Question 3.4: Let X be a r.v. with p.d.f. f (x) = e^−x, x > 0. Then: (i......
Let X be a r.v. with p.d.f. f(x)=e^{-x},\;x\gt 0. Then:
(i) Find the m.g.f. M_{X}(t) for the t’s for which it is finite.
(ii) Using M_{X}, obtain the quantities: E X,E X^{2},\,\mathrm{and}\,V\!\!ar(X).
(iii) If the r.v. Y is defined by:~Y=2-3X, determine M_{Y}(t) for the t’s for which it is finite.
(i) By Definition 3,
M_{X}(t)=E e^{t X}=\int_{0}^{\infty}e^{t x}\cdot e^{-x}d x=\int_{0}^{\infty}e^{-(1-t)x}\,d x
=-\frac{1}{1-t}e^{-(1-t)x}|_{0}^{\infty}\;\;\;\;\mathrm{(provided}\;t\neq1)
=-{\frac{1}{1-t}}(0-1)={\frac{1}{1-t}}\quad({\mathrm{provided}}\,1-t\gt 0{\mathrm{~or~}}t\lt 1).
Thus, M_{X}(t)={\frac{1}{1-t}},\,t\lt 1.
(ii) By (13),
\left.\frac{d}{d t}M_{X}(t)\right|_{t=0}=E X~~\mathrm{~and~}~~\left.\frac{d^{n}}{d t^{n}}M_{X}(t)\right|_{t=0}=E X^{n},~~~~n=2,3,….\qquad (13)
\frac{d}{d t}M_{X}({t})|_{t=0}=\frac{d}{d t}(\frac{1}{1-t})|_{t=0}=\frac{1}{(1-t)^{2}}|_{t=0}=1=E X,\frac{d^{2}}{d t^{2}}M_{X}(t)|_{t=0}=\frac{d}{d t}(\frac{1}{(1-t)^{2}})|_{t=0}=\frac{2}{(1-t)^{3}}|_{t=0}={2}=E X^{2}, so that, by (9),
V\!\!ar(X)=E X^{2}-(E X)^{2}.\qquad\qquad\qquad\qquad(9)V\!\!ar(X)=2-1^{2}=1.
(iii) By (12),