# Upper bound for sum of dependent normal variables

I am having difficulties with the following problem:

Assuming $$X$$ and $$Y$$ follow a bivariate normal distribution with $$\mu = 0$$ and $$\Sigma=\begin{pmatrix} 1 & \rho \\ \rho & 1 \end{pmatrix}$$ and $$U=X^2+Y^2$$, I want to prove that the inequality $$P(U>a)\leq \exp\left(-\frac{a}{2(1+|\rho|)}\right)$$ holds for all $$a$$.

So far, I tried to solve this using the Chernoff bound $$P(X\geq a)\leq e^{-ta}\cdot\mathbb{E}\left[e^{tX}\right]$$ but couldn't find or work out a closed form expression for the MGF $$M_U(t)$$.

Maybe one way to go about this would be to use a generalised Chi-squared random variable (https://math.stackexchange.com/questions/442472/sum-of-squares-of-dependent-gaussian-random-variables), which does not have a closed form MGF, but a CF that could be used to derive the MGF. However I could not find any solution this way.

Would be very glad if someone could even just guide me in the right direction!

• Perhaps reframing the question in terms of $A=X+Y$ and $B=X-Y$ might give you some ideas because $U=2(A^2+B^2)$ but now $A$ and $B$ are independent. Moreover, the event $U\gt a$ is a subset of the union of $A^2\gt a/4$ and $B^2\gt a/4,$ suggesting a way to exploit that independence.
– whuber
Commented May 16, 2023 at 19:32
• @whuber Many thanks for your input! Using $A=X+Y$ and $B=X-Y$, I get $U=\frac{1}{2}(A^2+B^2)$. From my understanding, we should have $$A,B\stackrel{i.i.d.}{\sim}N(0, 2(1+\rho)).$$ And since $A$ and $B$ are independent, so are $A^2$ and $B^2$ and $$A^2+B^2\sim\Gamma(1,4(1+\rho)).$$ I get $$P(U>a)=P(A^2+B^2>2a)=\exp\left(-\frac{a}{2(1+\rho)}\right).$$ What I am missing is the absolute of $\rho$ and why I would end up with an inequality. Am I on the right track? Commented May 17, 2023 at 15:03
• @Coach $A, B$ are not i.i.d. $A \sim N(0, 2(1 + \rho)), B \sim N(0, 2(1 - \rho))$. And even if $A^2 + B^2 \sim \Gamma(1, 4(1 + \rho))$ were true, how do you conclude its tail probability is $\exp(-a/(2(1 + \rho))$? The CDF of a Gamma r.v. does not have a closed-form. Commented May 17, 2023 at 15:04
• @Zhanxiong It does for integral shape parameters, especially for $1$! Coach: Without any loss of generality you may assume $\rho \gt 0$ because you can negate $Y$ without affecting $U,$ but that negates $\rho.$ Yet Zhanxiong is correct: although $A$ and $B$ are independent, when $\rho\ne 0$ they are not identically distributed. If you were to draw a picture of Zhanxiong's integration region, the effectiveness of that idea should become clear.
– whuber
Commented May 17, 2023 at 15:16
• I used this for the Gamma CDF. I got $P(A^2+B^2>2a)=1-P(A^2+B^2<2a)=1-\frac{\gamma\left(1,\frac{2a}{4(1+\rho)}\right)}{\Gamma(1)}$. As a result $P(A^2+B^2>2a)=1-\left(1-exp\left(-\frac{a}{2(1+\rho)}\right)\right)$ Commented May 17, 2023 at 15:19

We can first evaluate the probability $$P[U > a]$$ by noting $$X^2 + Y^2 \overset{d}{=} \xi^2 + \eta^2 + 2\rho\xi\eta$$, where $$\xi, \eta$$ i.i.d. $$\sim N(0, 1)$$. Therefore, for $$a > 0$$: \begin{align} & P[X^2 + Y^2 > a] = P[\xi^2 + \eta^2 + 2\rho\xi\eta > a] \\ =& \iint\limits_{[(x, y): x^2 + y^2 + 2\rho xy > a]}\frac{1}{2\pi}\exp\left(-\frac{1}{2}(x^2 + y^2)\right)dxdy. \tag{1} \end{align}
To evaluate this double integral, apply the polar coordinates transformation $$x = r\cos\theta, y = r\sin\theta$$ with $$r > 0, \theta \in [0, 2\pi)$$. The integral $$(1)$$ then becomes \begin{align} \iint\limits_{[(r, \theta): (1 + \rho\sin(2\theta))r^2 > a]}\frac{1}{2\pi}r\exp\left(-\frac{1}{2}r^2\right)drd\theta. \tag{2} \end{align} Note the region $$[(r, \theta): (1 + \rho\sin(2\theta))r^2 > a]$$ is contained in the region $$[(r, \theta): (1 + |\rho|)r^2 > a]$$ and the integrand is positive, hence the integral $$(2)$$ is bounded above by \begin{align} \iint\limits_{[(r, \theta): (1 + |\rho|)r^2 > a]}\frac{1}{2\pi}r\exp\left(-\frac{1}{2}r^2\right)drd\theta = \int_{\sqrt{\frac{a}{1 + |\rho|}}}^\infty e^{-r^2/2}rdr = \exp\left(-\frac{a}{2(1 + |\rho|)}\right). \end{align} This completes the proof.
To see why $$X^2 + Y^2 \overset{d}{=} \xi^2 + \eta^2 + 2\rho\xi\eta$$, note that if $$\begin{bmatrix} X \\ Y \end{bmatrix} \sim N_2(0, \Sigma)$$, then $$\begin{bmatrix} X \\ Y \end{bmatrix} \overset{d}{=} C\begin{bmatrix} \xi \\ \eta \end{bmatrix}$$, where $$\begin{bmatrix} \xi \\ \eta \end{bmatrix} \sim N_2(0, I_{(2)})$$ and $$C = \Sigma^{1/2}$$ is the square root matrix of $$\Sigma$$. It then follows that \begin{align} X^2 + Y^2 &= \begin{bmatrix} X & Y \end{bmatrix}\begin{bmatrix} X \\ Y \end{bmatrix} \\ & \overset{d}{=} \begin{bmatrix} \xi & \eta \end{bmatrix}C'C\begin{bmatrix} \xi \\ \eta \end{bmatrix} = \begin{bmatrix} \xi & \eta \end{bmatrix}\Sigma\begin{bmatrix} \xi \\ \eta \end{bmatrix} \\ &= \xi^2 + \eta^2 + 2\rho\xi\eta. \end{align}
• Great explanation, thank you! I suspected it could have something to do with polar notation or the Rayleigh distribution, but couldn't figure out how to get rid of the dependence. On that note, may I ask where the lemma $X^2+Y^2 \stackrel{d}{=}\xi^2+\eta^2+2\rho\xi\eta$ with $\xi, \eta \sim N(0,1)$ comes from? Commented May 17, 2023 at 13:19