Suppose that X is a random variable with the probability density function
f(x) = {e^-x if x ≥ 0, 0 otherwise
What is the probability density function of the random variable Y = ln(X)?
Math Help Please?
2018-07-25 1:13 pm
回答 (2)
2018-07-26 1:08 am
✔ 最佳答案
Let g(y) be the pdf for Y = ln X.Then since Y is 1-1, we have
g(y) = f(x) * (1/|dy/dx|)
.......= e^(-x) * (1/(1/x)), since x ≥ 0
.......= xe^(-x)
.......= e^y e^(-e^y), since x = e^y (from y = ln x)
.......= e^(y - e^y) for all y.
I hope this helps!
2018-07-25 1:46 pm
Y = ln(X)
X = e^Y
I hate to do this to you, but I have to run now, and I am giving you only the barest outline of my solution.
Let g be the probability density function of Y.
P(a < Y < b) = ∫[a,b] g(y) dy = ∫[e^a,e^b] f(x) dx
g(y) = e^(y - e^y)
My confidence in the solution is a bit shaky, but as I said, I have to run now.
X = e^Y
I hate to do this to you, but I have to run now, and I am giving you only the barest outline of my solution.
Let g be the probability density function of Y.
P(a < Y < b) = ∫[a,b] g(y) dy = ∫[e^a,e^b] f(x) dx
g(y) = e^(y - e^y)
My confidence in the solution is a bit shaky, but as I said, I have to run now.
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