probability question?

Question

I have to show that the density of the sum of 2 independent random variables is equal to:
f(x) = 1/(2sqrt(π)) * e^(-x^2/4)

Now I know that the formula needed is:
[-infinity to +infinity]∫f(x)(f(z-x)

And that a random variable X has a normal distribution if it has density:
f(x) = 1/(sqrt(2π)) *e^(-x^2/2)

This is what I’ve done so far:
(1/2 π) ∫ e^(-x^2/2) * e^(-(z-x)^2/2) dx

Now just focusing on the exponents:
(-2x^2 + 2xz)/2 – (z^2/2)
= - (x^2 – xz) – (z^2/2)
= - (x^2 – xz + z^2/4) – z^2/4

So the integral is now:
(1/2 π) * e^(-z^2/4) ∫ e^(-(x-1/2z)^2 dx

I said that the integral can be evaluated by subbing in u = x-1/2z.
du = dx.

So now I’m left with:
∫e^(-u^2) du.

I’m confused by how to get from this expression to 1/(2sqrt(π)) * e^(-x^2/4) because I thought that ∫e^(-u^2) du couldn't be computed.

Answer 1

You're just about there!

Remember that the integral you're doing is a definite integral; so you need, not an antiderivative for ∫e^(-u^2) du (which, as you say, can't be evaluated), but the value of the definite integral from -∞ to ∞ of ∫e^(-u^2) du This integral can be shown to equal √π, and you have the desired result.