In this case, $r = 0.10$, $\sigma = 0.45$, and $\Delta t = \frac{1}{12}$ (since the time interval is 1 month).
Plugging in these values, we get $u = e^{(0.10 - \frac{1}{2}(0.45)^2)\frac{1}{12} + 0.45\sqrt{\frac{1}{12}}} \approx 1.0513$.
The down factor, $d$, is
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