Return to Answer

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 $\rho = -1$.

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 $\rho = -1$.

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 the answer is quite simple. In particular, then $(X,Y)$ are bivariate Normal with correlation $\rho$ and joint pdf $f(x,y)$:

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 $\rho = -1$.

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 $\rho = -1$.

The best possible upper bound is the exact solution, and presumably the exact solution (and so the upper bound itself) will depend on the correlation $\rho$ between $X$ and $Y$.

We are given that $X \sim N(0,1)$ and $Y \sim N(0,1)$ where $X$ and $Y$ may be correlated, and let $\rho$ denote the correlation coefficient, $ -1 < \rho < 1$. This is all captured by a bivariate Normal pdf with joint density $f(x,y)$:

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 $\rho = -1$.

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 the answer is quite simple. In particular, then $(X,Y)$ are bivariate Normal with correlation $\rho$ and joint pdf $f(x,y)$:

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 $\rho = -1$.