Say $x_1$ and $x_2$ are normal random variables with known means and standard deviations and $C$ is a constant. If $y = \max(x_1,x_2,C)$, what is $\mathrm{Var}(y)$?

Well, I forgot to tell that $x_1$ and $x_2$ are independent.

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    $\begingroup$ You need to know the joint distribution function of $X_1$ and $X_2$. Are they given to be independent but you neglected to tell us so? $\endgroup$ – Dilip Sarwate Jul 30 '12 at 15:06
  • $\begingroup$ Presumably the $x_i$ are independent? $\endgroup$ – copper.hat Jul 30 '12 at 15:59
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    $\begingroup$ Answer: a complicated function of $C$ and of the means and standard deviations of $x_1$ and $x_2$. $\endgroup$ – Did Jul 30 '12 at 17:35

What I've got isn't pretty, but here goes (note: I use $\varphi(\cdot)$ and $\Phi(\cdot)$ to denote the standard normal density and CDF, resp.):

First, let's put everything in terms of standard normals:


The first key is drawing proper regions for the evaluation of the Max function:


With this we can see that:

$$\begin{eqnarray*} \mathbb{E}[Max\{X_{1},X_{2},C\}] & = & \intop\limits _{-\infty}^{C_{2}}\intop\limits _{-\infty}^{C_{1}}Cd\Phi(y_{1})d\Phi(y_{2})\\ & & +\intop\limits_{C_1}^{\infty} \intop\limits_{-\infty}^{y_1} y_1 d\Phi(y_2)d\Phi(y_1) \\ & & +\intop\limits_{C_2}^{\infty} \intop\limits_{-\infty}^{y_2} y_2 d\Phi(y_1)d\Phi(y_2) \\ \end{eqnarray*}$$

Simplifying what we can:

$\mathbb{E}[Max\{X_{1},X_{2},C\}] = C\Phi(C_1)\Phi(C_2)+(1-\Phi(C_1))\mathbb{E}[z\Phi(z) | Z>C_1]+(1-\Phi(C_2))\mathbb{E}[z\Phi(z) | Z>C_2]$

($Z$ distributed $N(0,1)$)

Similarly, we get:

$$\begin{eqnarray*} \mathbb{E}[(Max\{X_{1},X_{2},C\})^2] & = & \intop\limits _{-\infty}^{C_{2}}\intop\limits _{-\infty}^{C_{1}}C^2d\Phi(y_{1})d\Phi(y_{2})\\ & & +\intop\limits_{C_1}^{\infty} \intop\limits_{-\infty}^{y_1} y_1^2 d\Phi(y_2)d\Phi(y_1) \\ & & +\intop\limits_{C_2}^{\infty} \intop\limits_{-\infty}^{y_2} y_2^2 d\Phi(y_1)d\Phi(y_2) \\ \end{eqnarray*}$$

Which simplifies slightly:

$\mathbb{E}[(Max\{X_{1},X_{2},C\})^2] = C^2\Phi(C_1)\Phi(C_2)+(1-\Phi(C_1))\mathbb{E}[z^2\Phi(z)|Z>C_1]+(1-\Phi(C_2))\mathbb{E}[z^2\Phi(z)|Z>C_2]$

That's about as far as I've gotten.

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The solution is far from trivial, using Maple I obtained this:

enter image description here

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I do not know the actual expression for Var($y$), but to get started, $y$ will not follow a normal distribution and its cumulative probability distribution $F_Y(x)$ will be the product of the cumulative distributions $F_{X1}(x)$ and $F_{X2}(x)$, floored to zero for $x<C$.

For an estimation of Var($y$) you may find this other question helpful: Expectation of the maximum of gaussian random variables

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  • $\begingroup$ You re right. Actually, I already have the solution to expected value of max of two (or more) normal random variables. It is below. $\endgroup$ – user36831 Aug 1 '12 at 11:56

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