= Solution
The average log price is
$$
\frac1T\int_0^T\log S_tdt
=\log S_0+\frac12\left(r-\frac12\sigma^2\right)T
+\frac\sigma T I_T.
$$
Part (b) gives $\operatorname{var}(I_T)=T^3/3$, so the Gaussian variance of the final term is
$$
v=\frac13\sigma^2T.
$$
Rewriting the lognormal factor in the normalized form $e^{-v/2+\sqrt vZ}$ and discounting gives
$$
S_0e^{-(r/2+\sigma^2/12)T}
F\left(\frac13\sigma^2T,
\frac{Ke^{-(r/2-\sigma^2/12)T}}{S_0}\right).
$$
\b[Thus $\alpha=1/12$.]
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