Consider for one car owner the insurance policy with the following clauses:

  • Deductible: If the loss $X>d$, then the insurer pays only for loss above $d>0$.

  • Coverage Limit: If the loss $X>l$, then the insurer pays only for loss below $l>d$.

  • Distorted Distribution: The insurer may base the premium of the loss $X$ on the loss $Z$ with survival function $h(S_{x})$, where $S_{x}$ is the survival function of $X$ and $h:[0,1]\mapsto [0,1]$ is a continuous, strictly-increasing function.

Now, we are to assume that the potential loss $X$ to the owner is of Pareto distribution with density $\displaystyle f(x)=\frac{ab^a}{x^{a+1}}$, for $x \geq b$, $a>0$, $d>b>0$.

If we let $Y$ be the potential loss to the insurer, then $$Y = \begin{cases}0, & \text{if}\,b<x\leq d \\ (X-d)_{+}, & \text{if}\, d<x \leq l \\ l-d, & \text{if}\, x>l \end{cases} $$ I need to determine the induced distribution (measure) $\mu_{Y}$ and the corresponding probability distribution function $F_{Y}$ (which my professor never ceases to remind me are two different things).

For the first part, finding $\mathbf{\mu_{Y}}$, I have the following:

Let $A \in \mathcal{B}$ (the borel sets on the real line). Then, $$ \mu_{Y}(A) = P(Y \in A) = P(g(X) \in A)\\ (\text{where} Y=g(X),\,X\,\text{is has the distribution}\,\mu_{X}, \, \text{and}\,g:\mathbb{R}\to\mathbb{R})\\ = P(X\in g^{-1}(A)) = \mu_{X}(g^{-1}(A)) \\ = \mu_{X}(g^{-1}(A \cap \{0\}) + \mu_{X}(g^{-1}(A \cap (X-d)_{+}))+\mu_{X}(g^{-1}(A \cap (l-d))) \\ = \mu_{X}(g^{-1}(\{0\}) + \mu_{X}(g^{-1}(A \cap (X-d)_{+}))+\mu_{X}(g^{-1}(A\cap (l-d))) \\ = 1_{(\,\cdot \,)} \mu_{X}([b,d])+\mu_{X}(d,x] + 1_{(\, \cdot \,)} \mu_{X} (l, + \infty)$$

In the $(\,\cdot \,)$ for the indicator functions, I should have a set in each one in order to "switch on or off" those parts of the function as needed, but I'm not sure in either case what the set is supposed to be. So, my first question is, what is/are this/these set(s)?

Secondly, is this all I have to do for this part?

Thirdly, I am needing help calculating the probability distribution function $F_{Y}$ This is what I have so far:

  • for any $y<0$, $P(Y \leq y) = P(X < b) = 0$
  • for $y = 0$, we have $\displaystyle P(Y=0) = P(b \geq X \leq d) = \int_{b}^{d} \frac{ab^{a}}{x^{a+1}}dx = 1 - \left(\frac{b}{d}\right)^{a}$
  • for $0<y \leq l-d$, we have $\displaystyle P(Y \in (0,y]) = P(d<X \leq l) = \int_{d}^{l} \frac{ab^{a}}{x^{a+1}}dx = \left(\frac{b}{d} \right)^{a} - \left(\frac{b}{l} \right)^{a}$ (or do I need to do it from $P(b<X \leq l)$ because when $Y \in (0,y]$, the owner is still reponsible for the deductible?)
  • for $y > l-d$, we have $\displaystyle P(Y \geq y) = P(d<X<\infty) = \int_{d}^{\infty}f(x)dx$? I am not sure where I'm supposed to integrate for this part.

So, essentially, I need help with figuring out what the indicator functions should be on the induced probability distribution measure part, and I need help figuring out the induced probability distribution function. There are some gaps in my knowledge of the technical aspects of handling these things, so the more detailed your answer is, the more instructive it will be to me.

I genuinely hope that someone can help me with this, because I have been stuck on this problem for a few days now, and this is as far as I have gotten.

I thank you ahead of time for your time and patience!

  • 1
    $\begingroup$ You should use a different variable for damage (which is distributed according to f(x)) and for the loss to the car owner (which is always less than the damage). $\endgroup$ – user25959 Nov 14 '17 at 17:18

You are making quite a mess of the part finding $\mu_Y$.

To keep things less complex I would rather write:

$$\begin{aligned}\mathsf P\left(Y\in A\right) & =\mathsf P\left(Y\in A\wedge X\leq d\right)+\mathsf P\left(Y\in A\wedge d<X\leq l\right)+\mathsf P\left(Y\in A\wedge X>l\right)\\ & =\mathsf P\left(0\in A\wedge X\leq d\right)+\mathsf P\left(X-d\in A\wedge d<X\leq l\right)+\mathsf P\left(l-d\in A\wedge X>l\right)\\ & =\mathsf P\left(0\in A\right)\mathsf P\left(X\leq d\right)+\mathsf P\left(X-d\in A\wedge d<X\leq l\right)+\mathsf P\left(l-d\in A\right)\mathsf P\left(X>l\right)\\ & =1_{A}\left(0\right)\mathsf P\left(X\leq d\right)+\mathsf P\left(X\in d+A\cap\left(0,l-d\right]\right)+1_{A}\left(l-d\right)\mathsf P\left(X>l\right) \end{aligned}\tag1 $$

Here $\mu_Y(A):=\mathsf P(Y\in A)$ and $\mu_X(A)=\mathsf P(X\in A)$.

Observe that a constant like $0$ can be looked at as a random variable and that $\mathsf P(0\in A)=1_A(0)$. These constant random variables are always independent wrt to any other random variable, so that: $$\mathsf P\left(0\in A\wedge X\leq d\right)=\mathsf P\left(0\in A\right)\mathsf P\left(X\leq d\right)=1_A(0)P\left(X\leq d\right)$$ To find CDF observe that: $$\mathsf F_{Y}\left(y\right)=\mathsf P\left(Y\in\left(-\infty,y\right]\right)$$ so to find it we must substitute $A=(-\infty,y]$ in $(1)$ leading to:

$$\begin{aligned}\mathsf F_{Y}\left(y\right) & =1_{\left(-\infty,y\right]}\left(0\right)\mathsf P\left(X\leq d\right)+\mathsf P\left(X\in d+\left(-\infty,y\right]\cap\left(0,l-d\right]\right)+1_{\left(-\infty,y\right]}\left(l-d\right)\mathsf P\left(X>l\right)\\ & =1_{\left[0,\infty\right)}\left(y\right)\mathsf P\left(X\leq d\right)+\mathsf P\left(X\in\left(d,\min\left(y+d,l\right)\right]\right)+1_{\left[l-d,\infty\right)}\left(y\right)\mathsf P\left(X>l\right)\\ & =1_{\left[0,\infty\right)}\left(y\right)\mathsf F_{X}\left(d\right)+\left[\mathsf F_{X}\left(\min\left(y+d,l\right)\right)-\mathsf F_{X}\left(d\right)\right]_{+}+1_{\left[l-d,\infty\right)}\left(y\right)\left(1-\mathsf F_{X}\left(l\right)\right) \end{aligned}\tag2 $$

Finding $\mathsf F_X$ and substituting in $(2)$ makes things complete now.

  • $\begingroup$ I mean it when I say that that was one of the most helpful and instructive answers I have ever received on MSE. Thank you. $\endgroup$ – ALannister Nov 17 '17 at 22:53
  • $\begingroup$ You are very welcome. $\endgroup$ – drhab Nov 18 '17 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.