Answer:
Let's break down the problem step by step:
Given:
- Compound Poisson process with \( \lambda = 2 \).
- Individual claim density function \( f(x) = e^{2x} + 2e^{3x} \), for \( x > 0 \).
(i) To find \( p_1 \), we need to calculate the probability of having exactly one claim during a given time period. Since the compound Poisson process follows a Poisson distribution, we can use the formula for a Poisson distribution:
\[ p_k = \frac{e^{-\lambda} \lambda^k}{k!} \]
For \( k = 1 \), we have:
\[ p_1 = \frac{e^{-2} \cdot 2^1}{1!} = 2e^{-2} \]
(ii) To find the premium rate \( c \) with a safety loading of 80%, we need to add 80% to the expected value of the total claim amount.
The expected value of the total claim amount \( E[X] \) is given by:
\[ E[X] = \lambda \int_0^\infty x f(x) \, dx \]
\[ = 2 \left( \int_0^\infty x e^{2x} \, dx + 2 \int_0^\infty x e^{3x} \, dx \right) \]
Once we find \( E[X] \), we add 80% to it to get the premium rate \( c \).
(iii) To find the moment generating function \( M_X(r) \), we use the formula for the moment generating function of a compound Poisson distribution:
\[ M_X(r) = E[e^{rX}] \]
\[ = \exp(\lambda \int_0^\infty (e^{rx} - 1) f(x) \, dx) \]
(iv) To find the probability of ruin \( u \), we need to calculate the probability that the total claim amount exceeds the total premium collected.
\[ u = P(X_1 + X_2 + \ldots > c) \]
Where \( X_1, X_2, \ldots \) are the individual claim amounts. We can use the moment generating function to calculate this probability.
These steps should help you solve the problem. If you need further assistance with any specific part, feel free to ask!
