Question

Suppose that the continuous random variable x has the probability density function

f(x) = 1/2e−|x| , −∞ < x < ∞

find the m.g.f of x and use it to find E(x) and V ar(x).

Answer 1

$$M(t)=\displaystyle\int_{-\infin}^{\infin}e^{tx}f(x)dx$$

$$=\dfrac{1}{2}\displaystyle\int_{-\infin}^{0}e^{tx}e^xdx+\dfrac{1}{2}\displaystyle\int_{0}^{\infin}e^{tx}e^{-x}dx$$

$$=\dfrac{1}{2(t+1)}[e^{(t+1)x}]\begin{smallmatrix} 0 \\ -\infin \end{smallmatrix}+\dfrac{1}{2(t-1)}[e^{(t-1)x}]\begin{smallmatrix} \infin \\ 0 \end{smallmatrix}$$

$$=\dfrac{1}{2(t+1)}(1-0)+\dfrac{1}{2(t-1)}(0-1)$$

$$=\dfrac{1}{1^2-t^2}$$

$$M(t)=\dfrac{1}{1^2-t^2}$$

Differentiate MGF with respect to t.

$$M'(t)=\dfrac{2t}{(1^2-t^2)^2}$$

Put $t=0$

$$E(X)=M'(0)=0$$

Find $M''(t)$

$$M''(t)=2\cdot\dfrac{(1^2-t^2)^2-2t(1^2-t^2)(-2t)}{(1^2-t^2)^4}$$

$$=2\cdot\dfrac{1^2+3t^2}{(1^2-t^2)^3}$$

Put $t=0$

$$E(X^2)=M''(0)=2$$

$$Var(X)=E(X^2)-(E(X))^2$$

$$=2-(0)^2=2$$

$M(t)=\dfrac{1}{1^2-t^2}$

$E(X)=0$

$Var(X)=2$