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

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.

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).

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:****Location:**Singapore

@Alan,

The bonus question you provided is interesting!

I have not googled but would the answer be 0?

The bonus question you provided is interesting!

I have not googled but would the answer be 0?

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.

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.

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.

- katastrofa
**Posts:**10256**Joined:****Location:**Alpha Centauri

Curious animal. [$]\sqrt{-2\ln(1-\frac{1}{\sqrt[n]{2}})}[$] would be the median?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.