Suggestion for answer to Question 6

Open nathanagmon opened this issue 3 years ago • 0 comments

Similar to generating functions, another useful trick for calculating moments of various distributions is to use Feynman's integral trick (differentiating under the integral sign). The main idea is to introduce a fictitious parameter and rewrite the integrand as some derivative with respect to this parameter. (Note that this trick can also be used when evaluating expectation values for discrete distributions). Let me illustrate this for moments of the Gaussian distribution, which appear in the primer. Consider the integral

$https://latex.codecogs.com/svg.image?\bg{white}I = \int_{-\infty}^{\infty} x^{2k} e^{- x^2}dx$

In order to evaluate the above expression, we introduce a parameter in the probability density function

$https://latex.codecogs.com/svg.image?\bg{white}I(a) = \int_{-\infty}^{\infty} x^{2k} e^{-ax^2}dx$

Notice that this integral can be rewritten as

$https://latex.codecogs.com/svg.image?\bg{white}I(a) = \frac{\partial^{k}}{\partial a^{k}}\int_{-\infty}^{\infty} e^{-ax^2}dx$

This is just a Gaussian integral, and so we find

$https://latex.codecogs.com/svg.image?\bg{white}I(a) = \frac{\partial^{k}}{\partial a^{k}} \sqrt{\frac{\pi}{a}} = \frac{(2k-1)!!}{2^{k}}\sqrt{\frac{\pi}{a^{2k+1}}}$

Finally, we recover the desired result by setting the parameter to 1,

$https://latex.codecogs.com/svg.image?\bg{white}I = I(1)= \frac{(2k-1)!!}{2^{k}}\sqrt{\pi}$

Jul 01 '22 18:07 nathanagmon