91Ó°ÊÓ

The m.g.f of a random variable X is:

\(\psi \left( t \right) = \frac{1}{4}\left( {3{e^t} + {e^{ - t}}} \right); - \infty < t < \infty \)

\(\psi \left( t \right) = \frac{1}{4}\left( {3{e^t} + {e^{ - t}}} \right)\)

The first-order derivative of\(\psi \left( t \right)\)is

\(\psi '\left( t \right) = \frac{1}{4}\left( {3{e^t} - {e^{ - t}}} \right)\)

The second-order derivative of\(\psi \left( t \right)\)is

\(\psi ''\left( t \right) = \frac{1}{4}\left( {3{e^t} + {e^{ - t}}} \right)\)

We can get the mean value of X by computing\(\psi '\left( 0 \right)\).

\(\begin{align}\mu &= \psi '\left( 0 \right)\\ &= \frac{1}{4}\left( {3{e^0} - {e^{ - 0}}} \right)\\ &= \frac{1}{4}\left( {3 - 1} \right)\\ &= \frac{2}{4}\\ &= \frac{1}{2}\end{align}\)

Let,\({\sigma ^2}\)be the variance of X.

\(\begin{align}{\sigma ^2} &= \psi ''\left( 0 \right) - {\mu ^2}\\ &= \frac{1}{4}\left( {3{e^0} + {e^{ - 0}}} \right) - \frac{1}{4}\\ &= \frac{1}{4}\left( {3 + 1 - 1} \right)\\ &= \frac{3}{4}\end{align}\)

Therefore, the mean of random variable X is \(\frac{1}{2}\) and variance is \(\frac{3}{4}\).