Question

Find the moment generating function of random variable which is uniformly distributed over
(–a, a). Evaluate E (X2n).

Answer 1

Random variable $\xi$ is uniformly distributed over $(-a, a)$.

Moment generating function equals to

$$M_{\xi}(t) = E\left(e^{t\xi}\right) = \int_{-\infty}^{\infty} e^{tx} f_{\xi}(x) dx = \int_{-a}^{a} \frac{e^{tx}}{2a} dx = \left. \frac{1}{2a} \frac{e^{tx}}{t} \right|_{x = -a}^{x = a} = \frac{1}{2at} \left(e^{ta} - e^{-ta}\right)$$ $$E(\xi^{2n}) = M_{\xi}^{(2n)}(0)$$

For arbitrary $a, b$ random variable $\eta$ uniformly distributed on $(c, d)$ has such moment:

$$m_k = \frac{1}{k + 1} \sum_{i=0}^{k} c^i d^{k-i}$$

In our case $k = 2n; c = -a; d = a$;

Thus

$$E(\xi^{2n}) = \frac{1}{2n + 1} \left(\sum_{i=0}^{2n} (-a)^i a^{2n-i}\right) = \frac{a^{2n}}{2n + 1} \sum_{i=0}^{2n} (-1)^i = \frac{a^{2n}}{2n + 1}$$