These notes are pretty usefulhttp://walrandpc.eecs.berkeley.edu/126notes.pdf

sorry I mean W at T1 and W at t2. well I mean if you consider the brownian motion except that after t1, Wt is moved back to the origin, then the probablity that both Wt1 and Wt2 > 0 is 1/4 which is bigger than 1/8. The event and process DevonFangs describe should be more probable right?

<t>QuoteOriginally posted by: ilikeprobabilityThe problem is to find P(W1>0 and W2>0). W2 could be greater than or equal to W1 or could be less than or equal W1. We don't need to worry about it. We just need to analyze the situation when W2>0 and W1>0 and this can be done by using the independent in...

<t>QuoteOriginally posted by: ruairihQuoteOriginally posted by: DevonFangsA more extended explanation would be nice (why did you edit your post?)Apologies, I edited it because I answered for the conditional probabilty instead of the joint prob.I like the quadrant area answers. I was asked the versio...

<t>QuoteOriginally posted by: DevonFangsA more extended explanation would be nice (why did you edit your post?)Apologies, I edited it because I answered for the conditional probabilty instead of the joint prob.I like the quadrant area answers. I was asked the version where |t1| = |t2| but the soluti...

I dont think it can be 1/8. If they were independent variable then it would be 1/4 that they were both +ve

Got asked this on a phone interview. Its basically just the integral of a normal by a cumulative normal so .5* cumulative normal squared

<t>I was concerned about whether my solution was both necessary and sufficient too, but here is another way of phrasing it, it's identical but seems clearer I think:Consider yourself at the party. Consider the pairs (i,j) st. i is in the set of your friends and j is in the set of friends of your fri...

The pattern is of perfect squares (even number of factors). eg 4, 9, 16 etc

<t>QuoteOriginally posted by: DevonFangsQuoteOriginally posted by: wileyswruairih, if friend/knowing is mutual, i.e., if A is a friend of/knows B, then B is a friend of/knows A, then the answer seems to be 100 people. my approach / hint is to count the groups of 3 people with only one pair not knowi...

<t>QuoteOriginally posted by: frenchXQuoteOriginally posted by: ruairihYes, Wileysw has the correct answer, I did it the same way after drawing a sort of schematic diagram. Im interest to see how this is done by a more brute force method too.If you are new, you will quickly learn that Wileysw has ve...

Yes, Wileysw has the correct answer, I did it the same way after drawing a sort of schematic diagram. Im interest to see how this is done by a more brute force method too.

Im interested to see where this is going. So is the number of nodes on the circle the answer?

Yes, its a tricky one alright, I think I must have spent two days thinking about it on and off before solving it. THe mutual constraint stops it being infinite like you said.

It cant be 12 because everybody in the room knows 22 other people in the room