Financial Risk Manager Part 2

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Explanation:

To find the total change in the interest rate from the current level (date 0) to the upper node at date 2, we need to calculate the change in the interest rate at each period (monthly, meaning $dt = 1/12$ ) and sum them up.

The change formula is: $dr = \lambda(t) dt + \sigma(t) dw$ Since we are looking at the upper node, $dw = +\sqrt{dt}$ . So, for an upward move: $dr = \lambda(t) dt + \sigma(t) \sqrt{dt}$ .

Step 1: Calculate the change for period 1 (date 0 to date 1) $dt = 1/12 \approx 0.083333$ $\sqrt{dt} = \sqrt{1/12} \approx 0.288675$

$dr_1 = 0.0018 \times (1/12) + 0.0050 \times \sqrt{1/12}$ $dr_1 = 0.000150 + 0.0050 \times 0.288675 = 0.000150 + 0.001443 = 0.001593$

Step 2: Calculate the change for period 2 (date 1 to date 2) $dr_2 = 0.0030 \times (1/12) + 0.0080 \times \sqrt{1/12}$ $dr_2 = 0.000250 + 0.0080 \times 0.288675 = 0.000250 + 0.002309 = 0.002559$

Step 3: Calculate the total change Total Change = $dr_1 + dr_2 = 0.001593 + 0.002559 = 0.004152$ Convert this to basis points (bps) by multiplying by 10,000: $0.004152 \times 10,000 = 41.52 \text{ bps}$

This is approximately 42 bps, making Option B the correct answer.

Explanation:

The change formula is: $dr = \lambda(t) dt + \sigma(t) dw$ Since we are looking at the upper node, $dw = +\sqrt{dt}$ . So, for an upward move: $dr = \lambda(t) dt + \sigma(t) \sqrt{dt}$ .

Step 1: Calculate the change for period 1 (date 0 to date 1) $dt = 1/12 \approx 0.083333$ $\sqrt{dt} = \sqrt{1/12} \approx 0.288675$

$dr_1 = 0.0018 \times (1/12) + 0.0050 \times \sqrt{1/12}$ $dr_1 = 0.000150 + 0.0050 \times 0.288675 = 0.000150 + 0.001443 = 0.001593$

Step 2: Calculate the change for period 2 (date 1 to date 2) $dr_2 = 0.0030 \times (1/12) + 0.0080 \times \sqrt{1/12}$ $dr_2 = 0.000250 + 0.0080 \times 0.288675 = 0.000250 + 0.002309 = 0.002559$

This is approximately 42 bps, making Option B the correct answer.

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Question An analyst at an investment bank is calculating the monthly changes in a short-term interest rate tree given that the bank uses the following model to describe the interest rate process:

$dr = \lambda(t)dt + \sigma(t)dw$
In this process, $\lambda(t)$ represents the drift at date $t$ , $\sigma(t)$ represents the volatility at date $t$ , and $dw$ is a normally distributed random variable with a mean of zero and a standard deviation of $\sqrt{dt}$ . The analyst uses the following inputs to make the calculations:

Current level (date 0) of short-term interest rate: 2.75%

Drift at date 1 ( $\lambda(1)$ ): 0.0018

Drift at date 2 ( $\lambda(2)$ ): 0.0030

Annualized volatility at date 1 ( $\sigma(1)$ ): 0.0050

Annualized volatility at date 2 ( $\sigma(2)$ ): 0.0080

Probability of an upward or downward move in interest rates: 0.50

What is the change in the interest rate from the current level (date 0) to the upper node at date 2?

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Last updated: July 5, 2026 at 07:28

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