Hello everyone,
It looked like there was still some confusion at the end of today's exercise session regarding the computation of the second moment in Ex. 2 a). Hope the information below helps to clarify things.
- Substitute, as before in the calculation of the first moment, u:= x - m
- Due to the binomial Theorem, we now end up with a sum of three terms corresponding to integrals over zero-mean Gaussian pdf's scaled by u^2, 2um and m^2, respectively
- Terms belonging to 2um and m^2 can be calculated as shown previously in the derivation of the first moment, respectively using the anti-symmetry of the integrand (hence the integral equals zero), and the fact that we integrate over a pdf (hence the integral equals one)
- Term scaled by u^2: integrate by parts, i.e., solve \int f(u) g'(u) du with f(u)=u and g'(u) =u e^{-u^2/(2 \sigma^2)}
- The step missing in today's presentation: in order to calculate the indefinite integral of g'(u), simply use the substitution s = -u^2/(2 \sigma^2) and hence ds = - u/ \sigma^2 du.
- Calculate the (now very simple) indefinite integral w.r.t. s and then resubstitute s = -u^2/(2 \sigma^2)
- Following these steps you should now be able to derive that the second moment equals \sigma^2 + m^2 !
