Let $x\_1,\dots,x\_n$ be i.i.d. drawn from $N(0,I\_{p\times p})$. Consider the sample covariance matrix $W(n,p)=\frac 1n \sum\_{i=1}^n x\_ix\_i^T$, a Wishart matrix.


For fixed $n,p$, what is the expected spectral norm of $W(n,p)$, $$\mathbb E \left[ \left\|\frac 1n \sum\_{i=1}^n x\_ix\_i^T\right\|\_2\right ] ?$$


I'm familiar with the asymptotic Bai/Silverstein result, but not sure whether that can be exploited in the finite case. If the exact expression is unwieldy, a reasonable upper bound may suffice.



Expected value of the spectral norm of a Wishart matrix?
I don't think there's an exact expression, but the Bai–Yin result does give the right prediction. It's a little easier to state nice-looking results for a $p \times n$ matrix $X$ with independent standard Gaussian entries, so that what you're asking about is $n^{-1} \mathbb{E} \| X \|\_2^2$. The nicest result is that
$$
\| X \|\_2 \le \sqrt{n} + \sqrt{p} + t
$$
with probability at least $1 - 2 e^{-t^2/2}$ (see [Theorem II.13 of this paper](http://www.cwru.edu/artsci/math/szarek/TeX/DavSzaHB.pdf) or [Theorem 5.35 of this paper](https://arxiv.org/abs/1011.3027)), which implies in particular that
$$
\mathbb E \left[ \left\|\frac 1n \sum\_{i=1}^n x\_ix\_i^T\right\|\_2\right ] 
\le \left(1 + \sqrt{\frac{p}{n}}\right)^2 + \frac{4}{n} e^{-(\sqrt{n}+\sqrt{p})^2/2}
$$
assuming I did the integrals right. I don't recall offhand a similarly precise result for the lower bound, though there may be one out there. Chances are, though, that the upper bound is what you really need.


