# 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 $$W$$?

• It is not the same filtration. Knowing the sum of two components is not same as knowing the individual parts. Sep 13 at 11:41
• I see. Thank you for this inuitive comment. How could I proof that? Sep 13 at 12:52
• Prove that $\sigma\{(W_s^1,W_s^2)\,s\leq t \} \neq \sigma\{W_s^1+W_s^2\,s\leq t \}$ Sep 13 at 13:41
• How could I do that? I do not know exatly an element of the set will look like? Sep 13 at 13:46

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.