Expectation of maximum element of a vector whose expectation is zero Informal statement:
In short what I am asking is that we have a 2d vector whose elements are drawn i.i.d from $\mathcal{N}(0, 1)$. Then, we pick the maximum of the two. Once the index of the maximum is specified, we multiply it by an element of another vector $a$ whose index is the same as the index of maximum. Is the expectation of the new random variable zero?
Formal statement:
Let $\mathbf{b} \in \mathbb{R}^2$ where $b_1,b_2 \sim \mathcal{N}(0, 1)$ and $\mathbf{a} \in \mathbb{R}^2$ where $a_1 \neq0$ and $a_2 \neq0$. Also, let $i^{*} \in \arg\max_{i=1,2} b_i$ (both $b_1$ and $b_2$ may be same that is why we have $\in$ not $=$) and $X=a_{i^*}b_{i^*}$? Is $\mathbb{E}_{\mathbf{b}}[X]=\mathbb{E}_{\mathbf{b}}[a_{i^*(\mathbf{b})}b_{i^*(\mathbf{b})}]$?
My try:
Although, the index of maximum, i.e., $i^*$, is a function of $\mathbf{b}$, it is either 1 or 2. This $i^*$ is a discrete random variable. As soon as we know $i^*$, we know the mean corresponding $b_{i^*}$ is zero and no matter what multiplier gets multiplied to it, the expectation of the product is zero. This is my intuition. However, I am looking for a rigorous proof not an intuition. I think I need to consider some distribution on the indices and then convert the expectation to two expectations and solve it.
Note:
I am looking for a very precise and rigorous proof.
 A: What you are getting at is the order statistics of a random variable. That is, given $n$ i.i.d. random variables, what is the distribution of the $k$-th largest value among them? It will typically not be the same as the original distribution, which makes sense, e.g. the first order statistic should be biased to the left.
The normal distribution does not have nice closed formulas for its order statistics as far as I know, but the second order statistic among two i.i.d. normal random variables definitely has expectation greater than $0$. To see this formally, let $X_1, X_2 \sim \mathcal{N}(0, 1)$, and let $X_{(1)}, X_{(2)}$ be the first and second order statistics respectively. Then $E(X_{(1)} + X_{(2)}) = E(X_1 + X_2) = 0$. However, $X_{(2)} > X_{(1)}$ with probability $1$, so it follows that $E(X_{(2)}) > E(X_{(1)})$. Since these expectations must add to zero, we must have $E(X_{(2)})$ positive and $E(X_{(1)})$ negative, as desired.
A: Let $Z=i^{*}$. Then
$$ E[X] = E [ E [X | Z]]=P(Z=1)E[X|Z=1]+P(Z=2)E[X|Z=2]
$$
Now, $E[X|Z=1]= a_1 E[b_1 | b_1 > b_2]$
The expectation of the maximum of two standard iid normal is not trivial:
$$\frac{1}{\sqrt{\pi}} = 0.5642\cdots$$
Then
$$E[X] = \frac12 a_1 \frac{1}{\sqrt{\pi}} + \frac12 a_2  \frac{1}{\sqrt{\pi}} = \frac{1}{\sqrt{\pi}} ( a_1 + a_2)$$
