If a stock price can change up and down by a dollar a day and the probability of a unit increase is a 0.5, what is the probability that the stock price will increase by $12 before increasing by $57.

Question

anonymous · Answer

I'm not sure that I understand the question. If it can only change by a dollar at a time, then it would seem that the probability of it increasing by a total of $12 before it increases by a total of $57 would be 1. It can't increase by $57 without increasing by $12 first.

anonymous · Answer

Reference point/ base point of increase according to my colleagues is a point xi such that it can increase to 12 from xi. Similarly, they think it has to increase to 57 from this base point. It is a random walk problem. Hope this helps...

anonymous · Answer

Okay, I can understand the question as probability of a random walk to $57, but the $12 part is still a bit confusing to me. Any random walk to $57 will pass through $12.

anonymous · Answer

Lets assume then that the 12 bucks is not there, i think probability of the random walk one step at a time given 57 successes to reach 57 dollars would be root of 57.  Since this is the total distance from point of origin...apparently i have not considered failures and i dont know how to do this. How would u go about it?

anonymous · Answer

Well, do you have any constraints? Given an infinite amount of time, the stock price will cross every point an infinite number of times. That's why I'm still wondering about this $12 thing, I'm wondering if that's some way for them to put in a constraint, but I don't see how it would work.

dumbcow · Answer

http://en.wikipedia.org/wiki/Random_walk#One-dimensional_random_walk how many steps are we talking about? yeah what @nbouscal said

anonymous · Answer

Oh, okay, I think I get it. It's asking what the odds are of having a series of exactly 12 consecutive +1s before a series of exactly 57 consecutive +1s.

dumbcow · Answer

oh thats what i initially thought butit didn't seem to fit with random walks

anonymous · Answer

exactly, but are we considering failures or are we ignoring them completely? checking out that link...

anonymous · Answer

Well it seems to basically boil down to the probability of $12 and not $57, because after the $12 does happen, the probability of the $57 happening is 1. As for how to calculate that, I haven't the slightest idea, I haven't studied random walks hardly at all.

anonymous · Answer

To clarify, I mean P($12)∧¬P($57) (Which are both 1 given infinite time so this problem's definitely confusing).