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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!

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    $\begingroup$ 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. $\endgroup$
    – whuber
    Commented May 16, 2023 at 19:32
  • $\begingroup$ @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? $\endgroup$
    – Coach
    Commented May 17, 2023 at 15:03
  • $\begingroup$ @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. $\endgroup$
    – Zhanxiong
    Commented May 17, 2023 at 15:04
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    $\begingroup$ @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. $\endgroup$
    – whuber
    Commented May 17, 2023 at 15:16
  • $\begingroup$ 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 $P(A^2+B^2>2a)=1-\left(1-exp\left(-\frac{a}{2(1+\rho)}\right)\right)$ $\endgroup$
    – Coach
    Commented May 17, 2023 at 15:19

1 Answer 1

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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}

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  • $\begingroup$ 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? $\endgroup$
    – Coach
    Commented May 17, 2023 at 13:19
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    $\begingroup$ @Coach I added more details. Please check the updated answer. $\endgroup$
    – Zhanxiong
    Commented May 17, 2023 at 13:52

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