91Ó°ÊÓ

X has the pdf:

\(f\left( x \right) = \left\{ {\begin{aligned}{{}{}}{{x^{ - 2}};{\rm{ if }}x > 1}\\{0;{\rm{ otherwise}}}\end{aligned}} \right.\)

\(\begin{aligned}{}\psi \left( t \right) = E\left( {{e^{tx}}} \right)\\ = \int\limits_1^\infty {{e^{tx}}{x^{ - 2}}dx} \\ = {x^{ - 2}}\int\limits_1^\infty {{e^{tx}}dx} - \int\limits_1^\infty { - 2{x^{ - 3}}\left( {\int\limits_1^\infty {{e^{tx}}dx} } \right)dx} .\end{aligned}\)

Now, the integral \(\int\limits_1^\infty {{e^{tx}}dx} \) is undefined for \(t > 0\).

Hence, the m.g.f. of X does not exist for \(t > 0\).

Now consider that \(t \le 0\). Let \(t = - a;a \ge 0\). Then,

\(\begin{aligned}{}\int\limits_1^\infty {{e^{tx}}dx} = \int\limits_1^\infty {{e^{ - ax}}dx} \\ = - \frac{1}{a}\left( {{e^{ - ax}}} \right)_1^\infty \\ = \frac{1}{{a{e^a}}}.\end{aligned}\)

Using the above value for the integral, evaluate the m.g.f. of X as shown below:

\(\begin{aligned}{}\psi \left( t \right) = {x^{ - 2}}\int\limits_1^\infty {{e^{tx}}dx} - \int\limits_1^\infty { - 2{x^{ - 3}}\left( {\int\limits_1^\infty {{e^{tx}}dx} } \right)dx} \\ = {x^{ - 2}}\int\limits_1^\infty {{e^{ - ax}}dx} - \int\limits_1^\infty { - 2{x^{ - 3}}\left( {\int\limits_1^\infty {{e^{ - ax}}dx} } \right)dx} {\rm{ }}\left( {{\rm{Since, }}t = - a;a \ge 0} \right)\\ = {x^{ - 2}}\frac{1}{{a{e^a}}} - \int\limits_1^\infty { - 2{x^{ - 3}}\frac{1}{{a{e^a}}}dx} \\ = \frac{{{x^{ - 2}}}}{{a{e^a}}} + \frac{2}{{a{e^a}}}\left( {\frac{{{x^{ - 2}}}}{{ - 2}}} \right)_1^\infty \\ = \frac{{{x^{ - 2}}}}{{a{e^a}}} + \frac{1}{{a{e^a}}}\\ = - \frac{{1 + {x^{ - 2}}}}{{t{e^t}}}\left( {{\rm{replacing }}a{\rm{ by }}t} \right)\end{aligned}\)

Therefore, for \(t \le 0\), the m.g.f. of Xexists and is finite as shown below.

\(\psi \left( t \right) = - \frac{{1 + {x^{ - 2}}}}{{t{e^t}}}.\)