Why Combining Brownian Motions Does Not Preserve the Market Filtration

# standard/brownian market with different brownian motion


# standard/brownian market with different brownian motion












Consider for simplicity the following brownian market:

$$dS^0_t= r S^0_tdt$$

$$dS^1_t= S^1_t(r dt + dW^1_t + dW^2_t) $$

where the filtration is generated by $W^1,W^2$

Consider now $W_t:= \frac{1}{2}(W^1_t + W^2_t)$, which is a Brownian motion, too.

When I substitute $W_t$ in the above financial market, I would have $1$ Brownian motion and $1$ risky asset, so I could conclude that the market is complete. But that is not correct, since completness depends on martingale representation, which depends again on the underlying filtration.

But why do I have to pick another filtration in this context, since the filtration generated by $W^1,W^2$ is the same as being generated by $W$?

## Answer by algebruh (score 1)

https://quant.stackexchange.com/a/67933

tl;dr: informally: You cannot know the value of the difference of two random variables by knowing their sum.

Consider the following set $A =\{ \omega \in \Omega | W_1(t)(\omega) - W_2(t)(\omega) \in [0,1] \}$ This set is of course in the Filtration generated by $W_1$ and $W_2$ since the addition of measurable functions is measurable. Is this set in the (augmented) generated Filtration of $W$? For that either $P(A) = 0$ (because we consider the augmented filtration) or it exists a borel set $B$ (a set that can be build from intervals through countable set operations) s.t. $$\{ \omega \in \Omega | W(\omega)(t) \in B \} =A \ \mathbb{P}-a.s.$$ (you don't need to consider past time points since the brownian filtration is increasing ).

$P(A) > 0 $ since $W_1(t)- W_2(t)\sim \mathcal{N}(0,2t) .$

Assume it exists such a $B$. Then

$$ \{ \omega \in \Omega |W_1(t)(\omega) + W_2(t)(\omega)) \in B \}=\{ \omega \in \Omega | W_1(t)(\omega) - W_2(t)(\omega) \in [0,1] \} $$ (we absorb the 0.5 factor into the set B) Which is equivalent to:

$$ X + Y \in B \Leftrightarrow X-Y \in [0,1] \ \mathbb{P}-a.s. \ (\#)$$ where $X,Y$ are iid normally distributed. This is equivalent to

$$ Y \in B-X \Leftrightarrow Y \in [-1,0] \ \mathbb{P}-a.s.$$ Thus:

$$ B-X =[-1,0], \mathbb{P}-a.s.$$

inputting into LHS of (#):

$$ Y\in [-1,0] \ \mathbb{P}-a.s.$$

Which is wrong since $ Y$ is normally distributed.

Thus the filtrations are not equal

As an example for incompleteness: The derivative $H(t) = \mathbb{1}_{A}\mathbb{1}_{t}$ cannot be replicated.