Question

Let X be a random variable with pdf
f( x) = theta e ^(- theta x); theta > 0, x greater than equal to 0.
Find the moment generating function of X,
and hence find first three moments about
origin.

Answer 1

Solution

$$M(t) = E(exp(tx))$$

$$=\int exp(tx) \theta exp(-\theta x) dx$$

$$=\theta\int exp(-x(\theta - t)) dx$$

Integrating the above function from 0 to infinity yields:

$$\theta /( \theta - t)$$

Which is the mgf of f(x)

Answer: M(t)=(θ)/(θ−t)

1st moment is obtained by differentiating M(t) with respect to t and setting t to 0:

By division rule, we obtain:

$$dM(t) /(dt) = \theta / (\theta - t)^2$$

Setting t=0 yields:

$$\theta / (\theta)^2 = 1/ \theta$$

Which is the 1st moment

2nd moment is obtained by finding the second derivative of the mgf with respect to t and setting t to 0:

By division rule, we obtain:

$$(dM(t))^2/ d^2t = 2 \theta / (\theta - t)^3$$

as the second derivative.

Setting t to 0 yields:

$$2\theta / (\theta)^3 = 2/ \theta ^2$$

Which is the second moment

The 3rd moment is similarly obtained by getting the third derivative of the mgf with respect to t and setting t to 0:

By division rule, we obtain

$$(dM(t)) ^3/ dt^3 = 6\theta /(\theta-t) ^4$$

as the third derivative.

Setting t to 0 yields :

$$6\theta / \theta ^4 = 6 /\theta^3$$

Which is the third moment of f(x)