# Expectation

• Nov 8th 2012, 06:14 AM
Juju
Expectation
Hey,

I have given a normally distributed random variable $X\stackrel{d}{=}\mathcal{N}(\mu,\sigma^2)$ and want to compute the following expectation

$\mathbb{E}[X \mathbb{1}_{\{X\geq 1\}}]$

where $\mathbb{1}$ denotes the indicator function.

I think that it is not possible to compute the corresponding integral explicitly. Am I right?

• Nov 8th 2012, 09:16 PM
chiro
Re: Expectation
Hey Juju.

Can you find the random variable corresponding to X >= 1 where X has that Normal distribution?

Hint: Use the conditional distribution of X|X>=1 where P(A|B) = P(A and B)/P(B).
• Nov 8th 2012, 10:41 PM
Juju
Re: Expectation
unfortunately I do not understand what you mean by "corresponding random variable". Of course I can determine the distribution function of $X \mathds{1}_{\{X\geq 1\}}$ and the expectation is given by
$\int\limits_1^\infty (2 \pi \sigma^2)^{-\frac{1}{2}}xe^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}}dx$

But I cannot give this integral in a closed form.
• Nov 9th 2012, 03:58 PM
chiro
Re: Expectation
You should be able to evaluate the integral: (Hint: use a substitution v = (x-mu)^2 and change from dx to dv).

Also when I say corresponding random variable I just mean the random variable that is represented algebraically with respect to the limits of the integrand.

For example I can have Y = X + 3 and from this specify a conditional probability of P(Y|X) if I know the density function for the random variable X.

Now similarly, you can have limits in the integration that correspond with a particular algebraic expression but they may not have an easy interpretation algebraically (i.e. might be a lot more complex than the relation Y = X + 3).