Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 43 / 3 / iii / Solution

Yes: different adapted volatilities can give perfectly correlated terminal prices. The counterexample in part ii already establishes this with one nonconstant volatility. Both can be nonconstant as well. Put $h(x)=1+\frac{1}{2}\tanh x$ and takeThe Doléans-Dade exponential has volatility $h(B_t)$, bounded between $1/2$ and $3/2$. The second stock has volatility $h(B_t)S_t/(1+S_t)$, also bounded and nonconstant. Their squared volatility difference integrates toThe Novikov condition and boundedness give true square-integrable martingales, and the Itô isometry gives positive variance for each $t>0$. Yet $S_t^{\prime }=(1+S_t)/2$ gives correlation one. The mechanism is positive affine dependence, not equality of the multiplicative volatility coefficients; it also explains the failure of the literal part-ii premise.