2
$\begingroup$

Suppose I have two mean zero, independent Gaussian random variables

$X \sim \mathcal{N}(0,\sigma_1^2)$ and $Y \sim \mathcal{N}(0,\sigma_2^2)$.

Can I say something about the conditional expectation $E(X^2| k \geq|X+Y|)$?

I think the expectation should be given by the double integral

$$ E(X^2| k \geq|X+Y|) =\frac{\int_{y=-\infty}^\infty \int_{x= y - k}^{k-y} x^2 e^{-\frac{x^2}{2\sigma^2_1}}e^{-\frac{y^2}{2\sigma^2_2}} dx dy}{\int_{y=-\infty}^\infty \int_{x= y - k}^{k-y} e^{-\frac{x^2}{2\sigma^2_1}}e^{-\frac{y^2}{2\sigma^2_2}} dx dy} \,.$$

Is it possible to get an exact expression or a lower bound for this expectation?

Edit: Based on your comments I was able to get an intermediate expression for the nominator and the denominator.

Denominator:

It is well known that if $X \sim \mathcal{N}(0,\sigma_1^2)$ and $Y \sim \mathcal{N}(0,\sigma_2^2)$ and $ X \perp Y$, then $X + Y \sim \mathcal{N}(0,\sigma_1^2 + \sigma_2^2)$ and therefore \begin{equation*} \begin{aligned} \Pr(|X+Y| \leq k) &= \Phi \left( \frac{k}{\sigma_1 + \sigma_2} \right) - \Phi \left( \frac{-k}{\sigma_1 + \sigma_2} \right) \end{aligned} \end{equation*} so that \begin{equation*} \begin{aligned} \int_{y=-\infty}^\infty \int_{x= y - k}^{k-y} e^{-\frac{x^2}{2\sigma^2_1}}e^{-\frac{y^2}{2\sigma^2_2}} dx dy &= 2 \pi \sigma_1 \sigma_2 \Pr(|X+Y| \leq k) \\ &= 2 \pi \sigma_1 \sigma_2 \left\{\Phi \left( \frac{k}{\sigma_1 + \sigma_2} \right) - \Phi \left( \frac{-k}{\sigma_1 + \sigma_2} \right) \right\} \end{aligned} \end{equation*} Nominator: \begin{equation*} \begin{aligned} \int_{y=-\infty}^\infty \int_{x= y - k}^{k-y} x^2 e^{-\frac{x^2}{2\sigma^2_1}}e^{-\frac{y^2}{2\sigma^2_2}} dx dy & = \int_{y=-\infty}^\infty 2\int_{x= 0}^{k-y} x^2 e^{-\frac{x^2}{2\sigma^2_1}}e^{-\frac{y^2}{2\sigma^2_2}} dx dy, \quad (-x)^2 = x^2 \\  & = \int_{y=-\infty}^\infty (2\sigma_1)^{\frac{3}{2}}\int_{u= 0}^{\frac{(k-y)^2}{2 \sigma_1^2}} u^{\frac{3}{2}-1} e^{-u }e^{-\frac{y^2}{2\sigma^2_2}} du dy, \quad u = \frac{x^2}{2 \sigma_1^2} \\  & = \int_{y=-\infty}^\infty (2\sigma_1)^{\frac{3}{2}}\Gamma\left(\frac{3}{2},\frac{(k-y)^2}{2 \sigma_1^2} \right) e^{-\frac{y^2}{2\sigma^2_2}} dy, \quad \Gamma(s,x) = \int_0^x t^{s-1} e^t dt \\  & = (2\sigma_1)^{\frac{3}{2}} \sqrt{2}\sigma_2 \int_{v=-\infty}^\infty \Gamma\left(\frac{3}{2},\frac{\sigma_2^2}{\sigma_1^2}\left(\frac{k}{\sqrt{2}\sigma_2}-v\right)^2 \right) e^{-v^2} dv, \quad v = \frac{y}{\sqrt{2}\sigma_2} \\  & \geq 4 \sigma_1^{\frac{3}{2}}\sigma_2 \Gamma\left(\frac{3}{2}\right) \int_{v=-\infty}^\infty\left(1 + \frac{2}{3}\frac{\sigma_2^2}{\sigma_1^2}\left(\frac{k}{\sqrt{2}\sigma_2}-v\right)^2 \right)^{\frac{1}{2}} e^{-v(v+1)} dv \end{aligned} \end{equation*} where the last inequality uses the bound from this post. Any ideas how ti simplify this further to get a nontrivial lower bound on the conditional expectation $E(X^2| k \geq|X+Y|)$ are much appreciated.

