# Showing that $P (|X_1| < \Gamma , |X_2| < \Gamma)$ is increasing in $|\rho|$

Assume $$(X_1,X_2 )^T$$ is mean $$0$$ bivariate normal distributed with covariance matrix $$\Sigma = \left (\begin{matrix} 1 & \rho \\ \rho & 1 \end{matrix} \right)$$ and let $$\Gamma > 0$$ a positive constant. Then i would like to show that $$P (|X_1| < \Gamma , |X_2| < \Gamma)$$ is increasing in $$|\rho|$$.

Any tips? I already tried to simply use the integral representation of the probability, but could not show it.

• Are we meant to assume that $X_1,X_2$ are independent variables? Mar 23, 2020 at 11:02
• no they are bivariate normal distributed. Mar 23, 2020 at 11:04
• The bivariate normality is implied in the question. Mar 23, 2020 at 11:10
• @J.Field Sorry I missed that, thanks for clarifying Mar 23, 2020 at 11:10
• @StubbornAtom saying that a vector is normally distributed can either mean that its entries are iid normally distributed or that the whole vector has a multivariate normal distribution Mar 23, 2020 at 11:11

This follows from Mehler's Formula, in the form $$f(x_1,x_2)=\phi(x_1)\phi(x_2)\sum_{k=0}^\infty\frac {\rho^n}{n!} He_n(x_1)He_n(x_2),$$ where $$f$$ is the joint density of $$(X_1,X_2)$$, $$\phi$$ is the marginal density of the $$X_i$$, and $$He_n(x)$$ is a "probabilist's" Hermite polynomial. Integrating over $$[-\Gamma,\Gamma]\times[-\Gamma,\Gamma]$$ yields an expression of form $$I(\Gamma,\rho) := P(|X_1|<\Gamma, |X_2|<\Gamma) = \sum_{k=0}^\infty\frac {\rho^n}{n!}\left(\int_{-\Gamma}^\Gamma \phi(x)He_n(x)\,dx\right)^2,\tag 1$$ all of whose terms are non-negative. In fact $$He_n$$ is an odd polynomial if $$n$$ is odd, so the odd terms in (1) vanish, so we see $$I(\Gamma,\rho)$$ is given by a power series in $$\rho^2$$ with non-negative coefficients.

• Nice trick. Thanks for the answer. Mar 23, 2020 at 14:49

Note that the pdf $$f(x_1,x_2)$$ satisfies $$\frac{\partial f}{\partial \rho} > 0$$. From there, it suffices to observe that $$\frac{d P}{d \rho} = \frac{d}{d\rho} \iint_R f(x_1,x_2) dx_1 dx_2= \ \iint_R \frac{\partial f}{\partial \rho}\,(x_1,x_2) \,dx_1 dx_2 < 0,$$ where $$R$$ denotes the rectangle $$[-\Gamma,\Gamma]\times [-\Gamma,\Gamma]$$. After computing $$\frac{\partial f}{\partial \rho}$$, you should find that one of the resulting integrals can be solved as an iterated integral; solve the inner-integral with the substitution $$u_2 = x_2^2$$.

• Is it a well known fact, that the derivate suffices the inequality? I checked it myself, but it was kind of lenghty. Mar 23, 2020 at 14:51
• Yes: it is well known that to show that a function $I(\rho)$ is increasing, it suffices to show that $dI/d\rho$ is positive Mar 23, 2020 at 15:10
• @J.Field Honestly, I prefer the other answer Mar 23, 2020 at 15:10
• No i meant if it is well known, that the inequality $\frac{\partial f}{\partial p} > 0$ holds,but i guess i´ll take the other solution. Mar 23, 2020 at 19:31

Let $$f_{\rho}$$ be the density of $$(X_1,X_2)$$.

An interesting identity I stumbled upon in this connection is

$$\frac{\partial}{\partial\rho}f_{\rho}(x,y)=\frac{\partial^2}{\partial x\partial y}f_{\rho}(x,y)$$

This is part of a general result due to Plackett (1954).

Then differentiating under the integral sign,

\begin{align} \frac{d}{d\rho}P(|X_1|

That is, for some $$c>0$$ we have

$$\frac{d}{d\rho}P(|X_1|

Note that $$\rho > 0\implies -\frac1{1+\rho}>-\frac1{1-\rho}$$, so that right hand side of the above equation is positive for $$\rho\in (0,1)$$. Similarly it is negative for $$\rho\in (-1,0)$$. Hence the probability is increasing in $$|\rho|$$.