Okay I see that. I don't understand the gamma function but I take it from wikipedia that (-1/2)! is equal to Γ(1/2)
So I guess it comes down to Γ(1/2)μσ(2)^(1/2) which is
μσ(2)^{1/2}\int_{0}^\infty t^{-1/2}e^{-t}\,dt
I also don't know where to go about this integral, by parts I get left with...
okay so to double check so far:
E[x] = \int_{-\infty}^\infty xe^{(-(x-μ)^2/2σ^2)}\,dx
u=x-μdu=dx
E[x] = \int_{-\infty}^\infty ue^{(-u^2/2σ^2)}\,du + μ\int_{-\infty}^\infty e^{(-u^2/2σ^2)}\,du
These look like 2 of the examples on wikipedia but if I just solve them I get something like...
Oh! That makes a lot more sense.
E[x] = \int_{-\infty}^\infty xe^{(-(x-μ)^2/2σ^2)}\,dx
is the integral I'm left with. I see it's similar to the an example on the gaussian integral wiki page where \int_{-\infty}^\infty x^{2n}e^{-ax^2}\,dx
So I figured I'd go off and see if I could solve it...
So do I just take this
E[g(X)] = \int_{-\infty}^\infty g(x) f(x)\,dx
and use the function, f(x) = e^{(-(x-μ)^2/2σ^2)} as f(x) and would g(x) be the Gaussian function? Not really sure what g(x) is unless that guess is correct.
Okay so the Gaussian integral is just the integral from -to+infinity of e^(-x^2). I see what they did through to get the answer. Where does the Gaussian fit into that other integral you listed?
Thanks for the help. I was pretty clueless when this was presented in class because I haven't seen...
Not exactly. My professor ran very very quickly through the gaussian integral once before but I don't quite have the math background to have taken it in (in multivariable calculus currently).
1. What is the expectation value, <x>, for the
given distribution over the interval from – to + infinity of the function: f(x)=e^(-.5(x-mu)^2(sigma^-2))
2. This is a statistics problem i think. I just need to know how this type of problem is worked out because it is relevant to my...