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## Tuesday, September 8, 2009

### Calling out to idiot America

I would just like to make the following point.

If P \left( A \right) = \dfrac{ e^r }{1+e^r}, then, obviously, P \left( \neg A \right) = \dfrac{ 1 }{1+e^r}.

Last night I couldn't figure this out (yeah, I'm awesome), so I wrote a fancy function which has the result of outputting exactly the above. In somewhat simplified form, when A is the event that Y=1 and Y is a binary variable,

 P \left( Y = y \right) = \dfrac{ e^{y r} }{1 + e^r }


It takes a really special sort of blindness to write a fancy function to output an obvious fact when that fact is not itself understood.

bk drinkwater said...

It's my considered opinion that the wheel is not such a perfect machine that it can't stand to be reinvented periodically. Maybe, someone will chance across a better one.

At least, this is how I rationalize it when I do stuff like that.

fibby said...

I rewrote the function a little bit. What I had yesterday didn't fully capture the impressive blindness of what I'd actually done.

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