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Claims can be of two types: "severe" or "light"

$$\text{severe claims ~ } \exp(\lambda = 3)$$ $$\text{light claims ~ } \exp(\lambda = 5)$$

The insurer has established that two thirds of all claims are "light" and the rest are "severe". Let $f(x)$ denote the probability density function of any claim payment.

Show that:

$$f_X(x) = e^{-3x} + \frac{10}{3} e^{-5x}, x>0$$

Now suppose:

A = severe claims

B = light claims

$$f_A(a) = 3e^{-3a}, a>0$$ $$f_B(b) = 5e^{-5b}, a>0$$

It appears that they have just taken:

$$f_X(x) = \frac{1}{3}f_A(a) + \frac{2}{3}f_B(b)$$

I cannot see where this comes from. Here is what I tried:

Let $X= \frac{1}{3}A + \frac{2}{3}B$

$$F_X(x) = P[X \leq x]$$

$$=P[\frac{1}{3}A + \frac{2}{3}B \leq x]$$

$$=P[A \leq 3x - 2B]$$

Now using the Law of Total Probability

$$F_X(x)=\int_0 ^{\infty} P[A = 3x - 2B | B=b] P[B=b] db$$

$$=\int_0 ^{\infty} F_A(3x-2b) f_B(b) db$$

$$=\int_0 ^{\infty} [1 - e^{-3(3x-2b)}] 5e^{-5b} db$$

$$F_X(x)=5\int_0 ^{\infty} [e^{-5b} - e^{b - 9x}]db$$

Now, regardless of whether I try to integrate this or differentiate first with respect to x, I always have the problem that:

$$\lim_{b \to \infty} e^b = \infty$$

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    $\begingroup$ please add the self-study tag (and read its wiki) $\endgroup$ Sep 21, 2017 at 8:45

1 Answer 1

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You should integrate

$\int_{b=0}^{b=1.5x}$

instead of $\infty$, since the probability is zero for $b>1.5x$.


Aside from that issue, I believe you are adding the claims and probability of claims wrongly. $P[X \leq x] = P[severe]P[A \leq x] + P[light]P[B \leq x] = \frac{1}{3}*P[A \leq x] + \frac{2}{3}P[light]P[B \leq x]$.

But nice work on your convolution operation. What you were trying to calculate is the probability that the weighted sum of a severe claim and a light claim is less than or equal to X.


comparison with throwing dice

Imagine as comparison a random throw of 6 sided (light) dices and 12 (severe) dices. What is the probability of particular throw being $X$ if the probability of a light dice is 2/3 and the probability of a severe dice is 1/3? The answer is of course 2/3 p(dice_light=X) + 1/3 p(dice_severe=X).

What you were trying to calculate is the probability of the weighted sum ($\frac{1}{3}$ and $\frac{2}{3}$) of two such dices is a. For such calculation you have to do that integral thingy you are working out. (or with dices it is easier and you count the cases)

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  • $\begingroup$ In addition. Your integral stuff with the $3x-2b$ term as function parameters won't work out (you need to use it as the integration boundaries). You should evaluate the following $\int_{b=0}^{b=2x} \left( \int_{a=0}^{a=3x-2b} 5 e^{-5a} 3 e^{-3b} da \right) db$. This is the integral $\int \int 5 e^{-5a} 3 e^{-3b} da db$ on the domain 1/3a+2/3b<x. If you place this '1/3a+2/3b<x'-stuff as parameters then the integration constants 5 and 3 are not correct anymore. $\endgroup$ Sep 21, 2017 at 11:51

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