Assume that both $ B_t $ and $ Z_t $ are standard Brownian motions and that $ Corr(B,Z)=1 $.
How do those properties imply that $ B_t = Z_t $ a.s. ?
$Cov(B,Z) = t $ is as far as I can get, and I have no idea on how to proceed further. Any help, reading reference will be gladly welcome!
Background information on the question, not sure if relevant but might help.
Assume there are two processes:
$$ \frac{dP}{P} = \mu dt + \sigma dB_t $$ $$ \frac{dS}{S} = v dt + \eta dZ_t $$
Furthermore assume that $ Corr(B,Z)=1 $ then
$$ \frac{dS}{S} = v dt + \eta dB_t $$
I found this claim in section about complete markets on page 7 in paper valuation of claims on nontraded assets using utility maximization by Vicky Henderson
I guess it is assumed to be straightforward, but I can't get my head around that.
It seems that I have found the answer. I post it below the question because of 8h restriction.
To prove the above statement first one need to show that perfectly correlated RV of finite variance are linearly dependent. Assume $ Corr(X,Y) = 1$ and put
$$ Y': = \frac{\sigma_Y}{\sigma_X}X $$
Investigate variance of $Y-Y'$
\begin{align} Var(Y-Y')&=\mathbb{E}(Y^2) - 2\mathbb{E}(YY') + \mathbb{E}(Y'^2) - \mathbb{E}^2(Y-Y')\\ &=\sigma^2_Y + \frac{\sigma^2_Y}{\sigma^2_X}\sigma^2_X - 2Cov(Y,Y')\\ &=\sigma^2_Y + \frac{\sigma^2_Y}{\sigma^2_X}\sigma^2_X - 2\sigma_Y\frac{\sigma_Y}{\sigma_X}\sigma_X\\ &= 0 \end{align}
Hence $Y- Y' = b$ for some constant $b$. Now taking expectation on both sides yields $$ b =\mathbb{E}(Y) - \frac{\sigma_Y}{\sigma_X}\mathbb{E}(X) $$ Together $$ Y = \frac{\sigma_Y}{\sigma_X}X + \mathbb{E}(Y) - \frac{\sigma_Y}{\sigma_X}\mathbb{E}(X) $$
Since in case of Brownian motion $ \mathbb{E}(B_t) = \mathbb{E}(Z_t) = 0 $ and $ \frac{\sigma_{B_t}}{\sigma_{Z_t}}=1$ we obtain desired result $$ B_t = Z_t $$
Since $U$ and $V$ have unit variance, $\text{Corr}(U,V)=1$ means that $\text{Cov}(U,V)=1$, in particular, $E[(U-V)^2]=E[U^2]-2\text{Cov}(U,V)+E[V^2]=1-2\cdot1+1=0$, hence $U=V$ almost surely.
Applying this to $U=B_t/\sqrt{t}$ and $V=Z_t/\sqrt{t}$ yields $B_t=Z_t$ with full probability, for every $t$. Since $B$ and $Z$ are almost surely path-continuous, this implies that $B=Z$ with full probability.