# Question

There is a stick of length 1. We break it at a random spot and take the leftmost part as a stick of length x. Then we break THAT stick at a random spot and take the leftmost part as a stick of length y. Find the PDF of Y.

# Answer

f(x) = 1 for 0 < x < 1

f(x) = 0 otherwise

*This is the graph of f _{X}(x):*

We also know that

f_{Y|X}(y|x) = 1/x for 0 < y < x

f_{Y|X}(y|x) = 0 otherwise

*This is the graph of f _{Y|X}(y|x):*

Using the theorem of total probability for continuous RVs, we have that

$ f_{Y}(y) = \int_{-\infty}^{\infty}f_{Y|X}(y|x)f_{X}(x) dx $

$ = \int_{0}^{1}f_{Y|X}(y|x)(1) dx $

because we are doing an integral of x, and the probability that x < y or x > 1 is 0, the limits of integration become from y to 1:

$ = \int_{y}^{1}\frac{1}{x} dx $

$ = ln(\frac{1}{y}), 0 < y < 1 $