Question

Question 4: Find the variance of the exponential function using the moment generating function. f(x; 1) = {xe-x x>0 10 otherwise

Answer 1

Step 1:

Probability Density Function of Exponential Distribution is:

$f(x)=\lambda e^{-\lambda x}$,    x $\geq$ 0

       = 0, otherwise

The Moment Generating Function of Exponential Distribution is:

M(t) etter dc = e-3(1-1), dr 10

$=-\lambda \times \frac{e^{-x(\lambda -t)}}{\lambda -t}$

between limits 0 to $\infty$ .

Applying limits, we get:

$M(t)=\frac{\lambda }{\lambda -t}$

Step 2:

Differentiating M(t) we get:

$M'(t)=\frac{\lambda }{(\lambda -t)^{2}}$

So,

$E(X)=M'(0)=\frac{\lambda }{\lambda^{2}}=\frac{1}{\lambda }$

Step 3:

Differentiating , we get:

$M''(t)=\frac{2\lambda }{(\lambda -t)^{3}}$

So,

$E(X^{2})=M''(0)=\frac{2\lambda }{\lambda ^{3}}=\frac{2}{\lambda ^{2}}$

Step 4:

Variance of of Exponential Distribution is:

$Var(X)=E(X^{2})-(E(X))^{2}=\frac{2}{\lambda ^{2}}-\frac{1}{\lambda ^{2}}=\frac{1}{\lambda ^{2}}$

So,

Answer is:

$\frac{1}{\lambda ^{2}}$