TITLE: Maximum of two normal random variables
QUESTION [14 upvotes]: The main purpose of the following question is to get some intuition and deeper understanding why the presented method works which would hopefully help me in trying to adapt it to the setting I am dealing with in my research.
Let $X,Y\sim N(0,1)$, not necessarily independent. Suppose we want to find an upper bound for $\mathbb{E}\max(X,Y)$.
The most obvious approach would be something like the following
$$\mathbb{E}\max(X,Y)\leq \mathbb{E}|X|+\mathbb{E}|Y|=2\sqrt{2/\pi}\approx 1.59$$
However, I've found the trick in the literature that uses Laplace transform to get something better. Although the idea is much less obvious, details are still easy. For any $\lambda >0$, Jensen's inequality gives us the following
$$\mathbb{E}\max(X,Y) \leq \frac1{\lambda}\log\left(\mathbb{E}e^{\lambda\max(X,Y)}\right)\leq \frac1{\lambda}\log\left(\mathbb{E}e^{\lambda X}+\mathbb{E}e^{\lambda Y}\right) = \frac{\log(2e^{\lambda^2/2})}{\lambda}.$$
Minimizing this gives us the upper bound $\log(4) \approx 1.17$, which is better than the previous approach.
Now, my question is, heuristically/intuitively, why is second method better? Or to put in a different way, is there some easy way to see that the second method should give a better bound even before doing the actual calculations that confirm this?
At this stage, I don't have any intuition for why this works, and I am certainly not fine with that's the standard trick researchers in the field use.

REPLY [4 votes]: Given: $X \sim N(0,1)$ and $Y \sim N(0,1)$ where $X$ and $Y$ may be correlated.

If we assume that the underlying Normals are jointly Normal, then (a) the exact answer is quite simple, and (b) we can do much better, conditioning the maximum bound on the correlation coefficient. In particular, if $(X,Y)$ are bivariate Normal with correlation $\rho$, then the joint pdf $f(x,y)$ is:

Then, $E\big[Max[X,Y]\big]$ is:

where I am using the Expect function from the mathStatica package for Mathematica to automate the nitty gritties. Plainly, the expected maximum is contingent on $\rho$, as a quick plot illustrates:

As @oferzeitouni noted, the maximum possible value is $\sqrt{\frac{2}{\pi}}$, which is attained when $\rho = -1$.