91Ó°ÊÓ

The moment generating function of a uniform random variable that is given in Table

If $X~U(a,b)$

Then $M_x\left(t\right)=E\left[e^{tx}\right]$

$={∫}_a^b e^{tx}\frac{1}{b−a}dx$

$=\frac{1}{b−a}{\left[\frac{e^{tx}}{t}\right]}_a^b$

$=\frac{e^{bt}−e^{at}}{t(b−a)}$

Then it has been verified

Now,

${\mathrm{M}}_{\mathrm{x}}\left(\mathrm{t}\right)=\frac{1}{(b−a)}\left[\frac{e^{bt}−e^{at}}{t}\right]$

$=\frac{1}{(b−a)}\left[\frac{\left(1+bt+\frac{(bt{)}^2}{2!}+\frac{(bt{)}^3}{3!}+…….\right)−\left(1+at+\frac{(at{)}^2}{2!}+\frac{(at{)}^3}{3!}+……\right)}{t}\right]$

$=\frac{1}{(b−a)}\left[\frac{(b−a)t}{t}+\frac{\left(b^2−a^2\right)t^2}{2!t}+\frac{\left(b^3−a^3\right)t^3}{3!t}+……\right]$

$=1+\frac{(b+a)}{2}t+\frac{\left(b^2+a^2+ab\right)}{6}{t}^2+………$

Now,

${\mathrm{M}}_{\mathrm{x}}^{â‹\dots }\left(t\right)=\frac{(b+a)}{2}+\frac{\left(b^2+a^2+ab\right)}{6}2t+………$

${\mathrm{M}}_{\mathrm{X}}^{∗}\left(\mathrm{t}\right)=\frac{b^2+a^2+ab}{3}+o\left(t\right);o\left(t\right)=$Terms having higher power of t

So,

$\mathrm{E}\left(\mathrm{X}\right)={\left.{\mathrm{M}}_{\mathrm{X}}^{\mathrm{′}}\left(\mathrm{t}\right)\right|}_{t=0}=\frac{(b+a)}{2}$

$\mathrm{E}\left({\mathrm{X}}^2\right)={\left.{\mathrm{M}}_{\mathrm{X}}^{∗}\left(t\right)\right|}_{t=0}=\frac{\left(b^2+a^2+ab\right)}{3}$

$⇒V\left(X\right)=E\left(X^2\right)−\left[E\right(X){]}^2$

$=\frac{b^2+a^2+ab}{3}−\frac{b^2+a^2+2ab}{4}$

$=\frac{4\left(b^2+a^2+ab\right)−3\left(b^2+a^2+2ab\right)}{12}$

$=\frac{(b−a{)}^2}{12}$

It has been verify the formula for the moment generating function of a uniform random variable that is given in Table as$\frac{(b−a{)}^2}{12}$.