Financial Risk Manager Part 1

For a Poisson distribution, the probability mass function is given by:

$p(x) = \frac{e^{-\lambda} \lambda^x}{x!}$

Given: $p(1) = \frac{e^{-\lambda} \lambda^1}{1!} = 0.09$ $p(2) = \frac{e^{-\lambda} \lambda^2}{2!} = 0.0045$

We can solve for $\lambda$ by taking the ratio: $\frac{p(2)}{p(1)} = \frac{\frac{e^{-\lambda} \lambda^2}{2!}}{\frac{e^{-\lambda} \lambda}{1!}} = \frac{\lambda}{2}$

So: $\frac{0.0045}{0.09} = \frac{\lambda}{2}$ $0.05 = \frac{\lambda}{2}$ $\lambda = 0.10$

For a Poisson distribution, the variance equals the mean: $\text{Var}(X) = \lambda = 0.10$

Therefore, the correct answer is 0.1.