2
$\begingroup$

$X$ and $Y$ are independent standard normal variables. If we generate $n$ samples $(x,y),$ what is the correlation for samples $(x, y)$ with $x+y\gt0$?

$\endgroup$
5
  • 2
    $\begingroup$ The generation as such has no impact on the correlation. The pair $(X,Y)$ is distributed from a bivariate $N(0_2,I_2)$ distribution restricted to the half space $x+y>0$ and the correlation is defined as usual by$$\int_{x+y>0} (xy-\mathbb E[X]^2) c \exp\{-(x^2+y^2)/2\}\,\text dx\text dy$$where $c$ is the proper normalising constant. $\endgroup$
    – Xi'an
    Nov 17, 2021 at 7:05
  • $\begingroup$ thanks, so it means that the correlation still be zero? $\endgroup$
    – qwerty1010
    Nov 17, 2021 at 7:34
  • $\begingroup$ You have to compute this integral to check whether or not it is zero. $\endgroup$
    – Xi'an
    Nov 17, 2021 at 7:50
  • 1
    $\begingroup$ A plot of this truncated density strongly suggests the correlation ought to be negative, not zero. $\endgroup$
    – whuber
    Nov 17, 2021 at 15:07
  • $\begingroup$ You can easily approximate this quantity by simulationx=rnorm(T);y=rnorm(T);mean(x[x+y>0]*y[x+y>0])-mean(x[x+y>0])^2 $\endgroup$
    – Xi'an
    Nov 17, 2021 at 15:17

1 Answer 1

3
$\begingroup$

The question asks about the correlation coefficient of a truncated standard bivariate Normal distribution, where it is limited to the half-plane $X+Y\gt 0.$ (See the figure below.)

To work with this, it's convenient to let $U=X+Y$ (truncated to $U\gt 0$) and $V=X-Y$ (not truncated at all). Evidently

  • $U/\sqrt 2$ has a Half-normal distribution,

  • $V/\sqrt 2$ has a standard Normal distribution, and

  • $(U,V)$ is independent.

The last observation follows directly from the independence of $(X+Y, X-Y).$

It is immediate that

  1. $E[U/\sqrt 2] = \sqrt{2/\pi},$

  2. $E[V] = 0,$

  3. $E[U^2/2] = 1,$

  4. $E[V^2/2] = 1,$ and

  5. $E[UV] = E[U]E[V] = 0.$

Consequently

  1. $\operatorname{Var}(U) = E[U^2]-E[U]^2 = 2 - 2/\pi,$

  2. $\operatorname{Var}(V) = E[V^2]-E[V]^2 = 2,$ and

  3. $\operatorname{Cov}(U,V) = E[UV] - E[U]E[V] = 0.$

The question asks about the correlation of $((U+V)/2, (U-V)/2),$ whose formula is

$$\begin{aligned} \rho = \operatorname{Cor}((U+V)/2, (U-V)/2) &= \operatorname{Cor}(U+V, U-V) \\ &=\frac{\operatorname{Cov}(U+V,U-V)} {\sqrt{\operatorname{Var}(U+V)\operatorname{Var}(U-V)}}. \end{aligned}$$

The standard rules of covariance (namely, its bilinarity) reduce these expressions to

  1. $\operatorname{Cov}(U+V,U-V) = \operatorname{Var}(U)- \operatorname{Var}(V) = -2/\pi,$
  2. $\operatorname{Var}(U\pm V) = \operatorname{Var}(U) + \operatorname{Var}(V) \pm 2\operatorname{Cov}(U,V) = 2-2/\pi.$

The formula for $\rho$ thereby simplifies to

$$\rho = \frac{-2/\pi}{\sqrt{(2 - 2/\pi)^2}} = \frac{\pi}{1-\pi} \approx -0.4669422.$$

Figure

This plot shows 2000 independent $(X,Y)$ values. Those with $X+Y\gt 0$ are darker. Their least-squares fit appears in red: it is a line expected to have slope $\pi/(1-\pi)$ passing through the point $(\sqrt{1/\pi}, \sqrt{1/\pi}).$ It comes very close to that in this sample, helping to check the correctness of the formulas.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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