CM2 chapter 9 Practice Question 9.4 (iii)

[George Philip]

Q. Let Bt ( t >= 0 ) be a standard Brownian motion process starting with B0 = 0 .
Show that the probability that B1 and B2 both take positive values is 3/8.

Could someone explain the integration that was used to solve this question in the study material?
Thanks in advance.
.

 

[Steve Hales]

The integral is essentially summing up all of the probabilities of the two Brownian motions satisfying the given condition. The phi(x)dx term is the probability of X lying between x and x+dx, and the phi(y)dy term is the probability of Y lying between y and y+dy. Since these two things are independent we can multiply them together to get the joint probability phi(x)phi(y)dydx.
Notice how the limits on the integral match the requirements of X and Y.
Let me know if that wasn't what you were after.
Thanks

 

[Bill SD]

The integral is essentially summing up all of the probabilities of the two Brownian motions satisfying the given condition. The phi(x)dx term is the probability of X lying between x and x+dx, and the phi(y)dy term is the probability of Y lying between y and y+dy. Since these two things are independent we can multiply them together to get the joint probability phi(x)phi(y)dydx.
Notice how the limits on the integral match the requirements of X and Y.
Let me know if that wasn't what you were after.

Thanks Steve for explaining what the phi(x)dx and phi(y)dy terms represent. I have one clarification question on your response and two additional questions on other parts of this question (so thought easiest to keep in same thread).

1st Question: When you say 'given condition' (bold in quote above) are you referring to the the fact that both B1 and B2 take positive values or the fact that Y>-X or something else?

2nd Question: The solution to Part (i) of this question (9.4) says B2 has a zero probability of being exactly zero.
Is this because there is an almost infinite amount of possible values of B2 and therefore there is a zero probability that B2=any exact positive, negative or zero value. Or a different reason.

3rd Question: The solution to Part (iv) of this question (9.4) says "Bt will almost certainly take a negative value at some point close to t=0."
Why is Bt more likely to take a negative value close to t=0 rather than close to t=2?

Thanks in advance.

 

[Steve Hales]

1st Solution: The conditions are that X>0 and that Y>-X. This will ensure that B1>0 and B2>0.

2nd Solution: You've got the right reason. The same point can be illustrated with a game: if you pick and integer between 1 and 10 then I've got a 10% chance of guessing it correctly. I you pick any real number between 1 and 10 then I've got a zero chance of guessing it.

3rd Solution: At time t=0 the standard Brownian motion is prescribed to equal zero, so B0=0. Given that the process wobbles away from the origin with increments distributed like N(0,dt) where dt is infinitesimally small, it will almost surely go below zero because all it needs is a negative increment.

 

[Tushar Gora]

Tracing back to origin of the thread, I had a doubt regarding the (iii) part. I'm clear up till:
p = int(0;inf): phi(x) PHI(x) dx

I used the property that differentiation of CDF gives PDF. This gives me,

p = int(0;inf): PHI(x) dPHI(x)
= [ PHI(x)^2/2 ] (0;inf)
= ( PHI(inf)^2 - PHI(0)^2 ) / 2
= (1 - 0.5^2) / 2
= 3/8

I'm not sure whether the limits would remain same when I apply the transformation. Is my method correct? If not can anyone please help me with this?

 

[Steve Hales]

That's a neat way of handling the integral!
The limits do indeed remain the same. It might be clearer to say p=int(x=0;x=inf): PHI(x) dPHI(x).
This emphasises the point that x is the variable that's running from zero to infinity, even though you're integrating PHI(x) with respect to PHI(x).

 

[Tushar Gora]

Oh yes! that makes so much sense. Thank you so much.