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Chapter 9 practice question 9.4 iii

A

Alexa

Member
Can someone explain the method around 9.4(iii), generally on the approach and these points specifically:

1. Why we need X =B1-B0 and Y =B2-B2 instead of using B1 and B2 on their own since B1 and B2 are already normally distributed
2. I'm not sure why I need the probability of Y > -X and not Y > 0
3. The last line of the answer, why is this not treated as an integration by parts?

Thanks a ton
 
Hi
I'm having difficulty matching your query to the practice questions in the Course Notes. Maybe you could confirm the page number of the 2025 Course Notes that you're looking at, or post the question here?
Thanks
 
Hi Steve, sorry I may not be referring to the 2025 edition. The question is from the 2021 edition - not sure if you still have a copy of that.

9.4(iii): Let Bt where t >=0 be a standard Brownian motion process starting with B0=0.
Show tha the probability that B1 and B2 both take positive values is 3/8
 
OK. Thanks.
1. B1 and B2 are normally distributed, but they're not independent, and so can't be treated separately. B2 depends on B1 - remember that from the definition of standard Brownian motion, cov(B1,B2)=1.
2. B2 depends on B1 but X and Y are independent. Independent random variables are a lot easier to work with. If the process has reached a value of X at time 1, what constraint must be placed upon the next increment Y such that X+Y is also positive? Y>-X.
3. The integral is of the form \int f'(x)f(x)dx, which equals 1/2*f^2(x) via a reverse application of the chain rule. Integration by parts would of course work, but it's not as efficient or intuitive.
Hope that helps.
 
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