# Does $E[\hat \Sigma^{-1}] \to \Sigma^{-1}$ still hold for samples drawn from a non-normal population?

For a sample of observations $$\{x_i\}_{i=1}^n$$ where $$x_i=(x_{i1},\dots,x_{ik})^T$$ of a population random vector $$X=(X_1,\dots,X_k)^T$$, the population covariance is $$\Sigma = E[(X-E[X])(X-E[X])^T],$$ and the sample covariance is $$\hat \Sigma = \frac{1}{n} \sum_{i=1}^n (x_i-\bar{x})(x_i-\bar{x})^T.$$

Now, while $$\hat \Sigma$$ is an unbiased estimator of $$\Sigma$$, the same is not true for the inverse. If the samples are drawn from a multivariate normal distribution it can be shown that $$E[\hat \Sigma^{-1}] = \frac{n}{n-k-2}\Sigma^{-1}.$$ So the sample covariance is biased, but at least as $$n$$ gets large the bias will decrease and thus $$\lim_{n \to \infty }E[\hat \Sigma^{-1}] \to \Sigma^{-1}.$$

Now, what about the case when the samples are drawn from a different continuous distribution. Is there anything concrete we can say about $$\hat \Sigma^{-1}$$ as an estimator of $$\Sigma^{-1}$$?

In particular, will $$E[\hat \Sigma^{-1}] \to \Sigma^{-1}$$ still hold? If it doesn't hold in general, for what class of random variables does it still hold? Or does it really only hold in the case of a multivariate normal distribution?

• At a minimum, you need to assume the distribution is continuous, for otherwise $E[\hat\Sigma^{-1}]$ does not exist. (There is a nonzero chance that $\hat \Sigma$ is singular.)
– whuber
Commented Mar 19, 2021 at 13:28
• Ok I added that condition in. Commented Mar 19, 2021 at 13:29