Derive the BSM call formula by completing the following steps. Let \(F=F_{A}(0, T)\) for notation ease. We have

\[\begin{align*}C(0) e^{r T} & =E[\max (0, A(T)-K)] \\& =F \times E\left[\max \left(0, \frac{A(T)}{F}-\frac{K}{F}ight)ight] \\& =F \times \int_{-\infty}^{\infty} \max \left(0, x-\frac{K}{F}ight) f_{X}(x) d x \\& =F \times \int_{K / F}^{\infty}\left(x-\frac{K}{F}ight) f_{X}(x) d x \\& =F \times \int_{K / F}^{\infty} x f_{X}(x) d x-K \times \int_{K / F}^{\infty} f_{X}(x) d x\tag{6.22}\end{align*}\]where \(f_{X}(\cdot)\) is the pdf of \(A(T) / F \sim L N\left(-\frac{1}{2} \sigma^{2} T, \sigma^{2} Tight)\)\[f_{X}(x)=\frac{1}{x \sqrt{2 \pi \sigma^{2} T}} e^{-\frac{\left(\ln x+\sigma^{2} T / 2ight)^{2}}{2 \sigma^{2} T}} .\]

(a) The integral in the second term of Formula 6.22 is the area under the pdf of a \(L N\left(-\sigma^{2} T / 2, \sigma^{2} Tight)\) random variable, which is just the probability of falling in that region

\[I_{2}=\int_{K / F}^{\infty} f_{X}(x) d x=P\left[L N\left(-\frac{1}{2} \sigma^{2} T, \sigma^{2} Tight) \geq K / Fight]\]

Compute \(I_{2}\) in the above expression.

(b) To compute the first term in Formula 6.22, do a change of variable:

\[z=\frac{\ln x+\sigma^{2} T / 2}{\sigma \sqrt{T}}\]

hence

\[d x=\sigma \sqrt{T} e^{\sigma \sqrt{T} z-\sigma^{2} T / 2} d z\]

and complete the following steps

\[\begin{aligned}I_{1}=\int_{K / F}^{\infty} x f_{X}(x) d x & =\int_{K / F}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^{2} T}} e^{-\frac{\left(\ln x+\sigma^{2} T / 2ight)^{2}}{2 \sigma^{2} T}} d x \\& =\int_{\frac{\ln K / F}{\sigma \sqrt{T}}+\sigma \sqrt{T} / 2}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{-\frac{(z-\sigma \sqrt{T})^{2}}{2}} d z \\& =P\left[N(\sigma \sqrt{T}, 1)>\frac{\ln K / F}{\sigma \sqrt{T}}+\frac{1}{2} \sigma \sqrt{T}ight]\end{aligned}\]

Compute \(I_{1}\) in the above expression to arrive at the BSM formula: \(C(0)=e^{-r T}\left(F \times I_{1}-K \times I_{2}ight)\).