If I draw X uniformly from [0,1], and then draw Y uniformly from [X, 1], what’s the PDF of Y?

Question

It seems like there should be some nice symmetry trick to exploit without calculus but I can’t see it

Answer 1

The cdf is \begin{align} P(Y \le y) &= \int_0^1 P(Y \le y \mid X=x) \mathrm{d}x \\ &= \int_0^y \frac{y-x}{1-x} \mathrm{d}x \\ &= \int_0^y \left(\frac{y-1}{1-x} + 1\right) \mathrm{d}x \\ &= \left. \left((1-y)\ln(1-x) + x\right) \right|_0^y \\ &= (1-y)\ln(1-y) + y \end{align} So the pdf is $$\frac{d}{dy}\left((1-y)\ln(1-y) + y\right) = -\frac{1-y}{1-y}-\ln(1-y)+1 = -\ln(1-y)$$