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.


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.

  • $\begingroup$ 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


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))
  • $\begingroup$ 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

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