91Ó°ÊÓ

Let X be a given random variable.

The m.g.f. is given by,

\(\psi \left( t \right) = {e^{{t^2} + 3t}}; - \infty < t < \infty \)

\(\psi \left( t \right) = {e^{{t^2} + 3t}}\)

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

\(\psi '\left( t \right) = \left( {2t + 3} \right){e^{{t^2} + 3t}}\)

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

\(\psi ''\left( t \right) = {\left( {2t + 3} \right)^2}{e^{{t^2} + 3t}} + 2{e^{{t^2} + 3t}}\)

One can get the mean value of X finding\(\psi '\left( 0 \right)\).

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

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

\(\begin{align}{\sigma ^2} &= \psi ''\left( 0 \right) - {\mu ^2}\\ &= {\left( {2 \times 0 + 3} \right)^2}{e^0} + 2{e^0} - {3^2}\\ &= 9 + 2 - 9\\ &= 2\end{align}\)

Therefore, the mean value of X is 3 and the variance is 2.