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LisaMS
Member
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.
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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!
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!
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LisaMS
Member
Thank you very much - that helps a lot!
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
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