91Ó°ÊÓ

Let X follows a uniform distribution.

The p.d.f of X is

\(f\left( x \right) = \frac{1}{{b - a}};a < x < b\)

Now,

\(\begin{align}\psi \left( t \right) &= \int\limits_a^b {{e^{tx}}f\left( x \right)dx} \\ &= \int\limits_a^b {{e^{tx}}\frac{1}{{b - a}}dx} \end{align}\)

For,\(t \ne 0\),

\(\begin{align}\psi \left( t \right) &= \left. {\frac{{{e^{tx}}}}{{t\left( {b - a} \right)}}} \right|_a^b\\ &= \frac{{{e^{t\left( {b - a} \right)}}}}{{t\left( {b - a} \right)}}\\ &= \frac{{{e^{tb}} - {e^{ta}}}}{{t\left( {b - a} \right)}}\end{align}\)

\(\psi \left( 0 \right) = 1\)

Therefore, the m.g.f of X is:

\(\psi \left( 0 \right) = \left\{ {\begin{align}{}{\frac{{{e^{tb}} - {e^{ta}}}}{{t\left( {b - a} \right)}}{\rm{ for t}} \ne {\rm{0}}}\\{1{\rm{ for t = 0}}}\end{align}} \right.\)