Financial Risk Manager Part 2

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Explanation:

From the above information, we have that:
$k = 0.42$ , $r_\infty = 5.286\%$ , $\sigma = 2.07\%$ , $r_0 = 4.256\%$ and $\lambda = 0.256\%$ .

Thus the expected change in the short rate is:

\mathrm{d}r = K(r_\infty - R)\mathrm{d}t + \lambda\mathrm{d}t + \sigma\mathrm{d}w = k\left([r_\infty + \frac{\lambda}{k}] - r\right)\mathrm{d}t + \sigma\mathrm{d}w

Note that:

\theta \equiv r + \lambda/k

\therefore \theta \equiv 5.286\% + \left(\frac{0.256\%}{0.42}\right) = 5.9\%

\mathrm{d}r = 0.42 \times (5.9\% - 4.256\%) \frac{1}{12} + 0.0207 \times 0.16 = 0.0038874 \text{ or } 38.874 \text{ basis points}

The volatility over the next month is:

$`$ 207` \times \sqrt{\frac{1}{12}} = 59.76 \text{ basis points}

Explanation:

From the above information, we have that:
$k = 0.42$ , $r_\infty = 5.286\%$ , $\sigma = 2.07\%$ , $r_0 = 4.256\%$ and $\lambda = 0.256\%$ .

Thus the expected change in the short rate is:

\mathrm{d}r = K(r_\infty - R)\mathrm{d}t + \lambda\mathrm{d}t + \sigma\mathrm{d}w = k\left([r_\infty + \frac{\lambda}{k}] - r\right)\mathrm{d}t + \sigma\mathrm{d}w

Note that:

\theta \equiv r + \lambda/k

\therefore \theta \equiv 5.286\% + \left(\frac{0.256\%}{0.42}\right) = 5.9\%

\mathrm{d}r = 0.42 \times (5.9\% - 4.256\%) \frac{1}{12} + 0.0207 \times 0.16 = 0.0038874 \text{ or } 38.874 \text{ basis points}

The volatility over the next month is:

$`$ 207` \times \sqrt{\frac{1}{12}} = 59.76 \text{ basis points}

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Last updated: July 5, 2026 at 07:08

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