Derive the MGF of binomial distribution and hence finds it's mean and variance.

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written 6.2 years ago by

teamques10 ★ 69k

MGF for binomial distribution

By defination, MGF about origin is

$ M_o(t) = E (e^{tx}) = \sum{p_ie^{txi}}$

$ = \sum{{{^h}_c}_xp^xq^{n-x}e^{tx}}$

$ = \sum{{{^h}_c}_xq^{n-x}(pe^t)^x}$

$ M_o(t) = (q + pe^t)^n -------------1$

To find mean, differentiate $ M_o(t)$ and put t = 0

$E(X) = \frac{d}{dt} [M_o(t)]_{t=0} =\frac{d}{dt} [(q + pe^t)^n]_{t=0} $

$ =[n(q + pe^t)^{n-1}(0 + pe^t)]_{t=0} $

$E(X) = [npe^t(q + pe^t)^{n-1}]_{t=0} $

$E(X) = np [e^t(q + pe^t)^{n-1}]_{t=0} $

$E(X) = np [e^0(q + pe0)^{n-1}]_{t=0} $

$E(X) = np [q + p]^{n-1} ----\rightarrow e^0 = 1$

since q + p =1

$ \therefore np[1]^{n-1}$

$ \therefore E(X) = np = mean ---------2 $

$To find variance i.e \sigma^2 = E(X)^2- [E(X)]^2 -----------3 $

To find $ [E(X)]^2$ differentiate 1 twice w.r.t 't' and put t = 0

$ [E(X)]^2 = \frac{d^2}{dt^2} [M_o(t)]_{t=0} =\frac{d^2}{dt^2} [(q + pe^t)^n]_{t=0} $

$ =\frac{d}{dt} [n(q + pe^t)^{n-1}(0 + pe^t)]_{t=0} $

$E(X) = \frac{d}{dt} [np[e^t(q + pe^t)^{n-1}]]_{t=0} $

$ = [np[e^t(n-1)(e^t(q + pe^t)^{n-1-1})](0 + pe^t) + (e^t(q + pe^t)^{n-1}) ]_{t=0} $

Put t= 0

$ = [np[e^0(n-1)(e^0(q + pe^0)^{n-2})](pe^0) + (e^0(q + pe^0)^{n-1}) ] $

$ = [np[(n-1)(q + p)^{n-2}](p) + (q + p)^{n-1} ]-------\ but \ q+p=1 $

$ = np[(n-1)(1)^{n-2}](p) + (1)^{n-1} $

$ = np[(n-1)(1)](p) + (1) $

$ = np-p+1$

$ = np[np + (1-p)] ----------[since q= 1-p] $

$ = np [np + q] $

$E(X^2) = n^2p^2 + qnp -----------4 $

put 2 & 4 in 3

$ \sigma^2 = n^2p^2 + qnp - [np]^2 $

$ \sigma^2 = n^2p^2 + qnp - n^2p^2 $

$ \sigma^2 = variance = npq $

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