Conditional Expectation of a normal random variable.

Question

I have been asked to derive the expression for $E[X\mid X>a]$ where $X$ is a normal random variable with mean $\mu$ and variance $\sigma^2$ and $a$ is some constant belonging to $R$. It is my understanding that I need to integrate the product of $X$ and pdf of normal from $a<x<\infty$. But I am unable to figure out this integral. Can anyone help with that? If possible, kindly also guide how we can solve it using the MGF of normal random variable.

Answer 1

The conditional distribution $f(x|x>a)$ is given by

$$f_{X|X>a}(x)=\frac{1}{1-F_X(a)}f_X(x)\mathbb{1}_{(a;\infty)}(x)$$

The integral involved to calculate the expectation can be easy solved as it is in the form

$$\int x e^{-\frac{x^2}{2}}dx$$

and as you note, $x$ is something like the exponent derivative...

---

Example:

$X\sim N(1;1)$

Calculate $\mathbb{E}(X|X>1)$

$$\mathbb{E}(X|X>1)=\frac{1}{1-F(1)}\int_1^{\infty}\frac{1}{\sqrt{2\pi}}x e^{-\frac{(x-1)^2}{2}}dx=$$

Set $x-1=y$

$$2\Bigg[\int_0^{\infty}\frac{1}{\sqrt{2\pi}}ye^{-\frac{y^2}{2}}dy+\underbrace{\int_0^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dy}_{=\frac{1}{2}}\Bigg]$$

$$=2\Bigg\{-\frac{1}{\sqrt{2\pi}}\Bigg[e^{-\frac{y^2}{2}}\Bigg]_0^{\infty}+\frac{1}{2}\Bigg\}=\frac{2}{\sqrt{2\pi}}+1\approx1.797885$$

---

General example

$X\sim N(\mu;\sigma^2)$; Calculate $\mathbb{E}(X|X>a)$

$$\mathbb{E}(X|X>a)=\frac{1}{1-\Phi\Big(\frac{a-\mu}{\sigma}\Big)}\int_a^{\infty}\frac{1}{\sigma\sqrt{2\pi}}x e^{-\frac{1}{2}\Big(\frac{x-\mu}{\sigma}\Big)^2}$$

Set $\frac{x-\mu}{\sigma}=y$

$$\frac{1}{1-\Phi\Big(\frac{a-\mu}{\sigma}\Big)}\int_{\frac{a-\mu}{\sigma}}^{\infty}\frac{1}{\sqrt{2\pi}}(\sigma y+\mu) e^{-\frac{1}{2}y^2}dy$$

$$\frac{1}{1-\Phi\Big(\frac{a-\mu}{\sigma}\Big)}\Bigg\{\int_{\frac{a-\mu}{\sigma}}^{\infty}\frac{\sigma y}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}dy+\mu\int_{\frac{a-\mu}{\sigma}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2}dy\Bigg\}$$

$$\frac{1}{1-\Phi\Big(\frac{a-\mu}{\sigma}\Big)}\Bigg\{\frac{\sigma}{\sqrt{2\pi}}\Big[ -e^{-\frac{1}{2}y^2}\Big]_{\frac{a-\mu}{\sigma}}^{\infty}+\mu\Big[1-\Phi\Big(\frac{a-\mu}{\sigma}\Big] \Bigg\}$$

$$ \bbox[5px,border:2px solid red] { \mathbb{E}(X|X>a)=\frac{1}{1-\Phi\Big(\frac{a-\mu}{\sigma}\Big)}\Bigg\{\frac{\sigma}{\sqrt{2\pi}}e^{-\frac{1}{2}\Big(\frac{a-\mu}{\sigma}\Big)^2}+\mu\Big[1-\Phi\Big(\frac{a-\mu}{\sigma}\Big] \Bigg\} \ } $$