The amount of a claim that a car insurance company pays out follows an exponential distribution. By imposing a deductible of $d$, the insurance company reduced the expected claim payment by $10\%$. Calculate the percentage reduction on the variance of the claim payment.
This question was previously discussed here. Based on this discussion, I came up with this solution:
Let $X$ be the random variable for the claim amount before the deductible, d, is applied and $Y$ be the random variable for the claim amount after the deductible, d, is applied. Then, $Y = X-d$ if $d \le X <\infty$ and $Y = 0$ otherwise.
Now, $E[Y] = E[X] – 0.1 E[X] \implies E[Y] = 0.9E[X] = E[0.9X] \implies Y = 0.9X$.
So, $Var[Y] = Var[0.9X] = 0.9^2Var[X] = 0.81 Var[X]$
which means that the variance has been reduced by approximately $19\%$ but the correct answer is $1\%$. What have I done incorrectly?
Best Answer
Note that $$ \mathbb{E}[Y] \overset{\star}= \mathbb{E}[0.9 X] $$ does not in general imply $Y = 0.9X$. And what about the deductible? As you already noted, we actually have the following relationship between $X$ and $Y$: \begin{align*} Y = \boldsymbol{1}_{\{X\geq d\}} (X-d). \end{align*} The idea is to use $\star$ to determine $d$, and then calculate $\text{Var}[Y]$ more or less directly.
Proof
Note that $X \sim \text{exp}(\lambda)$, so that in particular $\mathbb{E}[X] = \lambda$. To simplify matters, let us first note that conditionally on $(X \geq d)$, the random variable $Y$ also follows an $\text{exp}(\lambda)$-distribution. This may be seen from the calculation \begin{align*} \mathbb{P}(Y \leq y \, | \, X \geq d) &= \mathbb{P}(X - d \leq y \, | \, X \geq d) \\ &= \frac{\mathbb{P}(d \leq X \leq y + d)}{\mathbb{P}(d \leq X)} \\ &= \frac{e^{-\frac{d}{\lambda}}-e^{-\frac{y+d}{\lambda}}}{e^{-\frac{d}{\lambda}}} \\ &= 1 - e^{-\frac{y}{\lambda}} \end{align*} for $y > 0$. It is actually a direct consequence of the so-called memoryless property of the exponential distribution.
For $k\in\mathbb{N}$, we obtain from the law of total expectation that \begin{align*} \mathbb{E}[Y^k] &= \mathbb{E}[Y^k \, | \, X \geq d]\cdot\mathbb{P}(X \geq d) + \mathbb{E}[Y^k \, | \, X < d]\cdot\mathbb{P}(X < d) \\ &= \mathbb{E}[Y^k \, | \, X \geq d]\cdot\mathbb{P}(X \geq d) \\ &= k! \lambda^k e^{-\frac{d}{\lambda}}, \end{align*} since the $n$'th moment of an exponential distribution with mean $\lambda$ is $n!\lambda^n$.
In particular, \begin{align*} \mathbb{E}[Y] &= \lambda e^{-\frac{d}{\lambda}}. \end{align*} Thus imposing $\star$, the deductible $d$ is given as the solution to \begin{align*} \lambda e^{-\frac{d}{\lambda}} = 0.9\mathbb{E}[X] = 0.9\lambda, \end{align*} which yields $d = -\log(0.9)\lambda \approx 0.11 \lambda$. (You may think about why $d>0.1 \lambda$ is reasonable.)
Using this value of $d$, which exactly yields $e^{-\frac{d}{\lambda}} = 0.9$, we find that \begin{align*} \text{Var}[Y] &= \mathbb{E}[Y^2] - \mathbb{E}[Y]^2 \\ &= 2\lambda^2 \cdot 0.9 - (0.9 \lambda)^2 \\ &= 0.99 \lambda^2 \\ &= 0.99 \text{Var}[X], \end{align*} as desired.