# Expected norm of sum of random orthogonal matrices

Somehow I got wondering about the following question today:

Suppose $Q_1,\ldots,Q_n$ are random (uniformly sampled) $d \times d$ orthogonal matrices.

What is the expected value of the quantity $\|\sum_i Q_i\|$?

Additionally, suppose I actually generate random skew-symmetric matrices $S_1,\ldots,S_n$, and then obtain corresponding orthogonal matrices via the matrix exponential, $e^{S_i}$.

What is the expected value of the quantity $\|\sum_i e^{S_i}\|$

EDIT: From the comments (and from Mikael's answer) it seems like this is a tough question. But already the case with large $d$ is quite useful.

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Are you interested in an asymptotic regime (e.g. n fixed, d large; or d fixed, n large)? I doubt there will be an exact formula otherwise; the operator norm is not algebraic enough for an algebraic miracle (as opposed to, say, the Frobenius norm). In the regime where n is fixed and d is large, free probability tools should in principle answer your question (or at least predict the answer). – Terry Tao Nov 5 '11 at 19:47
Conversely, in the regime where d is fixed and n is large, the classical central limit theorem should give good results, since the sum is going to be asymptotically a gaussian matrix (possibly shifted by a multiple of the identity, in the second question). – Terry Tao Nov 5 '11 at 19:49
@Terry: Ideally, non asymptotic results would be the best. I am, however, still curious about the limited asymptotic regimes where only one of $n$ or $d$ is allowed to be large (feels like the case of large $n$ should be easier). The Frobenius norm definitely seems much friendlier. – Suvrit Nov 5 '11 at 19:52

EDIT: My answer only deals with the $d \to \infty$ regime.
This question is not too naive (or at least the answer is hard). I am almost sure that for fixed $d$ there is no exact formula. For the limit as $d \to \infty$ I think that one expects that the norm of $\sum_1^n Q_i$ almost surely converges to $2 \sqrt{n-1}$, but I don't know if a proof exists yet (my guess is that everything works the same way as for unitaries, see below).
Second edit: I deleted a remark where I said that the norm of $\sum_1^N e^{S_i}$ is of order $\sqrt n$. This would be true if $E(e^{S_i})$ was $0$, which is not the case as Terry Tao points out in his comments to your question. In fact it is the norm of $\sum (e^{S_i} - E(e^{S_i}))$ which is of order $\sqrt n$.