Question

Use the Maclaurin's series expansion of the moment generating function of the standard normal distribution to show that

(a) μᵣ=0 when r is odd;
(b) μᵣ= r!/2ʳ/²(r/2)!when r is even.

Answer 1

The Maclaurin series of the moment generating function of the standard normal distribution demonstrates that the r-th moment is zero for odd r, and follows the formula r!/2^(r/2)(r/2)! for even r.

Step-by-step explanation:

Moments of the Standard Normal Distribution

To show that μr=0 when r is odd and μr= r!/2r/2(r/2)! when r is even using Maclaurin series expansion, we refer to the moment generating function (MGF) of the standard normal distribution, which is defined as M(t)=et2/2. The r-th moment of the distribution, μr, is the coefficient of tr/r! in the Maclaurin series expansion of M(t).

For r odd, since the MGF of a standard normal distribution is an even function, its series expansion contains only even powers of t. Thus, all coefficients corresponding to odd powers, including the r-th moment when r is odd, are zero. Hence, μr = 0 when r is odd.

For r even, the coefficient of the tr term in the expansion of M(t) can be calculated using the derivatives of M(t) at t=0, which yields μr= r!/2r/2(r/2)! for even r. The exponent in the factorial comes from the r/2 times each factor of t2 contributes two derivatives in the series expansion.