Financial Risk Manager Part 1

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Explanation:

To find the probability that the bond price is between $8.00 and $9.00, we need to integrate the probability density function over the interval [8, 9]:

P(8 \leq X \leq 9) = \int_{8}^{9} \left(\frac{x}{8} - 0.75\right) dx

First, let's integrate:

\int \left(\frac{x}{8} - 0.75\right) dx = \frac{x^2}{16} - 0.75x + C

Now evaluate from 8 to 9:

\left[\frac{9^2}{16} - 0.75 \times 9\right] - \left[\frac{8^2}{16} - 0.75 \times 8\right]

= \left[\frac{81}{16} - 6.75\right] - \left[\frac{64}{16} - 6\right]

= \left[5.0625 - 6.75\right] - \left[4 - 6\right]

= [-1.6875] - [-2] = -1.6875 + 2 = 0.3125

Converting to percentage: 0.3125 × 100% = 31.25%

Therefore, the probability is 31.250%, which corresponds to option C.

Explanation:

To find the probability that the bond price is between $8.00 and $9.00, we need to integrate the probability density function over the interval [8, 9]:

P(8 \leq X \leq 9) = \int_{8}^{9} \left(\frac{x}{8} - 0.75\right) dx

First, let's integrate:

\int \left(\frac{x}{8} - 0.75\right) dx = \frac{x^2}{16} - 0.75x + C

Now evaluate from 8 to 9:

\left[\frac{9^2}{16} - 0.75 \times 9\right] - \left[\frac{8^2}{16} - 0.75 \times 8\right]

= \left[\frac{81}{16} - 6.75\right] - \left[\frac{64}{16} - 6\right]

= \left[5.0625 - 6.75\right] - \left[4 - 6\right]

= [-1.6875] - [-2] = -1.6875 + 2 = 0.3125

Converting to percentage: 0.3125 × 100% = 31.25%

Therefore, the probability is 31.250%, which corresponds to option C.

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Assume the probability density function (pdf) of a zero-coupon bond with a notional value of `$10.00` is given by $f(x) = \frac{x}{8} - 0.75$ on the domain [6,10] where x is the price of the bond:
$f(x) = \frac{x}{8} - 0.75 \quad \text{s.t. } 6 \leq x \leq 10 \text{ where } x = \text{bond price}$
What is the probability that the price of the bond is between `$8.00` and `$9.00`?

Exam-Like

25.750%

0.0%

28.300%

0.0%

31.250%

100.0%

44.667%