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I have been reading this paper about the maximum and minimum of two normal distributed variables.

Inside the paper there is the formula for the expectation of this the maximum of the two variables. It is given below:

$$ \DeclareMathOperator{\E}{\mathbb{E}} \E(X) = \mu_1 \Phi\left( \frac{\mu_1-\mu_2}{\theta} \right) + \mu_2 \Phi\left( \frac{\mu_2-\mu_1}{\theta} \right) + \theta \phi\left( \frac{\mu_1-\mu_2}{\theta} \right) $$

Where:

  • $\mu_1$ is the mean of the first normally distributed random variable.
  • $\mu_2$ is the mean of the second normally distributed random variable.
  • $\Phi$ is the CDF of the standard normal.
  • $\phi$ is the PDF of the standard normal.
  • $\rho$ is the correlation co-efficient between the variables.
  • $\theta$ is $\sqrt{\sigma_{1}^2 + \sigma_{2}^2 + 2\rho\sigma_{1}\sigma_{2}}$

I am trying to use this to get the mean of the new distribution but I run in to what feels like an obvious problem.

If $\mu_{1} \neq \mu_2$ then the argument to the $\Phi$ function in either the first or second term of the sum will be negative. The standard CDF does not allow negative values‽

I feel like I have fundamentally misunderstood something here. Where have I gone wrong?

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    $\begingroup$ By definition, the CDF evaluated at any real number $x$ gives the chance that the variable (a standard Normal variable in this case) is less than or equal to $x:$ that is a probability and therefore lies between $0$ and $1.$ It cannot be negative. $\endgroup$
    – whuber ♦
    Commented Feb 12, 2019 at 20:33
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    $\begingroup$ Just a hint here Max(X,Y) = 1/2((X+Y)-|X-Y|). So you just have to calculate E[|Z|] where Z is a normal random variable with mean mu and std sigma. Then apply the formula to X-Y and of course E[X+Y]=E[X]+E[Y] $\endgroup$
    – Daniel Parry
    Commented Feb 12, 2019 at 20:48

1 Answer 1

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The CDF does not yield negative results (it only yields numbers between 0 and 1), but it can certainly take negative arguments.

> pnorm(-1,0,1)
[1] 0.1586553
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  • $\begingroup$ Thanks! Stupid mistake on my part. In Excel I was using the Norm.Inv function and then slapped my forehead when I saw this answer. $\endgroup$
    – Simon Johnson
    Commented Feb 12, 2019 at 20:42

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