Question

Find the moment generating function,mean and variance when the pdf of x is defined by `f(x)=(1/3)(2/3)^x. ; x=0,1,2,3,....` ?

Answer 1

• `M_X(t)=1/(3-2e^t)`
• mean=2
• variance = 10

Explanation:

The momentum generating function is

`M_X(t) = E(e^{tX})`

so, for the given pdf, we have

`M_X(t) = sum_{x=0}^oo e^{tx}f(x)`
`qquadqquadquad = sum_{x=0}^oo e^{tx}1/3(2/3)^x`
`qquadqquadquad = 1/3 sum_{x=0}^oo (2/3e^t)^x`
`qquadqquadquad = 1/3 1/(1-2/3e^t)`
`qquadqquadquad = 1/(3-2e^t)`

The mean is

`m_1 = M_X^((1))(0) = ((2e^t)/(3-2e^t)^2)_{t=0}=2`

and variance

`m_2 = M_X^((2))(0) = ((2e^t(3+2e^t))/(3-2e^t)^3)_{t=0}=10`