Help me integrate this?

Question

Okay, this is a tough one:

1/√(2*π) * ∫ (z^2 * exp(-(1/2)*z^2) dz)
-∞ to +∞

This is related to probability. Why is it that the area under the curve is 1? I understand that the area under a normal distribution pdf is 1, but I don't see how multiplying it by z^2 affects it. You don't need to prove it mathematically if you can explain it intuitively.
Thanks

Answer 1

f(z) =1/√(2*π) * exp(-(1/2)*z^2

is the pdf of the standard normal curve with mean 0 and s.d. 1

since the mean is zero,
the integral you have given is for the variance of the p.d.f which is 1.

if you want to actually evaluate the integral, change to twice the integral with 0 to ∞ limits since the integrand is even and use the substitution t = z^2 or z =√t and you end up with

{1/√(2*π) } * 2* gamma(1/2)

= 1

since gamma(1/2) = √π

Answer 2

Do an integration by parts with u=z and dv=zexp(-(1/2)z^2). You will get an integral that doesn't have the z^2 multiplied by the exponential. Now use what you know about the normal distribution. This is the trick for all even powers of z multiplied by the exponential.

As for why the integral for the normal distribution gives 1, it is easiest to multiply it by itself with two different variables and convert the resulting double integral into polar coordinates.

Answer 3

Good luck, Bud!!

Answer 4

oh god not those!!! bye bye!