# derivation of a gap related to extreme value theory

I have an expression to evaluate as follow:

$\mathbb{E}\left[\sum_{k=1}^K s_k f(x_k)\Big|s_k=s_k^{\ast} \right]$

where $\{s_k^\ast\}$ can be treated as a ${policy}$ which is defined as follows:

\begin{align} {s}_k^\ast=\left\{ \begin{aligned} &1, \quad \text{if } X_k=\max_j \{X_j\}>0 \\ &0, \quad otherwise \end{aligned}\right. \end{align} with \begin{align} X_k=a_k\left[\log (a_k x_k) \right]^+ -\Big(a_k-\frac{1}{x_k}\Big)^+ \end{align} where $a_k>0$ for all $k$ are constants, and $x_k>0$ is a random variable following $\exp(1)$ distribution. Let me define another simple policy: \begin{align} \widetilde{s}_k^\ast=\left\{ \begin{aligned} &1, \quad \text{if } x_k=\max_j \{x_j\} \\ &0, \quad otherwise \end{aligned}\right. \end{align}

According to the extreme value theory, we know the growth rate result: $x_{k^\ast}=\mathcal{O}(\log K)$, where $k^\ast=\arg\max_k \{x_k\}$, and $K$ is the number of random variables. Based on this result and the definitions of ${s}_k^\ast$ and $\widetilde{s}_k^\ast$, we know that \begin{align} {s}_k^\ast \rightarrow \widetilde{s}_k^\ast, \quad as \ K \rightarrow \infty \end{align} Therefore, we must have \begin{align} \mathbb{E}\left[\sum_{k=1}^K s_k f(x_k)\Big|s_k=s_k^{\ast} \right] \rightarrow \mathbb{E}\left[\sum_{k=1}^K s_k f(x_k)\Big|s_k=\widetilde{s}_k^{\ast} \right], \quad as \ K \rightarrow \infty \end{align} My question is whether we can quantify the gap w.r.t. $K$, i.e., \begin{align} \left|\mathbb{E}\left[\sum_{k=1}^K s_k f(x_k)\Big|s_k=s_k^{\ast} \right] -\mathbb{E}\left[\sum_{k=1}^K s_k f(x_k)\Big|s_k=\widetilde{s}_k^{\ast} \right]\right| \end{align} decrease at what order w.r.t. $K$ as $K$ increases?

• You need conditions on f to guarantee convergence. Please specify them. – John Jiang Mar 15 '16 at 17:45
• actually, in my case $f(x)=max[1-1/x,0]$. – Michael Fan Zhang Mar 16 '16 at 1:44
• OK. But I am confused as to what $s_k$ is, without the asterisk? – John Jiang Mar 17 '16 at 15:09
• Is the condition in the expectation a set of conditions for all k? In that case the expectation is simply E(f(x_k*)), where k* is the random index that of the largest x. This is just evaluating the expectation of some max of iid exp(1), which is pretty trivial. – John Jiang Mar 17 '16 at 15:12