Calculating Expected Loss in a Conditional that a Loss Event occur in a particular year.

Question

Regards, I found a question below and would like to test my view on this.

Question : The probability of a driver will not experience any loss event in one year ahead is $ 90 \%.$ In one year, it is known that the driver experience an event of loss of $X$ amount. The loss density function is below : $$ f(x) = 2 e^{-2x}, \:\: 0 < x < \infty $$ What is the expected loss at that year?

an Answer : I may see this as conditional expectation : $$ E[X | L] = \int_{0}^{ \infty } x f_{X|L}(x) \: dx $$ with $L$ is the event of experiencing any loss in that year. An argument would be that the conditional pdf is : $$ f_{X|L}(x) = \frac{ f(x) }{ P(L) } = \frac{f(x)}{0.1} $$ $$ E[X | L] = \int_{0}^{ \infty } x f_{X|L}(x) \: dx = 10 \int_{0}^{\infty} 2x e^{-2x} dx = 5 $$

But the official answer to this particular question was set to $0.05$. May I have some inputs on this question, thanks. All the best.

Answer 1

Let $X$ be the loss (amount), $X\sim \mathcal{E}xp(2)$ and $L$ a Bernoulli random variable with $p=0.1$, thus the expected loss can be calculated by the total expectation formula $$ E[X] = E[E[X|L]] = 0\times0.9 + 0.1\times E[X|L] = 0.1E[X] = 0.1\times\frac{1}{2} = 0.05. $$

Answer 2

Formally, if $X$ is the severity variable and $N$ the frequency variable, then $$X \mid N = 1 \sim \operatorname{Exponential}(\lambda = 2), \\ N \sim \operatorname{Bernoulli}(p = 0.1).$$ It follows from the law of total probability that $$\operatorname{E}[X] = \operatorname{E}[X \mid N = 1]\Pr[N = 1] + \operatorname{E}[X \mid N = 0]\Pr[N = 0] = \operatorname{E}[X \mid N = 1](0.1) + 0(0.9).$$ Since $$\operatorname{E}[X \mid N = 1] = \frac{1}{\lambda} = \frac{1}{2},$$ the result follows.

Note that the unconditional (marginal) distribution of $X$ is "mixed" in the sense that it contains a discrete part and a continuous part. Thus the CDF is discontinuous: $$F_X(x) = \Pr[X \le x] = \begin{cases} 0, & x < 0 \\ 1 - \tfrac{1}{10} e^{-2x}, & x \ge 0. \end{cases}$$ This is because when $x \ge 0$, we have $$\Pr[X \le x] = \Pr[X \le x \mid N = 0]\Pr[N = 0] + \Pr[X \le x \mid N = 1]\Pr[N = 1] \\ = (1)(0.9) + (1 - e^{-2x})(0.1) \\ = 1 - \tfrac{1}{10}e^{-2x}.$$