Probability Density Function with an exponential random variable

[lizzyb]

The question is: if X is an exponential random variable with parameter $$\lambda = 1$$, compute the probability density function of the random variable Y defined by $$Y = \log X$$.

I did $$F_Y(y) = P \{ Y \leq y \} = P \{\log X \leq y \} = P \{ X \leq e^y \} = \int_{0}^{e^y} \lambda e^{- \lambda x} dx = \int_{0}^{e^y} e^{- x} dx = -e^{-x} \Big |_0^{e^y} = -e^{-e^y} + 1 = 1 - e^{-e^y}$$

so

$$\frac {d F_{|X|} (x) } {dx} = p(x) = e^x e^{-e^x}$$

Unfortunately the answer isn't in the back of the book. Does that look okay?

 

[HallsofIvy]

The question is: if X is an exponential random variable with parameter $$\lambda = 1$$, compute the probability density function of the random variable Y defined by $$Y = \log X$$.

I did $$F_Y(y) = P \{ Y \leq y \} = P \{\log X \leq y \} = P \{ X \leq e^y \}$$$$= \int_{0}^{e^y} \lambda e^{- \lambda x} dx = \int_{0}^{e^y} e^{- x} dx$$

This step is wrong. Obviously you can't just drop the $\lambda$ like that. I suspect what you did was a substitution but you kept the same variable. If you let $u= \lambda x$, then du= \lambda dx so the integrand becomes e^u du. But you need to change the limits of integration also. When x= 0, u= 0 but when x= e^y, u= \lambda e^y. The correct integral is now
$$\int_0^{\lambda e^y} e^{-u}du= -e^{-u}\right|_{0}^{\lambda e^y}$$

$$= -e^{-x} \Big |_0^{e^y} = -e^{-e^y} + 1 = 1 - e^{-e^y}$$

so

$$\frac {d F_{|X|} (x) } {dx} = p(x) = e^x e^{-e^x}$$

Unfortunately the answer isn't in the back of the book. Does that look okay?

 

[lizzyb]

Since $$\lambda = 1$$ I just sort of dropped/substituted it. Thanks :-)