Show that for the sum of two exponential distributions, $f(x) = e^{-3x} + \frac{10}{3} e^{-5x}, x>0$ 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:
$$\lim_{b \to \infty} e^b = \infty$$
 A: 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)
