Question 6.3: (Distribution for a linear transformation of a random variab......
(Distribution for a linear transformation of a random variable)
Let X be a continuous random variable with density function ƒ and let
Y = aX + b,
be another random variable that arises as a linear transformation of X; here, a > 0 and b are given real numbers.
(i) Find an expression for the density f_Y of Y in terms of ƒ.
(ii) Apply the result in (i) to the case when the density of X is
Let F and F_Y be the distribution functions of X and Y, respectively.
(i) We have
F_{Y}(y)=P(Y\leq y)=P(a X+b\leq y)=P\left(X\leq{\frac{y-b}{a}}\right)using, for the last step, the fact that a > 0. Thus, we have
F_{Y}(y)=F\left({\frac{y-b}{a}}\right)so that, upon differentiating both sides with respect to y, we derive
f_{Y}(y)=F_{Y}^{\prime}(y)=\left(F\left({\frac{y-b}{a}}\right)\right)^{\prime}=F^{\prime}\left({\frac{y-b}{a}}\right)\left({\frac{y-b}{a}}\right)^{\prime}=f\left({\frac{y-b}{a}}\right)\cdot{\frac{1}{a}}.Thus, we have shown that the two densities ƒ and f_Y are related by
f_{Y}(y)={\frac{1}{a}}\cdot f\left({\frac{y-b}{a}}\right).(ii) For ƒ as specified, we have
f(x)= \begin{cases} 0, \qquad x \lt 0,\\ \mathrm{e}^{-x}, \quad x \geq 0. \end{cases}Therefore, we see that the density f_Y will take the value zero if we have
{\frac{y-b}{a}}\lt 0,i.e. if y < b, while f_Y will take the value
{\frac{1}{a}}\cdot f\left({\frac{y-b}{a}}\right)={\frac{1}{a}}\cdot\mathrm{e}^{-(y-b)/a}for values of y greater than or equal to b. We thus see that f_Y is given by
f_Y(y)= \begin{cases} 0, \qquad \qquad \quad ~~ y \lt b,\\ {\frac{1}{a}}\cdot\mathrm{e}^{-(y-b)/a}, \quad y \geq b. \end{cases}In Figure 6.5, the probability density functions ƒ(x) and f_Y (y) are displayed graphically.
