# Quantum statistics expectation value

1. Sep 4, 2011

### bearries

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 quantum class.

3. I have absolutely no idea where to begin.

2. Sep 4, 2011

### vela

Staff Emeritus
The expectation value of a function of a random variable X is given by
$$E[g(X)] = \int_{-\infty}^\infty g(x) f(x)\,dx$$
Do you know how to integrate a Gaussian?

3. Sep 5, 2011

### bearries

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).

4. Sep 5, 2011

### vela

Staff Emeritus
5. Sep 5, 2011

### bearries

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 this math before so I apologize for how slow I am.

6. Sep 5, 2011

### vela

Staff Emeritus
Tell us what integral you need to calculate to find <x>.

7. Sep 5, 2011

### bearries

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.

8. Sep 5, 2011

### vela

Staff Emeritus
g(X) is what you're finding the expectation value of. In this case, g(X)=X.

9. Sep 5, 2011

### bearries

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 with just some sort of u substitution but the closest I can bring the integral i have to that is $$e^{-μ^2}\int_{-\infty}^\infty xe^{-x^2/2σ^2}e^{xμ/σ^2}\,dx$$ So I doubt it's that easy and probably requires some sort of harder method. How should I go about this?

10. Sep 5, 2011

### vela

Staff Emeritus
Try the substitution u = x-mu.

11. Sep 5, 2011

### bearries

okay so to double check so far:
$$E[x] = \int_{-\infty}^\infty xe^{(-(x-μ)^2/2σ^2)}\,dx$$
$$u=x-μ$$$$du=dx$$
$$E[x] = \int_{-\infty}^\infty ue^{(-u^2/2σ^2)}\,du + μ\int_{-\infty}^\infty e^{(-u^2/2σ^2)}\,du$$
These look like 2 of the examples on wikipedia but if I just solve them I get something like
$$σ^5(2π)^{1/2} + μσ(2)^{1/2}(-1/2)!$$
is that right? a concern i have is that I never had to substitute x-μ=u back in

12. Sep 5, 2011

### vela

Staff Emeritus
You should find the first integral is 0. You can integrate it using a substitution or simply note that the integrand is an odd function.

What's (-1/2)! equal to?

13. Sep 5, 2011

### bearries

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 something minus the gamma function of -.5 which doesn't seem to go anywhere.

14. Sep 5, 2011

### vela

Staff Emeritus
Look at the very first integral on that Wiki page I linked to earlier.

15. Sep 6, 2011

### bearries

Thanks for the help! I finished and got through it in time for class. Doubled checked with a peer.