Expected value of the conditional distribution?

Question

You have X from Normal(mu=0, st.dev=5). What is E(X|X>0) ?

Answer 1

Here's a rough explanation of E(X|X>0):

Form a large sample of Normal(mu=0, st.dev=5) variables.
Cast out all of the samples except those for which X>0.
Take the mean of the remaining population.
That's an estimate of E(X|X>0).

Answer 2

Integrate xf(x) over the interval of x > 0, and then divide by
P(X > 0).

int(0 to inf) 1/sqrt(50pi)*x*exp(-x^2/50)dx

Let u = x^2/50.
Then du = xdx/25.
So xdx = 25du
Limits of integration will be the same.

1/sqrt(50pi) * int(0 to inf) 25exp(-u)du

= 25/sqrt(50pi) * (-exp(-u))(0 to inf)

= 5/sqrt(2pi) * (0 - (-1))

= 5/sqrt(2pi)

So divide by P(X > 0), which will be 0.5, so

E(X|X>0) = 10/sqrt(2pi) = 3.9894228040143267793994605993438 approximately.

edit: This is backed up by the simulation that ksoileau suggests above. You will get means close to this.