Calculating percent reduction in variance after deductible

Question

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?

Answer 1

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.

Answer 2

The law of total expectation gives $$\begin{align} \operatorname{E}[Y] &= \operatorname{E}[(X - d)_+] \\ &= \operatorname{E}[X - d \mid X > d]\Pr[X > d] + \operatorname{E}[0 \mid X \le d]\Pr[X \le d] \\ &= \operatorname{E}[X]\Pr[X > d]. \end{align}$$ Note the last equality uses the memoryless property.

Similarly, $$\begin{align} \operatorname{E}[Y^2] &= \operatorname{E}[(X-d)^2 \mid X > d]\Pr[X > d] \\ &= \operatorname{E}[X^2]\Pr[X > d]. \end{align}$$ Since we know for an exponential distribution with mean $\mu$ has survival $S_X(x) = \Pr[X > x] = e^{-x/\mu}$ and second moment $\operatorname{E}[X^2] = 2\mu^2$, it follows that $$\operatorname{E}[Y] = \mu e^{-d/\mu}, \quad \frac{\operatorname{E}[Y]}{\operatorname{E}[X]} = e^{-d/\mu} = 0.9,$$ and $$\operatorname{Var}[Y] = 2\mu^2 e^{-d/\mu} - (\mu e^{-d/\mu})^2, \quad \frac{\operatorname{Var}[Y]}{\operatorname{Var}[X]} = 2e^{-d/\mu} - (e^{-d/\mu})^2 = 2(0.9) - (0.9)^2 = 0.99$$ and there is a $1\%$ reduction in the variance.

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No integration is needed. If you did not recall the second moment of the exponential distribution, you can get it from the variance which is $\mu^2$ and first moment $\mu$: $$\operatorname{E}[X^2] = \operatorname{Var}[X] + \operatorname{E}[X]^2 = \mu^2 + \mu^2 = 2\mu^2.$$