= Solution
{c}
Over the stated year, the true total <variance> is
$$
v_{\rm true}=\alpha^2\int_0^1(1+t)^2dt
=\frac73\alpha^2.
$$
The second trader uses $v_{\rm offered}=(3\alpha/2)^2=9\alpha^2/4$. Their difference is
$$
\boxed{v_{\rm true}-v_{\rm offered}=\frac{\alpha^2}{12}>0}
$$
when $\alpha\neq0$. A nondegenerate <European call option> has positive <option vega>, equivalently a price strictly increasing in total <variance>. The offered price is therefore below the model value, so \b[buy the option] in the stated comparison.
If the deterministic <volatility> path is indeed the true one and the frictionless Black-Scholes trading assumptions apply, buy the cheap option and short its <replicating portfolio> computed with that true path. The initial cash received is the positive price difference, while the two terminal payoffs cancel exactly. Investing the initial surplus yields a strictly positive terminal gain with no net claim liability, an <arbitrage> within this model. A mere uncertain <volatility> belief would not by itself provide that replication guarantee. At $\alpha=0$ both variances coincide and this price difference disappears.
\Image[/past-exam-of-the-martingales-and-financial-mathematics-course-of-the-university-of-oxford/2007-integrated-variance.png]
{title=Equal average <volatility> does not imply equal integrated <variance>}
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