# Is the result of comparing samples from 2 log normal distributions, uniform?

I have sample values coming from two truly random log normal distributions, A and B. They are continuous (analogue voltages actually). They are totally independent of one another, but theoretically identical in all aspects of size and shape. I then repetitively compare sample value A with sample value B, as in

IF valueA > valueB THEN RESULT = TRUE

IF valueA < valueB THEN RESULT = FALSE


The comparison is done with analogue circuitry (not digital) so A = B cannot happen. What is the distribution of RESULT? I think that it has to be uniform. I think that it's effectively a Von Neumann extractor that will remove all bias /asymmetry from the log normal input distributions. Does anyone concur?

• Hint: What happens to your situation when you interchange $A$ and $B$? How does that affect the results? – whuber Oct 28 '16 at 0:06
• @whuber I haven't built the thing yet as the decision to proceed rests with the answer to this question :-) – Paul Uszak Oct 28 '16 at 0:33
• No, whuber's hint means for you to use thought, not the actual device. You described assumed properties which - if they're true - should let you give an answer to that question without building the device. As a different form of the same hint, if you identified which was which using sticky notes (with "A" and "B" written on them, respectively) but then interchanged the sticky notes, what would be different about your description? – Glen_b Oct 28 '16 at 0:56
• @Glen_b I'm not entirely sure otherwise I wouldn't have asked, especially as some time and expense ride on the outcome of this enquiry. Is a clear answer really so difficult? Can the readers infer that what might be considered simple statistics questions (for experts) are not welcome here? – Paul Uszak Oct 28 '16 at 11:38
• No answer as yet. 6 months - still waiting for a simple yes /no by your leave... – Paul Uszak Apr 8 '17 at 1:25

You have two independent random variables $X, Y$ with identical distributions (we also assume the distribution is continuous, so exact equality has probability zero). You ask for $$\DeclareMathOperator{\P}{\mathbb{P}} p= \P (X < Y)$$ But the assumption of identical distributions tell you that $$1-p = \P (Y < X)$$ must be equal! So, $p=1-p$ which has the solution $p=\frac12$.
$$\P(X < Y) = \int \P(X < Y \mid Y=y) f(y)\; dy = \\ \int \P(X < y) f(y) \; dy = \int F(y) f(y) \; dy = \\ F(y) F(y) \bigg\rvert_{-\infty}^\infty - \int f(y) F(y)\; dy =\\ 1- \int F(y) f(y)\; dy$$ so we got the same equation, with the same solution. By the way, we didn't use the lognormal assumption at all. Moreover, our first argument neither did use independence, it only needs exchangeability, so the result is also valid with a symmetric kind of dependence. The second (calculus) argument did use independence, though, showing that the simple argument is really superior.