Financial Risk Manager Part 1

Step 1: Find the constant C

Since the probability density function must integrate to 1 over its entire domain:

$$\int_{-\infty}^{\infty} f(x)\,dx = 1$$ $$\int_{0}^{0.5} C e^{-x}\,dx = 1$$ $$C \left[-e^{-x}\right]_0^{0.5} = 1$$ $$C \left[-e^{-0.5} + e^{0}\right] = 1$$ $$C \left[-e^{-0.5} + 1\right] = 1$$ $$C (1 - e^{-0.5}) = 1$$ $$C = \frac{1}{1 - e^{-0.5}} \approx 2.5$$

Step 2: Find the 75th percentile

The 75th percentile $q_{75}$ satisfies:

$$P(X < q_{75}) = \int_0^{q_{75}} f_X(x)\,dx = 0.75$$ $$\int_0^{q_{75}} C e^{-x}\,dx = 0.75$$ $$C \left[-e^{-x}\right]_0^{q_{75}} = 0.75$$ $$C \left[-e^{-q_{75}} + 1\right] = 0.75$$ $$-e^{-q_{75}} + 1 = \frac{0.75}{C}$$ $$e^{-q_{75}} = 1 - \frac{0.75}{C}$$ $$-q_{75} = \ln\left(1 - \frac{0.75}{C}\right)$$ $$q_{75} = -\ln\left(1 - \frac{0.75}{C}\right)$$

Substituting $C = \frac{1}{1 - e^{-0.5}}$:

$$q_{75} = -\ln\left(1 - 0.75(1 - e^{-0.5})\right)$$ $$q_{75} = -\ln\left(1 - 0.75 + 0.75e^{-0.5}\right)$$ $$q_{75} = -\ln\left(0.25 + 0.75e^{-0.5}\right)$$ $$q_{75} \approx -\ln\left(0.25 + 0.75 \times 0.6065\right)$$ $$q_{75} \approx -\ln\left(0.25 + 0.4549\right)$$ $$q_{75} \approx -\ln(0.7049)$$ $$q_{75} \approx 0.36$$

Therefore, the 75th percentile is approximately 0.36.