Calculate the percentage reduction on the variance of the claim payment 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 reduces the expected claim payment by 10%. Calculate the percentage reduction on the variance of the claim payment.
My attempt
Let $X$ = claim payment before deductible and $Y$= claim payment after deductible
$X \sim Exp(\frac{1}{\theta}) \Rightarrow f(x)=\frac{1}{\theta}e^{-\frac{x}{\theta}}$ and $F(x)=P(X<x)=1-e^{-\frac{x}{\theta}}$, $x>0,\theta>0$.
$\mu_x=\theta, \sigma^2_x=\theta^2, \mathbb{E}(X^2)=2\theta^2$
$Y=\left\{
\begin{array}{lr}
       x-d, &  \hspace{2mm} X>d \\
        0, & \hspace{2mm} otherwise\\
     \end{array}
   \right.$
$G(Y|X>d)=P(Y<y|X>D)=\frac{P(X-d<y|X>d)}{P(X>d)}=\frac{P(d<X<y+d)}{P(X>d)}=\frac{F(y+d)-F(d)}{1-F(d)}=1-e^{-\frac{y}{\theta}}, y>0$.
Thus $g(y|X>d)=\frac{d}{dy}(1-e^{-\frac{y}{\theta}})=\frac{1}{\theta}e^{-\frac{y}{\theta}}, y>0$. This means $Y|X>d \sim Exp(\frac{1}{\theta})$.
Because $\mathbb{E}(Y)=(x-d)P(X>d)=0.9\mathbb{E}(X)=0.9\theta$, we have $P(X>d)=0.9$ since $0 \le P(X>d) \le 1$ and $y=x-d=\mathbb{E}(X)=\theta$?
Thus $\mathbb{E}(Y|X>d)=y \cdot P(X>d)= \theta \cdot 0.9$, which implies that $(\mathbb{E}(Y|X>d))^2=(0.9\theta)^2=0.81\theta^2$.
$\mathbb{E}(Y^2|X>d)=y^2 \cdot P(X>d)=\int_{d}^{\infty} y^2 f(x)dx=\int_{d}^{\infty} (x-d)^2 \frac{1}{\theta}e^{-\frac{x}{\theta}}dx.$
By letting $u=x-d$, I see that this integral equals $e^{-\frac{d}{\theta}}\mathbb{E}(X^2)=(1-F(d))\cdot 2\theta^2=0.9(2\theta^2)=1.8\theta^2$ and thus
$Var(Y|X>d)=1.8\theta^2-0.81\theta^2=0.99\theta^2=0.99\sigma^2_x$.
$ \therefore Var(Y|X>d$) is reduced by $1$%.
One solution I was looking at uses the law of total expectation as follows:
\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}}, k \in \mathbb{N}
\end{align*}
Why is this true?
 A: You wrote

Because $\mathbb E[Y] = (X-d) \Pr[X > d] = 0.9 \mathbb E[X]$

which is not correct.  You should write
$$\mathbb E[Y] = \mathbb E[X-d \mid X > d] \Pr[X > d] = \operatorname{E}[X]\Pr[X > d] = 0.9 \mathbb E[X];$$  that is to say, you have omitted the expectation operator, and expectation on the RHS is conditional on $X > d$; then since $X$ is memoryless, $(X - d \mid X > d) \sim X$.  This is what allows us to claim $\mathbb E[X - d \mid X > d] = \mathbb E[X]$, and ultimately, $\Pr[X > d] = 0.9$.  It is not necessary to do all the previous work.  If you wish to perform the computation explicitly, then
$$\begin{align}
\operatorname{E}[Y] &= \int_{x=0}^\infty \max(x - d, 0) f_X(x) \, dx \\
&= \int_{x=d}^\infty (x-d) \frac{1}{\theta} e^{-x/\theta} \, dx \\
&= \int_{y=0}^\infty y \frac{1}{\theta} e^{-(y+d)/\theta} \, dy \tag{$x = y + d$} \\
&= e^{-d/\theta} \int_{y=0}^\infty \frac{y}{\theta} e^{-y/\theta} \, dy \\
&= \theta e^{-d/\theta} \\
&= \mathbb E[X] \Pr[X > d].
\end{align}$$
The purpose of memorylessness is to avoid this computation, but either way, it is not difficult.
To calculate the variance of $Y$, we first compute the second moment in the same way as we did the first:
$$\mathbb E[Y^2] = \mathbb E[(X-d)^2 \mid X > d]\Pr[X > d] = \mathbb E[X^2] \Pr[X > d].$$  Again, we use the fact that $X$ is memoryless, hence $\left((X - d)^2 \mid X > d\right) \sim X^2$.  So $$\operatorname{E}[Y^2] = 2\theta^2 \Pr[X > d] = 1.8 \theta^2,$$ and
$$\operatorname{Var}[Y] = 1.8 \theta^2 - (0.9)^2 \theta^2 = 0.99 \theta^2.$$
It is easy to see in the general case that
$$\mathbb E[Y^k] = \mathbb E[(X - d)^k \mid X > d]\Pr[X > d] + \mathbb E[0 \mid X \le d]\Pr[X \le d] = \mathbb E[X^k] \Pr[X > d].$$  This is just a consequence of the memorylessness property.
Then the moments are simply $$\mathbb E[X^k] = \int_{x=0}^\infty x^k \frac{1}{\theta} e^{-x/\theta} \, dx = \theta^{k-1} \int_{x=0}^\infty (x/\theta)^k e^{-x/\theta} \, dx = \theta^k \int_{z=0}^\infty z^k e^{-z} \, dz = \theta^k k!.$$  Alternatively, we can reason that
$$M_X(t) = \mathbb E[e^{tX}] = \int_{x=0}^\infty \frac{1}{\theta} e^{tx} e^{-x/\theta} \, dx = \frac{1}{\theta(1/\theta - t)} \int_{x=0}^\infty (1/\theta - t) e^{-(1/\theta - t)x} \, dx = \frac{1}{1 - \theta t},$$ for $t < 1/\theta$.  But by series expansion and linearity of expectation, $$\mathbb E[e^{tX}] = \sum_{k=0}^\infty \mathbb E \left[\frac{(tX)^k}{k!}\right] = \sum_{k=0}^\infty \frac{\mathbb E[X^k]}{k!} t^k,$$ hence
$$\frac{1}{1 -\theta t} = \sum_{k=0}^\infty (\theta t)^k = \sum_{k=0}^\infty \frac{\mathbb E[X^k]}{k!} t^k,$$ and by comparing coefficients, we obtain
$$\mathbb E[X^k] = \theta^k k!.$$
A: If you are asking about the calculation in the solution you mention, assuming the claim when paid out by the insurance company (i.e. claim conditional on the deductible being exceeded) has exponential distribution with scale parameter $\lambda>0$, we have by iterated expectations
$$E[Y^k]=E[E[Y^k|{\bf1}_{X\geq d}]]\\
=E[Y^k|{\bf1}_{X\geq d}=1]P({\bf1}_{X\geq d}=1)+E[Y^k|{\bf1}_{X\geq d}=0]P({\bf1}_{X\geq d}=0)\\
=E[Y^k|X\geq d]P(X\geq d)+E[Y^k|X< d]P(X<d),$$
and the second term is zero since the insurance company doesn't make any payment when the deductible is not exceeded.
Their expression then follows from using two facts for any random variable $W$ that has an exponential distribution with scale parameter $\lambda>0$:
CDF: $P(W\leq w)=1-\exp(-w/\lambda),\quad w\geq 0,$
Moments: $E[W^k]=k!\lambda^k,\quad k\in\mathbb{N}.$
