covariance matrix is not a good estimator of its population counterpart \[@Marcenko:18, @Karoui1:16\]. To illustrate this point, consider $\Sigma_p=I_p$, so all the eigenvalues are $1$. A result from \[@Geman:12\] shows that if entries of $X_i$’s are i.i.d (let $X_i$’s have mean zero and variance 1) with a finite fourth moment and if $p/n \rightarrow \theta <1 $, then the largest sample eigenvalue $l_1$ satisfies: $$\begin{aligned}
l_1~ \rightarrow ~(1+\sqrt{\theta})^2, ~~~~~~ a.s\end{aligned}$$ This suggests that $l_1$ is not a consistent estimator of the largest eigenvalue $\sigma_1$ of population covariance matrix. In particular if $n=p$ then $l_1$ tends to $4$ whereas $\sigma_1$ is $1$. This is also evident in the eigenvalue plot in Figure 2.1. The distribution of $l_1$ also depends on the underlying structure of the true covariance matrix. From Figure 2.1, it is evident that the smaller sample eigenvalues tend to underestimate the true eigenvalues for large $p$ and small $n$. For more discussion on this topic, see @Karoui1:16.

To correct for this bias, a natural choice would be to shrink the sample eigenvalues towards some suitable constant to reduce the over-dispersion. For instance, @Stein:28 proposed an estimator of the form $\tilde{\Sigma}=\tilde{U} \Lambda (\tilde{\lambda}) \tilde{U}$, where $\Lambda (\tilde{\lambda})$ is a diagonal 