# What is the correlation of a truncated bivariate Normal distribution?

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

• 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. Nov 17, 2021 at 7:05
• thanks, so it means that the correlation still be zero? Nov 17, 2021 at 7:34
• You have to compute this integral to check whether or not it is zero. Nov 17, 2021 at 7:50
• A plot of this truncated density strongly suggests the correlation ought to be negative, not zero.
– whuber
Nov 17, 2021 at 15:07
• 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 Nov 17, 2021 at 15:17

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

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

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.