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 :-)