I do not understand why the volatility is equal to the square root of 2,5sigma in the second line. Would someone please provide an explanation for this.
Since you've jumped straight in to part (iii) of the question, I'm guessing you're happy with part (ii) The issue with part (iii) is that we can't use the Black-Scholes option pricing formula directly because the volatility changes throughout the life of the contract. The challenge is then to find a constant "average" volatility figure that would be applicable over the whole six months. If we call this \(\tilde{\sigma}\), then we'll want the stochastic term in the expression for \(S_{0.5}\) to look like \(\tilde{\sigma}Z_{0.5} \). To establish this figure we need to look at the volatility term of \(S_{0.5}\) in part (ii), ie \(\sigma(2Z_{0.5}-Z_{0.25})\). We now need to find the variance of this expression, which we can do by casting it in terms of the increment \(Z_{0.5}-Z_{0.25}\), ie \(\sigma(2Z_{0.5}-Z_{0.25}) = 2\sigma(Z_{0.5}-Z_{0.25}) + \sigma Z_{0.25}\). This has a variance of \(\frac{5}{4}\sigma^2\), which when equated to the variance of \(\tilde{\sigma}Z_{0.5} \) (which is \(\frac{1}{2}\tilde{\sigma}^2 \)) we find that \(\tilde{\sigma}=\sqrt{2.5}\sigma \). Hopefully that should do it - let me know if not!
Can some guidance to (ii) be given? I am struggling a bit ... Also why is the variance of 2*sigma(Z_0.5 - Z_0.25) + sigma*Z_0.25 = (5/4)*sigma^2 ? As well as the variance of sigma_tilde*Z_0.5 = (1/2)*(sigma_tilde)^2 ? Ultimately giving sigma_tilde = sqrt(2.5*sigma) Thanks in advance, James
