# Mixed Probability Distribution - Minimum of Pareto distributed r.v.'s

Suppose that the potential loss of a risk $X$ has a Pareto distribution and therefore probability density function $$f(x) = \frac{\alpha x_{0}^{\alpha}}{x^{\alpha + 1}},\,\, \text{for shape parameter}\,\alpha > 0\, \text{and location parameter}\, x_{0}>0.$$

Now, the coverage limit $l$ for a risk loss $X$ (with probability density $f$ as given above) is the largest amount to be claimed. I need to determine the probability distribution of the insurer's potential attained loss, which is $Y = \min\{ X, l\}$.

For some background, in my notes, I have the case when instead of the coverage limit, we consider a deductible $d > x_{0}$, where the insurer attains the potential loss $W = (X-d) = \max \{X-d,0\}$, and we have the following three cases:

• Case 1: $w<0$. Then, $P(W \leq w) = 0$
• Case 2: $w = 0$. Then, $\displaystyle P(W=0)=P(W \leq d) = \int_{x_{0}}^{d} f(x) dx = - \left[ \left(\frac{x_{0}}{x} \right)^{\alpha}\right ]_{x_{0}}^{d}$
• Case 3: $w > 0$. Then, $$P(W \leq w) = P(W = 0) + P(W \in (0,w]) = P(X \leq d) + P(X \in (d, d+w]) \\ =- \left[ \left(\frac{x_{0}}{x} \right)^{\alpha}\right ]_{x_{0}}^{d} + \int_{d}^{d+w}f(x)dx = - \left[ \left(\frac{x_{0}}{x} \right)^{\alpha}\right ]_{x_{0}}^{d} - \left[ \left(\frac{x_{0}}{x} \right)^{\alpha}\right ]_{d}^{d+w}$$ So $W$ has a distribution neither discrete nor absolutely continuous.

I'm supposed to do something similar here with $Y$, but I'm not sure how what we did with $W$ "translates" into dealing with mins instead of maxes. Could somebody please help me with this? Thank you.

It's very similar. We know $Y\le l$ so $P(Y\le l)=1.$ You will lose $l$ whenever the damage exceeds $l,$ so the distribution has a mass at $l$ of size
$$P(Y=l) = P(X\ge l) = \int_l^\infty f(x)dx.$$ Then we lose the amount of the damage provided it is less than $l,$ so for $y<l,$ $$P(Y\le y) = P(X\le y) = \int_{x_0}^y f(x)$$ so there is a continuous distribution of losses below $l.$
• As a pedantic side-note, it is probably a bad idea to call $x_0$ a location parameter, which has a technical meaning. It is a scale parameter. – spaceisdarkgreen Nov 10 '17 at 1:49
• is it discrete at $Y = l$? – user100463 Nov 10 '17 at 2:02
• no, your instructor just misspoke I think. And yes it's discrete at $l$ (generally I say it has an atom at $l$ or a mass at $l$), since $P(Y=l) > 0.$ – spaceisdarkgreen Nov 10 '17 at 2:14
• @ALannister The fact that there's a deductible just means you always subtract the deductible off of the insurer's loss, so doesn't have anything to do with the limits. In the case where the damage is greater than the cap, the insurer pays out the cap minus the deductible. So there a probability $\int_{cap}^\infty p(x)dx$ that the insurance company's loss is $cap-deductible.$ Then below that there will be $\int_{x_0}^y f(x)dx$ probability that the loss is less than $y-deducible.$ Then below that, there is a $\int_{x_0}^{deductible}f(x)dx$ probability that the insurer pays zero. – spaceisdarkgreen Nov 17 '17 at 5:24