Exponential conditional probability 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?
 A: 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.
$$
\Pr(\text{type 1}\mid 4000) = \frac{\Pr(\text{type 1})\cdot L(\text{type 1})}{\Pr(\text{type 1})\cdot L(\text{type 1}) + \Pr(\text{type 2})\cdot L(\text{type 2})}.
$$
A: 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
$$
\frac{50}{50 + e^{16/5}} \approx 0.670848.
$$
