(\hat{\sigma}_m^2 = \mathbb{E}\left[\frac{1}{m - 1}\sum_{i=1}^m \left(x^{(i)} - \hat{\mu}_m\right)^2\right] - \sigma^2 = 0$, so $\mathbb{E}(\hat{\sigma}_m^2) = \mathbb{E}\left[\frac{1}{m - 1}\sum_{i=1}^m \left(x^{(i)} - \hat{\mu}_m\right)^2\right] = \sigma^2 \not = \mathbb{E}\left[\frac{1}{m}\sum_{i=1}^m \left(x^{(i)} - \hat{\mu}_m\right)^2\right] = \mathbb{E}(\hat{\sigma'}_m^2)$, since $m-1 \not = m$.