Maximum of two normal random variables 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.
 A: What should be really done is as follows. Suppose that $X_1,\dots,X_n$ are any real random variables for which individual distribution laws are known but no information is given on the joint distribution law. Choose $t$ such that $\sum_jP(X_j>t)\le 1\le \sum_jP(X_j\ge t)$. Then $E\max(X_j)\le t+\sum_jE(X_j-t)_+$ and the bound is the best possible.
I wonder when we will finally realize that if somebody has trouble finding the antiderivative of $\frac{x+1}{x^2\sqrt{x^2+4}}$, this will, most likely, never affect his work in mathematics, but if somebody doesn't see things like this one, his education has been thoroughly screwed up. When shall we stop teaching calculus and start teaching analysis at least at the most basic level?
I apologize for the rant, but there is a limit to everything. I do not blame the students, but I'd like everyone who comes to the class with a book by Stewart in hand think of this thread a bit before coming to the blackboard.
A: 
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 $\rho = -1$.
A: First, an upper bound that beats your second bound is the following: use the equality 
$$\max(a,b)=(a+b+|a-b|)/2.$$
Then
$$E\max(X,Y)=E|X-Y|/2\leq \frac{1}{2} (E|X|+E|Y|)=E|X|=\sqrt{2/\pi}\sim 0.798$$
This bound cannot be improved as the case $Y=-X$ shows.
So you see that your second bound is better not because you used the exponential moments, but rather because your first bound controls the max function way too brutally - you lost a factor of $2$. The advantage of the second method over the little trick I showed above is that it generalizes better when you deal with the max of more than two variables.
