written 5.0 years ago by | • modified 5.0 years ago |
Solution:
$M_0(t) = E(e^{tx})$
=$\sum pi e^{tx_i}$
= $\sum (\frac{1}{2})^x.e^{tx}$
=$\sum (\frac{1}{2^x}).e^{tx}$
=$\sum (\frac{e^t}{2})^x$
since =1,2,3,...
=$\frac {e^t}{2}$+$(\frac {e^t}{2})^2$+($\frac {e^t}{2})^3$
$\frac{e^t}{2}\{1+(\frac{e^t}{2})+(\frac{e^t}{2})^2$
=$\frac {e^t}{2}[\frac{1}{(1-\frac {e^t}{2})}]$
= $\frac {e^t}{2}[\frac{1}{\frac{2-e^t}{2}}]$
= $\frac {e^t}{2}[\frac{2}{2-e^t}]$
=$m.g.f.= M_0(t)=[\frac{e^t}{2-e^t}]$-----(1)
mean , $\mu_1'=[\frac{d}{dt} M_0(t)]_{t=0}$
=$\mu_1'=[\frac{(2-e^t)(e^t)-e^t(0-e^t)}{(2-e^t)^2}]_{t=0}$
=$\mu_1'=[\frac{(2-e^0)(e^0)-e^0(0-e^0)}{(2-e^0)^2}]$
=$\mu_1'=[\frac{(2-1)(1)-1(0-1)}{(2-1)^2}]$
=$\mu_1' = mean =2 $
variance = $\mu_2'-\mu_1'^2$-----(2)
to find $\mu_2'$,
$\mu_2'=[\frac {d^2}{dt^2} M_0(t)]_{t=0}$
$\mu_2'=\frac {d^2}{dt^2}[\frac {e^t}{2-e^t}]_{t=0}$
=$\mu_2'=\frac{d}{dt}[\frac{(2-e^t)(e^t)-e^t(0-e^t)}{(2-e^t)^2}]_{t=0}$
=$\mu_2'=\frac{d}{dt}[\frac{(2e^t-e^{2t}+e^{2t}}{(2-e^t)^2}]_{t=0}$
=$\mu_2'=\frac{d}{dt}[\frac{2e^t}{(2-e^t)^2}]_{t=0}$
==$\mu_2'=[\frac{(2-e^t)^2(2e^t)-(2e^t.2(2-e^t)(0-e^t)}{(2-e^t)^4}]_{t=0}$
=$(2-e^t)[\frac {2(2-e^t)e^t-4e^t(-e^t)}{(2-e^t)^4}]_{t=0}$
=$[\frac {2(2-e^t)e^t+4e^{2t}}{(2-e^t)^3}]$
=$[\frac {2(2-1)1+4{1}}{(2-1)^3}]$
$\mu_2'=6$
variance = $\mu_2'-\mu_1'^2$
=$6-2^2$
Variance=2