An insurance policy is written to cover a loss, $X$, where $X$ has a uniform distribution on $[0,1000]$ At what level must a deductible be set in order for the expected payment to be $25$% of what it would be with no deductible?
We want to find a $D$ such that:
$X \in [0,D] \cup (D,1000]$
So:
$\frac{1}{4} 500 = \int_D^{1000} \frac{X-D}{1000}dx$
I figured since D is a constant we would treat it as such and after integrating we would get:
$\frac{x^2-x}{2000} \Big\vert_D^{1000}$
but instead it is supposed to come out as:
$\frac{(X-D)^2}{2000} \Big\vert_D^{1000}$
Why is this?
Write $Y=(X-D)\cdot\mathsf 1_{\{X>D\}}$, then for $0\leqslant t\leqslant 1000-D$ we have \begin{align} \mathbb P(Y\leqslant t) &= \mathbb P(Y\leqslant t,X\leqslant D) + \mathbb P(Y\leqslant t,X> D)\\ &= \mathbb P((X-D)\cdot\mathsf 1_{\{X>D\}}\leqslant t, X\leqslant D) + \mathbb P((X-D)\cdot\mathsf 1_{\{X>D\}}\leqslant t, X>D)\\ &= \mathbb P(X\leqslant D) + \mathbb P(D<X\leqslant t+D)\\ &= \frac D{1000} + \left(\frac{t+D}{1000}\right)\cdot\mathsf 1_{(0,1000-D]}(t)\\ &= \frac{t+D}{1000}. \end{align} It follows that \begin{align} \mathbb E[Y] &= \int_0^{1000-D} \mathbb P(Y>t)\ \mathsf dt\\ &= \int_0^{1000-D} \left(1-\frac{t+D}{1000}\right)\mathsf dt\\ &= \int_D^{1000} \left(1-\frac t{1000}\right)\mathsf dt\\ &= 1000 - \frac{(1000)^2}{2\cdot1000} - D + \frac{D^2}{2\cdot 1000}\\ &= 500 - D\left(1-\frac D{2000}\right). \end{align} With no deductible, the expected payment is $\mathbb E[X]=500$, and hence $$ \mathbb E[Y]= \frac14\cdot 500 \implies 500 - D\left(1-\frac D{2000}\right) = 125\implies D=500. $$