Question

See the link below for the question. It's a question for those familiar with actuarial science and statistics.

http://i.imgur.com/qcCRKio.png

Answer 1

Hm since \(LER=\dfrac{E (X\land d)}{E(X)}\) for ordinary deductible, but \(E( X\land d)\) includes dedutibles \(d\) when \(X > d\), so for a franchise deductible, I believe you'd get
\[ LER = \frac{E(X \land d)-d\cdot P(X > d)}{E(X)}\] since there is no deductible applied for \(X >d\) and you want to subtract away those deductibles from \(E(X \land d)\)

Since your have \(X\sim PAR(3, 800)\), and that the Pareto is a scale distribution, so you can apply inflation factor to the 2nd parameter \(\theta=800\) to get \(800 \cdot 1.08 = 864\).

Then you can fin, using the formula: \[ E(X \land d)=\int_0^d xf(x)\, dx+d\cdot S(d)\] where \(S(d)=1-F(d)\)

But the expectation of the Pareto has the easy formula, \[E(X)=\frac{\theta}{\alpha-1}=\frac{864}{3-1}=432\]

\[f(x)=\frac{\alpha\theta^{\alpha}}{(x+\theta)^{\alpha+1}}=\frac{3(864)^3}{(x+864)^{3+1}}\]
So, \[\int_{0}^{300}x \cdot \frac{3(864)^3}{(x+864)^4 }\, dx \\ =3(864)^3 \int_{0}^{300} \frac{x}{(x+864)^4}\, dx \\=3(864)^4 \int_{864}^{1164} \frac{u-864}{u^4 }\, du \\
=3(864)^3 \int_{864}^{1164}\left( u^{-3}-864u^{-4}\right) \, du \\
= 71.296\]

Now, \(P(X > d)=S(d)=1-F(d)=1-\left[ 1-\left(\frac{864}{300+864} \right)^3\right]=0.40896\)

So, \[ E(X \land d)=71.296+300(0.40896)=193.984\]

so, \[LER=\frac{193.984-300(0.40896)}{432}=0.1650 \]
I think there is a faster way to do this but I don't quite remember.

Answer 2

oh ya the integral I just used a simple substitution \(u=x+864\) and change the bounds accordingly