Subject: Applied Mathematics 2

Topic: Probability Distribution

Difficulty: High

By definition moment generating function m.g.f. about origin is

M$_0$(t) = E(e$^{tX}$) = ∑p$_i$e$^{tx_i}$ = ∑$\frac{(e^{-m}×m^x)}{x!}$.e$^{tx}$ = e$^{-m}$ ∑$\frac{(me^t )^x}{x!}$ = e$^{-m}$ .e$^{me^t}$

[Note = ∑$\frac{k^x}{x!}$ = 1 + k + $\frac{k^2}{2!}$ + $\frac{k^3}{3!}$....................................]

Differentiating M$_0$(t) and putting t =0 ,we get the required moments

Now, [$\frac{\mathrm{d} }{\mathrm{d} t} M_0(t)]_{t=0}$ = e$^{m(e^t-1)}$.me$^t$

[$\frac{\mathrm{d} }{\mathrm{d} t} M_0(t)]_{t=0}$ = m

∴ mean = μ$_1'$= m

[$\frac{\mathrm{d^2} }{\mathrm{d} t^2} M_0(t)]_{t=0}$ = $m[ e^t .e^{m(e^t-1)}. me^t + e^{m(e^t-1)}. e^t]$

[$\frac{\mathrm{d^2} }{\mathrm{d} t^2} M_0(t)]_{t=0}$ = m(m+1) = μ$_2'$

Therefore variance = μ$_2'$ - [μ$_1'$]$^2$ = m$^2$ +m - m$^2$ = m