- #1
lizzyb
- 168
- 0
The question is: if X is an exponential random variable with parameter [tex]\lambda = 1[/tex], compute the probability density function of the random variable Y defined by [tex]Y = \log X[/tex].
I did [tex]F_Y(y) = P \{ Y \leq y \} = P \{\log X \leq y \} = P \{ X \leq e^y \} = \int_{0}^{e^y} \lambda e^{- \lambda x} dx = \int_{0}^{e^y} e^{- x} dx = -e^{-x} \Big |_0^{e^y} = -e^{-e^y} + 1 = 1 - e^{-e^y}[/tex]
so
[tex] \frac {d F_{|X|} (x) } {dx} = p(x) = e^x e^{-e^x}[/tex]
Unfortunately the answer isn't in the back of the book. Does that look okay?
I did [tex]F_Y(y) = P \{ Y \leq y \} = P \{\log X \leq y \} = P \{ X \leq e^y \} = \int_{0}^{e^y} \lambda e^{- \lambda x} dx = \int_{0}^{e^y} e^{- x} dx = -e^{-x} \Big |_0^{e^y} = -e^{-e^y} + 1 = 1 - e^{-e^y}[/tex]
so
[tex] \frac {d F_{|X|} (x) } {dx} = p(x) = e^x e^{-e^x}[/tex]
Unfortunately the answer isn't in the back of the book. Does that look okay?