Suppose I have two independent normal random variables, $X$ and $Z$ with $\mu_x$, $\sigma^2_x$ and $\mu_z$, $\sigma^2_z$. Suppose I also know that $x+z\geq y$. How do I find the conditional distribution of $z$, that is $f(Z|x+z\geq y)$?

What I have so far is that the conditional distribution of $Z$ for any $\overline x$, such that $\overline x + z = y$ is a truncated normal: \begin{equation} f(Z|X=\overline{x})=\begin{cases} 0 & \text{if }z<y-\overline{x}\\ \frac{\phi(\mu_{z},\sigma_{z}^{2};z)}{1-\Phi(\mu_{z},\sigma_{z};z)} & \text{if }z\geq y-\overline{x} \end{cases} \end{equation}

My instinct tells me I need to integrate over every possible $\overline x$, but I'm not sure what function to integrate over, exactly. If I use the above conditional distribution, I get

\begin{equation} f(Z|X+z\geq y)=\int_{-\infty}^{\infty}\frac{\phi(\mu_{z},\sigma_{z}^{2};z)}{1-\Phi(\mu_{z},\sigma_{z};z)}d\overline{x} \end{equation}

but I'm not sure if this is correct. Am I on the right track here?

  • $\begingroup$ It looks like a good approach, but your expression for $f(X\mid X=\bar x)$ makes no sense: the left hand side depends only on $\bar x$ but the right hand side also explicitly depends on $z$ and $y$ (whatever you might mean by them). $\endgroup$
    – whuber
    Commented Jan 15, 2020 at 17:51
  • $\begingroup$ Sorry, that's supposed to be $f(Z|X=\overline x)$ (now fixed). $y$ is a constant (maybe I should have used $a$ to avoid confusion) and $\overline x$ is also constant, so the right hand side should only depend on $z$. $\endgroup$
    – thagomizer
    Commented Jan 15, 2020 at 19:31
  • $\begingroup$ Presuming $z$ is the argument of the density function, the right hand side must depend on $z$ and $\bar x.$ The symbol "$y$" isn't even defined. $\endgroup$
    – whuber
    Commented Jan 15, 2020 at 20:17

1 Answer 1


This post offers a solution in three manageable steps. They are presented in an order that streamlines the exposition but numbered in the order they would be carried out. All the necessary calculations are highlighted.

Step 3

Begin with a simplified version of this problem where variables $(U,V)$ are independently standard Normal (so each has CDF $\Phi$). Suppose you have three numbers $\alpha,$ $\beta,$ and $\xi\ge 0$ with $\alpha^2 + \beta^2 = 1$ and $\beta \ge 0.$ We wish to find the distribution of $\alpha U + \beta V$ conditional on $U \ge \xi.$

This is relatively easy to solve because the condition $U\ge \xi$ merely eliminates all the support of $(U,V)$ to the left of the vertical line $U=\xi.$ Thus the conditional density function at $(u,v),$ where $u \ge \xi,$ is still proportional to $\exp(-(u^2+v^2)/2)$ and the constant of proportionality is

$$C(\xi) = \frac{1}{2\pi\, (1 - \Phi(\xi))}.$$

When $\beta=0$ the solution is obvious ($V$ is no longer involved in the problem), so assume now that $\beta \gt 0.$ Changing variables from $(u,v)$ to $(s,t)$ with $s = \alpha u + \beta v$ and $t = -\beta u + \alpha v$ preserves $u^2+v^2$ amd the absolute differential $|\mathrm{d}u\,\mathrm{d}v|.$ (This occurs because $\alpha^2+\beta^2=1.$) The inequality $u \ge \xi$ becomes $t \le (\alpha s - \xi)/\beta.$ As usual, the marginal density of $S = \alpha U + \beta V$ is obtained by integrating out $t,$

$$\eqalign{f_{\alpha,\beta,\xi}(s) &= C(\xi) \int_{-\infty}^{ (\alpha s - \xi)/\beta} \exp\left(-(s^2+t^2)/2 \right)\mathrm{d}t \\ & = C(\xi)\sqrt{2\pi} \exp(-s^2/2)\, \Phi\left(\frac{\alpha s - \xi)}{\beta}\right) .}$$

Step 2

How, then, can we reduce the original question to this easy case? Start with the most general bivariate normal variables $(X,Z)$ with means $(\mu_X,\mu_Z),$ standard deviations $\sigma_X$ and $\sigma_Z,$ respectively, and correlation coefficient $\rho.$ (The question assumes $\rho=0$ but this offers no real simplification.) Re-express the problem in terms of the random variables

$$U = \frac{X-\mu_X}{\sigma_X},\quad W = \frac{Z - \mu_Z}{\sigma_Z}$$

and then set

$$V = \frac{W - \rho U}{\sqrt{1-\rho^2}}.$$

These are natural transformations to apply because the first two standardize the variables and the last removes their correlation, achieving the original simplified assumptions about $(U,V).$ See https://stats.stackexchange.com/a/71303/919 for more about these operations, including illustrations.

Step 1

The rest is algebraic mopping-up. The original condition $X+Z \ge y$ becomes

$$(\sigma_X + \rho \sigma_Z)\,U + \sigma_Z \sqrt{1-\rho^2}\, V\ \ge\ y - \mu_X - \mu_Z.$$

Divide both sides by the square root of the sum of squares of the coefficients

$$\Delta = \sqrt{(\sigma_X + \rho \sigma_Z)^2 + \sigma_Z^2 (1-\rho^2)}$$

to produce

$$\alpha U + \beta V \ge \xi$$ where

$$\alpha = \frac{(\sigma_X + \rho \sigma_Z)}{\Delta},\quad \beta = \frac{\sigma_Z \sqrt{1-\rho^2}}{\Delta},\quad \xi = \frac{y - \mu_X - \mu_Z}{\Delta}.$$

Because $\alpha^2 + \beta^2 = 1$ and $\beta \ge 0,$ we're done.

The same approach works to analyze any nontrivial condition of the form $\lambda X + \mu Z \ge y.$ In this case $\beta$ could end up being negative, but a preliminary change of variable from $(X,Z)$ to $(X,-Z)$ will take care of that problem while negating $\mu,$ $\mu_Z,$ and $\rho$ but otherwise leaving everything else the same.

  • 1
    $\begingroup$ This is exactly what I was looking for, thanks! $\endgroup$
    – thagomizer
    Commented Jan 16, 2020 at 15:19

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