Financial Risk Manager Part 1

Explanation

For a Poisson random variable, the probability mass function is:

[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}]

Given:

• [P(X = 1) = 0.09 = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-\lambda} \lambda] ...(i)
• [P(X = 2) = 0.0045 = \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^2}{2} = 0.0045]

From equation (ii): [\frac{e^{-\lambda} \lambda^2}{2} = 0.0045] [e^{-\lambda} \lambda^2 = 0.009] ...(ii)

Now, divide equation (ii) by equation (i): [\frac{e^{-\lambda} \lambda^2}{e^{-\lambda} \lambda} = \frac{0.009}{0.09}] [\lambda = 0.1]

For a Poisson distribution, the variance equals the mean: [\text{Var}(X) = \lambda = 0.1]

Therefore, the correct answer is D. 0.1