Gaussian conditional probability

SigmundFraud · June 18th, 2012, 9:35 am

X & Y are i.i.d. X~N(0,1) & Y~N(0,1).What is P(X>0 | X+Y>0) ?

babyQuant · June 18th, 2012, 10:29 am

if A and B are defined as A: X>0 B: X+Y>0then P(A|B)=P(A)*P(B|A)/P(B)P(A)=0.5P(B)=0.5P(B/A) = integral ( N(x) dx ) with integration limits -inf to +inf this comes from if x>0 then x+y>0 for y=-x to +inf

Traden4Alpha · June 18th, 2012, 11:23 am

3 out of 4 pieces of half a pie = 0.375! EDIT: ARGH ! It is 0.75 and I'm outsmarting myself. Must. Drink. More. Coffee.

anilmag · June 18th, 2012, 11:45 am

It is indeed 3/4. If you plot X and Y on the respective axis, any point above y=-x line satisfies X+Y>0, and this triangle is the sample space. So we need to check, within this sample space what is the area satisfying X>0, which is 3 out of 4 pieces.

yegulalp · June 18th, 2012, 12:26 pm

Answer of 3/4 also holds if we relax to X, Y iid and symmetric around 0 (but not necessarily normal). I.e. P(X>c) = P(X<-c) for any c.

Landscape · July 7th, 2012, 4:18 pm

ChicagoGuy · August 16th, 2012, 5:43 pm

Here is a related question:Let W be a Brownian motion. What is P(W_2>0|W_1)?

almostcutmyhair · August 20th, 2012, 3:19 am

QuoteOriginally posted by: ChicagoGuyHere is a related question:Let W be a Brownian motion. What is P(W_2>0|W_1)?

emac · August 23rd, 2012, 9:29 am

QuoteOriginally posted by: almostcutmyhairQuoteOriginally posted by: ChicagoGuyHere is a related question:Let W be a Brownian motion. What is P(W_2>0|W_1)?How can your answer possibly be correct if it doesn't depend on W_1?