Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : MAY 2015

For binomial distribution, P $(x =x) = \space ^n(xp^xq^{n-x})$

By definition, moment generating function about origin $M_0 (t) = E (e^{tx})$

$$=\sum\limits^n_{x=0}P_ie^{tx}$$

$=\sum\limits^n_{x=0} \space^n(_xp^xq^{n-x}e^{tx}\\ =\sum\limits^n_{x=0}\space ^n(_xpe^t)^xq^{n-x}\\ =(q+pe^t)^n \space \space \space \space \space \Big[\sum\limits^n_{x=0} \space ^n(_xa^xb^{n-x}=(a+b)^n\Big]\\ Now , \\ \mu_x^1=\Bigg[\dfrac {d^x}{dt^x}M_0(t)\Bigg]_{t=0}\\ \mu_1^1\Bigg[\dfrac d{dt}(q+pe^t)^n\Bigg]_{t=0}\\ =[n(q+pe^t)^n]_{t=0} \\ =n[(q+pe^t)^{n-1}pe^t]_{t=0}\\ =n[(q+pe^0)^{n-1}pe^0]\\ =n(q+p)^{n-1}p\\ =n.1^{n-1}.p \space \space \space \space [q+p=1]\\ =np\\ \mu_2^1=\Bigg[\dfrac {d^2}{dt^2}M_0(t)\Bigg]_{t=0}\\ =\Bigg[\dfrac {d^2}{dt^2}(q+pe^t)^{n}\Bigg]_{t=0}\\ =\Bigg[\dfrac d{dt}np(q+pe^t)^{n-1}.e^t\Bigg]_{t=0}\\ =[np[(q+pe^t)^{n+1}.e^t+e^t.(n-1)(q+pe^t)^{n-2}pe^t]]_{t=0}\\ =np[(q+p)^{n-1}.1+1.(n-1)(q+p)^{n-2}p.1]\\ =np[(1)^{n-1} + (n-1)(1)^{n-2}p] \hspace {1cm} [q+p=1] \\ =np[1+np-p]\\ =np[Q+np]\hspace {1cm} [1-p=q]\\ =npq +n^2p^2\\ Mean \space \space =\mu_1^1=np\\ Variance = \mu_2=\mu_2^1-(\mu_1^1)^2\\ =(npq+n^2p^2)-(np)^2\\ =npq$