Consider again the exchange option in Example 9.5. Now assume that \(W_{1}\) and
\(W_{2}\) are no longer independent, but that the local correlation is given by
\(d W_{1} \cdot d W_{2}=e d t\). (We still assume that both Wiener processes
have unit variance parameter, i.e. that \(\left.d \bar{W}_{1}^{2}=d
\bar{W}_{2}^{2}-d t .\right)\) How will this affect the Black-Scholes-type
formula given in the example?
