# Another Actuary Question

Here is another question I have while practicing for the P/! exam:

An automobile policy owner has a 10% chance of having an accident in a $3$ year period. Given that an accident occurs, the payout for the accident follows the following distribution

$$f(x) = 0.5e^{-0.5x}$$

with $x$ representing thousands of dollars. If no accident takes place, the driver receives a rebate of $\$100$. What is the standard deviation of the payout? Since$x$appears to be following an exponential distribution, I thought that$\mu$would be$2$(or since$x$is in thousands of dollars, the average payout to be$\$2000$ assuming there is an accident)

I know form the answer key that the answer is $\$851$, but I'm stuck as to why. • You have pretty much answered your own question: "the average payout [is]$2000 assuming there is an accident". But you also have to allow for the case in which there isn't an accident. Jan 5 '18 at 0:38
• David, my plan was to determine the variance then take the square root. Here's exactly how far I got: E(x) = .1($2000) + .9($100) = $290. However, when I go to calculate E(x^2), that's where I'm getting stuck. I thought to use the formula E(x^2)= 0.1(E((x_a)^2)) + 0.9(E((x_n)^2)) with x_a being the payout assuming there is an accident and x_n being the payout assuming there is no accident. Jan 5 '18 at 0:53 ## 2 Answers You have calculated the mean assuming there is an accident as$\$2000$. You could also calculate the standard deviation as $\$2000$assuming there is an accident, making$E[X\mid A]=2000$and$E[X^2\mid A]=2000^2+2000^2=8000000$Similarly you could calculate$E[X\mid A^c]=100$and$E[X^2\mid A^c]=0^2+100^2=10000$That then makes •$E[X]=0.1\times 2000 + 0.9\times 100=\$290$
• $E[X^2]=0.1\times 8000000 + 0.9\times 10000=809000$
• $\text{Var}(X)=809000-290^2=724900$
• the overall standard deviation $\sqrt{724900}\approx \$851$In this case, if accident then the payout has mean$2000$with variance$2000^2$. If no accident then the payout has mean$100$with variance$0. By the law of total variance, \begin{align*} \text{total var} &= \text{mean}(\text{var}) + \text{var}(\text{mean})\\ &= 0.1 \times 2000^2 + (0.9 \times 100^2 + 0.1 \times 2000^2 - 290^2)\\ &= 724900, \end{align*} and\sqrt{724900} = 851.41\$.

I just learned this in a probability class. I only vaguely understand it conceptually.

• Thank you! I had thought to try to set it up using this formula, but by the time I thought of it, I was so frustrated with this problem and so tired that I couldn't figure out how to set it up. Jan 5 '18 at 12:42