Question Bank 2.17

[Oxymoron]

I would like to know why the following method leads to an incorrect solution:

E[exp(-2Bt) | Fs]
= exp(-2Bs) * E[exp(-2(Bt-Bs)| Fs]

Now expanding the exponent, (I'm dropping | Fs since we are dealing with the increments and Fs is immaterial owing to the independence property)
= exp(-2Bs) * E[1 + -2(Bt-Bs)/1! + (-2*(Bt-Bs))^2/2! + ...]

Taking expectations inside the bracket, (Bt-Bs)^3 and higher orders is equal to 0. Also, E[Bt-Bs] = 0, so, it equates to:
= exp(-2Bs)*(1 + 4*(t-s)/2)
= exp(-2Bs)*(1 + 2*(t-s))

The solution gives the answer as : exp(-2Bs)*exp(2*(t-s))


Thanks!

 

[Mike Lewry]

I think there are a couple of reasons this doesn't work:

First, the expansion you're using is valid for small exponents only, but Bt-Bs could be huge.

Second, dBs^3 and higher orders is zero, but this applies to infinitesinal increments only. Bt-Bs is a finite increment and so this result doesn't apply. So the higher order terms can't just be ignored.

Does that explain it?

 

[manish.rex]

I would like to know why the following method leads to an incorrect solution:

E[exp(-2Bt) | Fs]
= exp(-2Bs) * E[exp(-2(Bt-Bs)| Fs]

Now expanding the exponent, (I'm dropping | Fs since we are dealing with the increments and Fs is immaterial owing to the independence property)
= exp(-2Bs) * E[1 + -2(Bt-Bs)/1! + (-2*(Bt-Bs))^2/2! + ...]

Taking expectations inside the bracket, (Bt-Bs)^3 and higher orders is equal to 0. Also, E[Bt-Bs] = 0, so, it equates to:
= exp(-2Bs)*(1 + 4*(t-s)/2)
= exp(-2Bs)*(1 + 2*(t-s))

The solution gives the answer as : exp(-2Bs)*exp(2*(t-s))


Thanks!

assuming the t-s is a very small quantity, (1+ (t-s)) ~ exp(t-s), hence your answer is the same as given in the text book

 

[Oxymoron]

I think there are a couple of reasons this doesn't work:

First, the expansion you're using is valid for small exponents only, but Bt-Bs could be huge.

Second, dBs^3 and higher orders is zero, but this applies to infinitesinal increments only. Bt-Bs is a finite increment and so this result doesn't apply. So the higher order terms can't just be ignored.

Does that explain it?

Yes Mike, thanks!