Exponential conditional probability

Question

There are two types of claims that are made to an insurance company. Let $N_i(t)$ denote the number of type $i$ claims made by time $t$, and suppose that $\{N_1(t), t \ge 0\}$ and $\{N_2(t), t \ge 0\}$ are independent Poisson processes with rates $λ_1 = 10$ and $λ_2 = 1$.

The amounts of successive type 1 claims are independent exponential random variables with mean 1000 dollars whereas the amounts from type 2 claims are independent exponential random variables with mean 5000 dollars. A claim for 4000 dollars has just been received; what is the probability it is a type 1 claim?

Here's my approach to the problem:

\begin{align} & P(\text{claim} = \text{type 1}\mid 4000) \\[10pt] = {} & \frac{P(4000\mid \text{claim} = \text{type 1})(\text{claim} = \text{type 1})}{P(4000\mid \text{claim} = \text{type 1})P(\text{claim} = \text{type 1}) + P(4000\mid\text{claim} = \text{type 2})P(\text{claim} = \text{type 2})} \end{align}

by Bayes' formula.

My question is, how do I calculate $P(4000\mid \text{claim} = \text{type 1})$ and $P(4000\mid \text{claim} = \text{type 2})$? The dollar value of claims is exponentially distributed, and since the distribution is continuous I can't fix a value to the pdf, I need a range. Any tips on how to calculate these two values?

Answer 1

Since the claim size has a continuous distribution, the probability that it is a particular amount is $0.$ One uses a probability density instead of a probability in such cases.

\begin{align} \Pr(\text{type 1}) & = \frac{10}{11} \\[10pt] \Pr(\text{type 2}) & = \frac 1 {11} \\[10pt] L(\text{type 1}) & = \frac 1 {1000} e^{-4000/1000} \\[10pt] L(\text{type 2}) & = \frac 1 {5000} e^{-4000/5000} \end{align} $L$ is the likelihood function.

$$ \Pr(\text{type 1}\mid 4000) = \frac{\Pr(\text{type 1})\cdot L(\text{type 1})}{\Pr(\text{type 1})\cdot L(\text{type 1}) + \Pr(\text{type 2})\cdot L(\text{type 2})}. $$

Answer 2

Let $T_{i,n}$ be the arrival times of $N_i(t)$ for and $C_{i,n}$ the claim values associated with these arrivals, for $i=1,2$. Let $\lambda_i$ be the rate of $N_i(t)$ and $1/\mu_i$ the mean of $C_{i,1}$, for $i=1,2$. Let $T_n$ be the superposition of $T_{1,n}$ and $T_{2,n}$ and $C_n$ the superposition of $C_{1,n}$ and $C_{2,n}$. Then by Bayes' rule we have for any $c>0$ $$ \mathbb P(T_1=T_{1,1}\mid C_1 = c) = \frac{\mathbb P(C_1=c\mid T_1=T_{1,1})\mathbb P(T_1=T_{1,1})}{\mathbb P(C_1=c)}. $$ Now, $$ \mathbb P(T_1=T_{i,1}) = \mathbb P(T_{i,1}<T_{j,1}) = \frac{\lambda_i}{\lambda_i+\lambda_j}, \quad (i,j)\in\{(1,2),(2,1)\}, $$ $$ \mathbb P(C_i=c\mid T_1 = T_{i,1}) = f_{C_{i,1}}(c) = \mu_i e^{-\mu_i c},\quad i\in\{1,2\}, $$ and $$ \mathbb P(C_1=c) = \mathbb P(C_1=c\mid T_1=T_{1,1}) + \mathbb P(C_1=c\mid T_1=T_{2,1}), $$ and hence \begin{align} \mathbb P(T_1=T_{i,1}) &= \frac{\mu_1 e^{-\mu_1 c}\left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)}{\mu_1 e^{-\mu_1 c}\left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right) + \mu_2 e^{-\mu_2 c}\left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)}\\ &= \frac{\lambda_1\mu_1 e^{-\mu_1 c}}{\lambda_1\mu_1 e^{-\mu_1 c} + \lambda_2\mu_2 e^{-\mu_2 c}}. \end{align} In this example, we have \begin{align} \lambda_1 &= 10\\ \lambda_2 &= 1\\ \mu_1 &= 1/1000\\ \mu_2 &= 1/5000\\ c &= 4000, \end{align} and substituting these values yields $$ \frac{50}{50 + e^{16/5}} \approx 0.670848. $$