1. Calculate the expected value and variance of the exponential distribution using the moment generating function.

Consider the binomial function

\[f_x(x) = \lambda e^{-\lambda x} \ if\ x>0\]

The Moment generating function is:

\[M_x(t) = \int_{0}^\infty\ e^{tx} \lambda e^{-\lambda x} dx \\ = \lambda \int_{0}^\infty\ e^{(t-\lambda)x}dx \\= \lambda \frac{1}{t-\lambda }e^{(t-\lambda)x} |_0^\infty\]

\[\lambda (0-\frac{1}{t-\lambda }) = \frac{\lambda}{\lambda -t}\]

Differentiate Moment generating function:

\[\frac{\text{d}M_x(t)}{\text{d}t} = \frac{\lambda}{(\lambda -t)^2}\]

Expectation - t = 0:

\[ \frac{\lambda}{(\lambda -0)^2} = \frac{\lambda}{(\lambda)^2} = \frac{1}{\lambda}\]

Differentiate second Moment generating function:

\[\frac{\text{d}^2M_x(t)}{\text{d}t^2} = \frac{2\lambda}{(\lambda-t)^3}\]

When t = 0:

\[ E(x^2) = \frac{2\lambda}{(\lambda-0)^3} = \frac{2\lambda}{(\lambda)^3} = \frac{2}{\lambda^2}\]

\[ V(x) = E(x^2) - E(x)^2 \\= \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}\]