$\endgroup$
2
  • $\begingroup$ You'll need to divide by the integral you posted but with the first $x^2$ removed as the probability of $k\geq |X+Y|$ is less than 1. $\endgroup$
    – JimB
    Nov 19, 2022 at 22:13
  • $\begingroup$ You can obtain an exact expression in terms of the standard Normal CDF and a Gamma$(3/2)$ CDF. Use the substitution $u = x^2/2$ to express $$\int_{-k}^k u^2 e^{-u^2/2}\,\mathrm{d}u = 2^{3/2}\int_0^k x^{3/2-1}e^{-x}\,\mathrm{d}x = \frac{2^{3/2}}{\Gamma(3/2)} \int_0^k x^{3/2-1}e^{-x}\,\mathrm{d}x.$$ $\endgroup$
    – whuber
    Nov 19, 2022 at 23:14

1 Answer 1

1
$\begingroup$

Let's simplify a little. Define

$$(U,V) = \frac{1}{\sqrt{\sigma_X^2+\sigma_Y^2}}\left(X+Y,\ \frac{\sigma_Y}{\sigma_X}X - \frac{\sigma_X}{\sigma_Y}Y\right).$$

You can readily check that $U$ and $V$ are uncorrelated standard Normal variables (whence they are independent). In terms of them,

$$X = \frac{\sigma_X}{\sqrt{\sigma_X^2 + \sigma_Y^2}} \left(\sigma_X U + \sigma_Y V\right) = \alpha U + \beta V$$

defines the coefficients of $X$ in terms of $(U,V).$ The question desires a formula for

$$E\left[X^2 \mid |X+Y|\ge k\right] = E\left[\left(\alpha U + \beta V\right)^2 \mid |U| \ge \lambda\right]$$

with $\lambda = k\sqrt{\sigma_X^2 + \sigma_Y^2} \ge 0.$

Expanding the square, we find

$$\begin{aligned} E\left[\left(\alpha U + \beta V\right)^2 \mid |U| \ge \lambda\right] &= \alpha^2E\left[U^2 \mid |U| \ge \lambda\right] \\&+ 2\alpha\beta E\left[UV \mid |U| \ge \lambda\right] \\&+ \beta^2 E\left[V^2 \mid |U| \ge \lambda\right]. \end{aligned}$$

The second term is zero because $E[V]=0$ and $V$ is independent of $U$. The third term is $\beta^2$ because the independence of $V$ and $U$ gives

$$E\left[V^2\mid |U|\ge \lambda\right] = E\left[V^2\right] = 1.$$

This leaves us to compute the first conditional expectation. The standard (elementary) formula expresses it as the fraction

$$E\left[U^2 \mid |U|\ge \lambda\right] = \frac{\left(2\pi\right)^{-1/2}\int_{|u|\ge \lambda} u^2 e^{-u^2/2}\,\mathrm{d}u}{\left(2\pi\right)^{-1/2}\int_{|u|\ge \lambda} e^{-u^2/2}\,\mathrm{d}u}$$

The denominator is $\Pr(|U|\ge \lambda) = 2\Phi(-\lambda)$ where $\Phi$ is the standard Normal distribution function.To compute the numerator, substitute $x = u^2/2$ to obtain

$$\frac{1}{\sqrt{2\pi}}\int_{|u|\ge \lambda}u^2 e^{-u^2/2}\,\mathrm{d}u = \frac{2^{3/2}}{\sqrt{2\pi}}\int_{\lambda^2/2}^\infty x^{3/2\,-1}\ e^{-x}\,\mathrm{d}x = \frac{1}{\Gamma(3/2)}\int_{\lambda^2/2}^\infty x^{3/2\,-1}\ e^{-x}\,\mathrm{d}x.$$

This equals $\Pr(Z\ge \lambda^2/2)$ where $Z$ has a Gamma$(3/2)$ distribution. It is a regularized incomplete gamma function, $P(3/2, \lambda^2/2).$ Consequently, with $\lambda \ge 0,$

$$E\left[\left(\alpha U + \beta V\right)^2 \mid |U| \ge \lambda\right] =\beta^2 + \frac{\alpha^2 P(3/2, \lambda^2/2)}{2 \Phi(-\lambda)}.$$

To illustrate, this R implementation of the conditional expectation (with a representing $\alpha,$ b representing $\beta,$ and $k$ representing $\lambda$) uses pnorm for $\Phi$ and pgamma for the Gamma distribution:

f <- function(a, b, k) { 
  b^2 + a^2 * pgamma(k^2/2, 3/2, lower.tail = FALSE) / (2 * pnorm(-k))
}
$\endgroup$
1
  • $\begingroup$ Brilliant! For completeness, I did ask about the complementary conditional expectation $E(X^2 | |X+Y| \leq k)$, would you mind changing this in your answer? Is it possible to generalize this idea to more general products $E(XY | |X + Y + Z| \leq k)$ for independent gaussian random variables? $\endgroup$
    – Peter Paul
    Nov 21, 2022 at 9:32

Your Answer

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

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