# Distribution of X-U(0,1) conditioned on sigma algebra of Y/X, where is Y is U(0,1)?

The question I have is:

Define X,Y to be two independent uniform(0,1) random variables and $Z:=\frac{Y}{X}$

Compute $P(X<x|\sigma(Z))$

The answer given apparently by "straightforward elementary computations" is for $x\geq 0$

$P(X<x|\sigma(Z))$=min$\{{x^2,1}\}\mathbb{I}_{Z\leq 1}$ $+$ min$\{{x^2 Z^2,1}\}\mathbb{I}_{Z\geq 1}$.

My idea was to condition on the $Z\leq 1$ and ${Z\geq 1}$ then compute using the joint density of X and Y, but this seems to work for the first term but doesn't for the second? Any help would be much appreciated.

• Is this a question from a textbook/assignment/exam? If so, please consider adding [self-study] tag and read its wiki. – T.E.G. - Reinstate Monica Jan 19 '17 at 23:00
• @whuber Is this actually a duplicate of the indicated post? That deals with the distribution of $Z$, but this doesn't seem to be asking for that. If the indicated thread does completely cover the answer to this, I think at least a brief comment here explaining why those very differently phrased questions are the same would be helpful. – Glen_b Jan 20 '17 at 0:17
• @Glen I interpreted this question as requesting the "straightforward elementary computations" needed to obtain the distribution of a ratio of independent uniform variables, and then saw those computations explicitly laid out in clear answers in the duplicate. I did not see any request, explicit or implicit, to re-interpret the more formal language of sigma algebras. If, on the other hand, the duplicate had been closed with a reference to this question, then I agree that might have required an explanation. – whuber Jan 20 '17 at 0:43
• @whuber this isn't a duplicate of that question, I'm not trying to work out the ratio distribution, I'm conditioning on the sigma algebra and then trying to obtain the above probability. – Dan Taylor Lewis Jan 20 '17 at 7:44
• Thank you Dan. After rereading your question carefully (prompted by @Glen_b) I recognized that difference and reopened it last night. – whuber Jan 20 '17 at 16:11