2
$\begingroup$

Given four independent, identically distributed Gaussian random variables with zero mean and unit variance $x_1$, $x_2$, $y_1$, $y_2$, consider

\begin{equation} u \equiv \max(x_1+C\, y_1, x_2+C \, y_2) - \max(x_1-C \, y_1, x_2-C \, y_2), \end{equation}

where $C$ is a real number.

Do you know how to compute the PDF of $u$, or a least its variance?

$\endgroup$
5
$\begingroup$

Using $\max(a,b) = \dfrac{a+b}{2} + \left| \dfrac{a-b}{2}\right|$, write $u = w_1 + |w_2| - |w_3|$ where $$ \eqalign{ w_1 &= C (y_1 + y_2) \cr w_2 &= \dfrac{1}{2} (x_2 - x_1 + C (y_2 - y_1))\cr w_3 &= \dfrac{1}{2} (x_2 - x_1 - C (y_2 - y_1))\cr}$$ are jointly normal with mean $0$ and covariance matrix $$ V = \pmatrix{2 C^2 & 0 & 0 \cr 0 & (1+C^2)/2 & (1-C^2)/2\cr 0 & (1-C^2)/2 & (1+C^2)/2\cr}$$ In particular, $w_1$ is independent of $w_2$ and $w_3$. Thus $\text{Cov}(w_1, |w_2|) = 0$ and $\text{Cov}(w_1, |w_3|) = 0$. The only nontrivial computation is $ \text{Cov}(|w_2|, |w_3|)$. If $|C| < 1$ get $$ \eqalign{ {\mathbb E} [|w_2| |w_3|] &= \dfrac{1-C^2}{\pi} \arctan\left(\frac{2|C|}{C^2-1}\right) + \dfrac{1-C^2}{2} + 2 \frac{|C|}{\pi}\cr {\mathbb E}|w_2| &= {\mathbb E} |w_3| = \sqrt{\dfrac{C^2+1}{\pi}}\cr \text{Cov}(|w_2|,|w_3|) &= \dfrac{1-C^2}{\pi} \arctan\left(\frac{2|C|}{C^2-1}\right) + \dfrac{1-C^2}{2} - \dfrac{(1-|C|)^2}{\pi} } $$ and then $$ \text{Var}(w_1 + |w_2| - |w_3|) = 2\dfrac{C^2-1}{\pi} \arctan\left(\frac{2|C|}{C^2-1}\right) + 4 C^2 - \frac{4|C|}{\pi} $$

EDIT: That formula was only for $|C|<1$ because of branch problems with the arctan. Here's one that should work for all $C \ne 0$:

$$ \text{Var}(w_1 + |w_2| - |w_3|) = 2 \dfrac{C^2-1}{\pi} \arctan \left(\frac{1-C^2}{2|C|}\right) + 3 C^2 - \frac{4|C|}{\pi} + 1 $$

$\endgroup$
  • $\begingroup$ Are you sure? $\mathtt{NExpectation[(Max[x + 2 y, u + 2 v] - Max[x - 2 y, u - 2 v])^2,}$ $\mathtt{\{x \approx NormalDistribution[], y \approx NormalDistribution[],}$ $\mathtt{u \approx NormalDistribution[], v \approx NormalDistribution[]\}]}$ gives 9.22454 while $\mathtt{ N[(c^2 - 1) (2/Pi) ArcTan[2 Abs[c]/(c^2 - 1)] + 4 c^2 - 4 Abs[c]/Pi] /. c -> 2}$ gives 15.2245. So it looks like there is a difference of at least 6. $\endgroup$ – Matt F. Mar 1 '16 at 3:38
  • $\begingroup$ You're taking $C = 2$, my formula was for $|C|<1$. They are different (due to the branch cut of arctan). $\endgroup$ – Robert Israel Mar 1 '16 at 3:55
  • $\begingroup$ That works! It'd be helpful to put the limitation at the front of the answer. $\endgroup$ – Matt F. Mar 1 '16 at 4:03
0
$\begingroup$

Using capital letters for random variables, let $V_{i}^{\pm} = X_{i} \pm CY_{i} \sim \mathcal{N}(0,1\pm C^{2})$, for $i=1,2$. Define matrix $ A_{\pm} = \begin{bmatrix} 1 & \pm C & 0 & 0\\ 0 & 0 & 1 & \pm C\end{bmatrix} $, notice that $\begin{pmatrix}V_{1}^{\pm}\\V_{2}^{\pm}\end{pmatrix} = A_{\pm}\begin{pmatrix}X_{1}\\Y_{1}\\X_{2}\\Y_{2}\end{pmatrix}$, and that $A_{\pm}A_{\pm}^{\top} = \begin{pmatrix}1\pm C^{2} & 0\\0 & 1\pm C^{2}\end{pmatrix}$ is diagonal, meaning the components $V_{1}^{\pm}$ and $V_{2}^{\pm}$ are respectively (for the same sign) independent (see here). Quantity of interest is $U = U^{+} - U^{-}$, where, by the independence argument, the random variables $U^{\pm} = \max(V_{1}^{\pm},V_{2}^{\pm})$ has respective CDF $F_{\pm}(x) = \Phi^{2}\left(\frac{x}{\sqrt{1\pm C^{2}}}\right)$, where $\Phi(.)$ is the standard normal CDF. From this, you may be able to compute the CDF and PDF of the difference. I will try to come back to the explicit calculation later.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.