Probability Density Function with an exponential random variable

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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?
 

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  • #2
HallsofIvy
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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 \}[/tex][tex] = \int_{0}^{e^y} \lambda e^{- \lambda x} dx = \int_{0}^{e^y} e^{- x} dx[/tex]
This step is wrong. Obviously you can't just drop the [itex]\lambda[/itex] like that. I suspect what you did was a substitution but you kept the same variable. If you let [itex]u= \lambda x[/itex], 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= ey, u= \lambda e^y. The correct integral is now
[tex]\int_0^{\lambda e^y} e^{-u}du= -e^{-u}\right|_{0}^{\lambda e^y}[/tex]

[tex] = -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?
 
  • #3
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Since [tex]\lambda = 1[/tex] I just sort of dropped/substituted it. Thanks :-)
 

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