Thread: The Distribution Technique

I'm working on a problem from Hogg and Tanis, "Probability and Statistical Inference", 8th ed., problem 5.1-7,

$5000 is invested at a rate R selected from a uniform distribution on the interval (.03, .07). Once R is selected the sum is compounded instantaneously for a year so that X=5000e^R by the end of the year.

Find the distribution function of X.

My solution was to say P(X < x) = P(5000e^R<x) = P(R<ln(x/5000)).

This I calculate as the integral from .03 to ln(x/5000) of 10. That come out as 10(ln x/5000-.03) which is not the answer in the back of the book. Help?

I've just seen at least some of the error of my way, though I'm not sure why the previous method is wrong.

Anyway, using the technique from that chapter, P(R<ln(x/5000)) =F(r) thus

f(r)=F'(r)=25*(5000/x)*(1/5000) = 25/x which is at least some of the way to the book's answer but not all the way there.