Answer to Question #258860 in Statistics and Probability for scholar

Question

Let x be a continuous random variable having density function fx(x)= 1/2 e^-|x| dx, - infinity< x<infinity. show that Mx(t)= (1-t^2)^-1 , -1<x<1

Answer 1

for $-1<x<1$ :

$M_X(t)=E(e^{tx})=\displaystyle{\int^{1}_{-1}}e^{tx}f_X(x)dx=\frac{1}{2}\displaystyle{\int^{0}_{-1}}e^{tx}\cdot e^{x}dx+\frac{1}{2}\displaystyle{\int^{1}_{0}}e^{tx}\cdot e^{-x}dx=$

$=\frac{e^{x(t+1)}}{2(t+1)}|^0_{-1}+\frac{e^{-x(1-t)}}{2(t-1)}|^{1}_{0}=\frac{1}{2(t+1)}-\frac{e^{-(t+1)}}{2(t+1)}+\frac{e^{-1(t+1)}}{2(t-1)}-\frac{1}{2(t-1)}=\frac{1}{1-t^2}$