# Latest content was relocated to https://bintanvictor.wordpress.com. This old blog will be shutdown soon.

## Friday, March 13, 2015

### BM: why Bt^3 isn't a martingale

Greg gave a hint: . Basically, for positive X, the average is higher, because the curve is convex.

Consider the next brief interval (long interval also fine, but a brief dt is the standard approach). dX will be Normal and symmetric. the +/- 0.001 stands for dX. For each positive outcome of dx like 0.001, there's an equally likely -0.001 outcome. We can just pick any pair and work out the contribution to E[(X+dX)3].

For a martingale, E[dY] = 0 i.e. E[Y+dY] = E[Y]. In our case, Y:=X3 , so E[(X+dX)3] need to equal E[X3] ....

Note that Bt3 is symmetric so mean = 0. It's 50/50 to be positive or negative, which does NOT make it a martingale. I think the paradox is filtration or "last revealed value".

Bt3 is symmetric only when predicting at time 0. Indeed, E[Bt3 | F_0] = 0 for any target time t. How about given X(t=2.187) = 4?

E[(4 + dX)^3] works out to be 4^3 + 3*4*E[dX^2] != 4^3