Question

Solve the following problem :

If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, find variance of X.

Answer 1

Given, P[X = 1] = 0.4, P[X = 2] = 0.2,

e^–1 = 0.3678

For Poisson distribution,

X ~ P(m)

The p.m.f. of X is given by

P[X = x] = ${\displaystyle \frac{{\text{e}}^{-\text{m}}{\text{m}}^x}{x!}}$

Now,
P[X = 1] = ${\displaystyle \frac{{\text{e}}^{-\text{m}}{\text{m}}^1}{1!}}$ = me^-m

∴ 0.4 = me^-m               ...(i)

P[X = 2] = ${\displaystyle \frac{{\text{e}}^{-\text{m}}{\text{m}}^2}{2!}}$ = ${\displaystyle \frac{{\text{m}}^2{\text{e}}^{-\text{m}}}{2}}$

∴ 0.2 = ${\displaystyle \frac{{\text{m}}^2{\text{e}}^{-\text{m}}}{2}}$

∴ 0.4 = m² e^–m          ...(ii)

∴ ${\displaystyle \frac{0.4}{0.4}=\frac{{\text{m}}^2{\text{e}}^{-\text{m}}}{{\text{me}}^{-\text{m}}}}$     ...[From (i) and (ii)]

∴ m = 1

∴ X ~ P(1)

∴ Var (X) = m = 1.