Question

4. For the random variable X with the pdf f(x) = 1 defined on [0,1]:

a. Obtain the pdf and cdf for the transformation Y=-2lnX
b. P(Y > e) =?

Answer 1

a) The cumulative distribution function of Y is

$F_Y(y) =P(Y\leq y) = P(-2ln X \leq y) = P( X > e^{- \frac{y}{2}})$

$\Rightarrow F_Y(y) =1 - P( X \leq e^{-\frac{y}{2}})$

$\Rightarrow F_Y(y) =1 - F_X(e^{-\frac{y}{2}})$

$\Rightarrow F_Y(y) =1 - e^{-\frac{y}{2}},y>0$

The pdf of Y will be

$f_Y(y) = \frac{\partial }{\partial y} F_Y(y)= \frac{\partial }{\partial y}[1 - e^{-\frac{y}{2}}] = \frac{1}{2}e^{-\frac{y}{2}},y>0$

$\therefore f_Y(y) = \frac{1}{2}e^{-\frac{y}{2}},y>0$

b) $P(Y>e) = 1 - P(Y\leq e) = 1 - F_Y(e) = 1 - e^{-\frac{e}{2}} = 0.743119$