Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : MAY 2014

For Poisson distribution, $p(X=x)=\dfrac {e^{-m}m^x}{x!}$ , where m is the Poisson parameter.

By definition moment generating function about origin $=E(e^{tx})\\ =\sum\limits_{x=0}^{\infty}P_te^{tx}\\ =\sum\limits_{x=0}^{\infty}\dfrac {e^{-m}m^x}{x!}.e^{tx}\\ =e^{-m}\sum\limits_{x=0}^{\infty}\dfrac {(me^t)^x}{x!}\\ =e^{-m}.e^{-me^t}\Bigg[ \sum\limits_{x=0}^{\infty} \dfrac {a^x}{x!}=e^a\Bigg]\\ =e^{-m+me^t}\\ =e^{e^{-m}(1-e^t)}$

Now,

$\mu_1^1=\Bigg[\dfrac {d\mu}{dt^{\mu}}\Bigg]_{t=0}\\ =\Bigg[\dfrac {d\mu}{dt^{\mu}}e^{-m(1-e^t)}\Bigg]_{t=0} \\ \therefore \mu_1^1=\Bigg[\dfrac d{dt}e^{-m(1-e^t)}\Bigg]_{t=0}\\ =\Big[e^{-m(1-e^t)}.(-m)(0-e^t)\Big]_{t=0}\\ =e^{-m(1-1)}.(-m).(0-1)\\ =e^0.m \\ =m\\ \therefore \mu_2^1=\Bigg[\dfrac {d^2}{dt^2}e^{-m(1-e^t)}\Bigg]_{t=0}\\ =\Bigg[\dfrac d{dt}e^{-m(1-e^t)}.me^t\Bigg]_{t=0}\\ =\Big\{m\Big[e^{-m(1-e^t)}.e^t+e^t.e^{-m(1-e^t)}-m.(0-e^t)\Big]\Big\}_{t=0}\\ =m\Big[e^{-m(1-1^t)}.1+1.e^{-m(1-1)}.m.1\Big]\\ =m[1+m]\\ =m+m^2\\ \therefore Mean =\mu_1^1=m\\ \therefore Variance =\mu_2^1-(\mu_1^1)^2\\ =(m+m^2)-(m)^2\\ =m$