Serving the Quantitative Finance Community

 
User avatar
kiann
Topic Author
Posts: 18
Joined: April 16th, 2008, 6:39 pm

Derivation for two independent brownian motions

March 30th, 2019, 10:19 pm

Suppose W1(t) and W2(t) are two independent standard Brownian motions. What is the probability that both processes are larger than 0.667 at t=1.0
 
User avatar
bearish
Posts: 6450
Joined: February 3rd, 2011, 2:19 pm

Re: Derivation for two independent brownian motions

March 31st, 2019, 2:47 am

So you have two independent random variables, W1(1) and W2(1), each of which are distributed as N(0,1). The probability of both being larger than x is going to be the product of each being greater than x. 
 
User avatar
Alan
Posts: 10615
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

Re: Derivation for two independent brownian motions

April 1st, 2019, 3:27 am

Here's a standard related problem that's a good follow-up for the OP to puzzle over. 

Consider a random sequence: the location of n independent BM's at time 1. 
What's the distribution function of the maximum of the sequence; 
i.e., what's [$]F(x)[$] where

[$]F(x) =  Prob \left( \max \left\{W_1(1), W_2(1), \cdots, W_n(1) \right\} < x \right)[$]  ?

Bonus: what's the expected value of that maximum as n becomes very large (but not infinite)?  

Hint: given bearish's comment, the first answer is easy and the second is hard (if you don't google).
 
galvinator
Posts: 12
Joined: August 8th, 2018, 9:10 am
Location: Singapore

Re: Derivation for two independent brownian motions

July 30th, 2019, 6:28 am

@Alan,
The bonus question you provided is interesting! 
I have not googled but would the answer be 0?
 
User avatar
Alan
Posts: 10615
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

Re: Derivation for two independent brownian motions

July 30th, 2019, 11:31 pm

Thanks.

No, first, I said "n is large but not infinite", so the puzzle is to provide an asymptotic formula that depends upon n. Keep thinking!

Second, if n were infinite, the expected max is [$]+\infty[$].

Hint: start from bearish's comment, generalize it, and start calculating. 
 
krs
Posts: 20
Joined: August 20th, 2019, 12:40 pm

Re: Derivation for two independent brownian motions

December 28th, 2020, 5:27 pm

is it not going to be max( F(x1), F(x2)....F(xn) )? when n is infinite, the value of the function will be equal to 1.  
 
User avatar
katastrofa
Posts: 10082
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

Re: Derivation for two independent brownian motions

January 1st, 2021, 10:34 pm

Thanks.

No, first, I said "n is large but not infinite", so the puzzle is to provide an asymptotic formula that depends upon n. Keep thinking!

Second, if n were infinite, the expected max is [$]+\infty[$].

Hint: start from bearish's comment, generalize it, and start calculating. 
Curious animal. [$]\sqrt{-2\ln(1-\frac{1}{\sqrt[n]{2}})}[$] would be the median?