Calculating variance of a claim payment I am very confused about this problem and would appreciate some enlightening.
Question: Losses, $X,$ under an insurance policy are exponentially distributed with mean $10.$ For each loss, the claim payment $Y$ is equal to the amount of the loss in excess of a deductible $d > 0.$ Calculate $\text{Var}(Y).$
My apparently incorrect reasoning:
$Y = X - d,$ and we have this theorem:
"For any constants $a$ and $b,$ $\text{Var}(aX + b) = a^2\text{Var}(X)$"
So $\text{Var}(Y) = \text{Var}(X - d) = \text{Var}(X) = 100$ since $X$ is an exponential RV. Where is my mistake?
The answer given is:
$100[2e^{−0.1d} − e^{−0.2d}].$
 A: Given the memoryless property of the exponential distribution, the distribution of $Y=\max(X-d,0)$ will have the same $\lambda=\frac 1 {10}.$
The expected probability that the insurance company will have to pay at all (claim larger than deductible) is given by the distribution of the exponential evaluated at $x>d$:
$$\Pr(X>d)= e^{-0.1 d}$$
Therefore $Y=0$ with probability $1-e^{-0.1d},$ and follow an exponential distribution with mean $10$ with probability $e^{-0.1 d}.$
The $E(Y)=0\cdot (1-e^{-0.1d}) + 10 \cdot e^{-0.1d}$
We are going to need the formula $\text{Var}=E(Y^2)-E(Y)^2,$ and the latter part after the minus is $100\cdot e^{-0.2d}.$
The $E(Y^2)$ requires some leg work. By LOTUS,
$$\begin{align}
E(Y^2) &= \int_0^\infty y^2 \cdot 0.1 e^{-0.1y}\;dy\\[2ex]
&=\left.y^2 \cdot (-e^{-0.1 y})\right|_{0}^{\infty} - \int_0^{\infty} (-e^{0.1 y}) \cdot 2y \; dy\\[2ex]
&=\frac{2}{0.1}\int_0^{\infty}y\; 0.1e^{0.1 y}  \; dy\\[2ex]
&=\frac{2}{0.1}E(Y) = \frac{2}{0.1}\frac{1}{0.1}=2\cdot 100
\end{align}$$
but before we apply the formula of the variance, we need to go back to the deductible, and the probability that the company actually has to make payments:
$$E(Y^2)=0\cdot (1-e^{-0.1d})+200\cdot e^{-0.1d}.$$
Now we obtained the desired result $100 \left(2 e^{-0.1d}- e^{-0.2d}\right).$
