The penultimate line of the solution to Assignment X2.4(ii) jumps from: E[Bt^4] + 6t*E[Bt^2] +3t^2 to: 3t^2 + 6t*t +3t^2 = 12t^2. Trying to understand why E[Bt^4] = 3t^2 and E[Bt^2] = t. Is this simply because we simplify the expressions in the solution to X2.4(i) to only include t and ignore Bs and s? So therefore: E[Bt^2|Fs] = Bs^2 +t-s = t AND E[Bt^4|Fs] = Bs^2 +6(t-s)*Bs^2 + 3(t-s)^2 = 3t^2. Bonus Q: Appreciate if you can signpost me to past CT8 exam questions on martingales. The only one in the Acted Revision booklet is from CT8 April 2017 Q6(i). And would there be marks in the exam reserved for proving this condition for Martingales (ie. that the absolute value is <infinity) or it's unnecessary.
Hi Bill, Yes, these results are the ones in (i), except with s=0 (as we are talking about the increment from 0 to t. Since Bt is standard Brownian motion, we know that B0 = 0). April 2014, Q3(ii) is another example of a CT8 exam question on martingales. There are other examples which link to other areas of the course eg Cameron-Martin-Girsanov theorem or equivalent martingale measure. There are also tricky questions where, instead, you show that a function is a martingale by showing that the dt term is zero (eg September 2005, Q8, September 2010, Q6(i)). The E[|Xt|] < infinity for all t condition is part of the martingale definition so we would expect you to state that condition in a question proving that a function is a martingale. However, considering that in the Core Reading it only states that E[|Wt|] < infinity since Wt < infinity almost surely, we would not expect much (if any) justification of this point. I hope this helps, please let me know if you have any followups. Alvin.
