Gaussian width of sparse balls The Gaussian width of a set $T\subset \mathbb{R}^n$ is defined as,
$$
G(T) = E\left[\sup_{\theta \in T} \sum_{i=1}^n \theta_i W_i\right],
$$
where, $\mathbf{W}=(W_1,\ldots,W_n)$ is a sequence of i.i.d. $N(0,1)$ random variables. I am interested in finding the value of $G(T)$ for 
$$
T(s) \equiv \{\theta\in\mathbb{R}^n: {\|\theta\|}_0 \leq s,{\|\theta\|}_2\leq 1\},
$$
the set of all $s$-sparse vectors within the unit ball, with $s\in\{1,\ldots,n\}$. This is an exercise problem in Wainwright's book on HD-Statistics. I have been able to show,
$$
G(T(s)) = E\max_{|S|=s} {\|\mathbf{W}_S\|}_2,
$$
and $S$ is a subset of $\{1,\ldots,n\}$, with cardinality $|S| = s$. Here the subscript $S$ denotes the components of $\mathbf{W}$ corresponding to $S$.
Then, using Gaussian concentration inequality and the union bound, I can get,
$$
P\left(\max_{|S|=s}{\|\mathbf{W}_S\|}_2 \geq \sqrt{s} + t\right)\leq \binom{n}{s} \exp\{-t^2/2\},\ \text{for all $t>0$.}
$$
I can use the bound,
$$
\binom{n}{s}\leq {\left(ne/s\right)}^s, \ \text{for all $s=1,\ldots,n$.}
$$
Finally, I need to integrate to obtain the bound on the expectation. I am unable to do it to get the desired upper bound (of the order),
$$
K\sqrt{s\log(en/s)},\ \text{where $K$ is some constant.}
$$
Any ideas would be helpful!
 A: Notation: $C$ below denotes (possibly different) absolute constants.
Recall that for $N$ sub-gaussian variable $X_i$ (independence not required) with $\max_i \| X_i\|_{\psi_2}\le K$, $E \max_{i\le N} X_i \le CK \sqrt{\log{N}}.$
For our problem, max in $E \max_{|S| \le s} |W_S|$ enumerates over $N:=\sum_{k=1}^s \binom{n}{k}$ different subsets of ${1,\dots,n}$. Also use Gaussian concentration inequality we obtain $\max_{|S|\le s} \| W_S-\sqrt{|S|}\|_{\psi_2}\le C$.
So we have $$E \max_{|S|\le s} (|W_S|-\sqrt{|S|}) \le C \sqrt{\log(\sum_{k=1}^s \binom{n}{k})} \le C \sqrt{s\log(en/s)}$$
where we used $\sum_{k=1}^s \binom{n}{k} \le (\frac{ne}{s})^s.$
Finally, this implies that (using $\sqrt{|S|}\le \sqrt{s}$ and move $\sqrt{s}$ to RHS)
$$E \max_{|S| \le s} W_S \le \sqrt{s}+C\sqrt{s\log(en/s)}\le C \sqrt{s\log(en/s)}$$
by $n>s$.
A: After some effort I obtained the following. Please comment. 
Let $T = \max_{|S|=s}{\|\mathbf{W}_S\|}_2$. Then, $P(T>\sqrt{s}+ t)\leq {(ne/s)}^s \exp\{-t^2/2\}$, for all $t>0$. Then, for some $a>\sqrt{s}$ (to be chosen later),
\begin{eqnarray*}
E(T) & = &\int_0^a P(T\geq t)~dt + \int_a^\infty P(T\geq t)~dt \\
     & \leq &  a +{(ne/s)}^s \sqrt{2\pi}\cdot P(Z > a-\sqrt{s}),\ \text{where, $Z\sim N(0,1)$,}\\
& \leq & a +{(ne/s)}^s \sqrt{2\pi}\cdot \exp\{-{(a-\sqrt{s})}^2/2\},\ \text{using a Subgaussian one-sided tail bound,}\\
& = & \frac{1}{2}\left[2e^{\log{(a)}} + 2\exp\left\{s\log{(ed/s)}+\log{\sqrt{2\pi}}-\frac{1}{2}{(a-\sqrt{s})}^2\right\}\right].
\end{eqnarray*}
This provides an upper bound on $E(T)$ and we will minimize this upper bound in terms of $a$. Using AM-GM inequality, this bound is minimized if the terms are equal. This is same as requiring,
$$
a = s\log{(ed/s)}+\log{\sqrt{2\pi}}-\frac{1}{2}{(a-\sqrt{s})}^2.
$$
With some additional work by re-writing the above equation, this is equivalent to requiring,
\begin{eqnarray*}
a &=& \sqrt{s}+\sqrt{2s\log{(ed/s)}}\cdot{\left[1+\frac{1+2\log{\sqrt{2\pi}}}{2s\log{(ed/s)}}-\frac{1}{2\sqrt{s}\log{(ed/s)}}\right]}^{1/2} - 1 \\
&\leq & K\sqrt{s\log{(ed/s)}},
\end{eqnarray*}
for some sufficiently large $K>0$, when $s$ is large enough. Since $a$ is the desired minimized upper bound on $E(T)$, the claim follows. 
Earlier, I had used the sharper Mills inequality tail bound instead of the sub-Gaussian bound and ended up with a very complicated expression.
