Consider an insurance company that has two types of policy: Policy $A$ and Policy $B$.
Total claims from the company arrive according to a Poisson process at a rate of $9$ per day.
A randomly selected claim has a $1/3$ chance that it is of policy $A$.
A randomly selected claim from policy $A$ has a $2/3$ probability of being over $10000$, while a randomly selected claim from policy $B$ has probability 2/9 of being over $10000$.
Determine the expected number of claims over $10000$ on a given day.
My attempt:
Label the Poisson Process that that claims are of policy $A$ and of policy $B$
$$N_A (t) \sim \mathrm{Poisson}(\frac{1}{3} \cdot t \cdot 9) = \mathrm{Poisson}(3t) \\ N_B(t) \sim \mathrm{Poisson}(\frac{2}{3}\cdot t \cdot 9) = \mathrm{Poisson}(6t)$$
respectively (so $N_A(t)$ is the random variable that counts events up to time $t$)
Now similarly, we "split" each poisson process into claims that are over $10000$ and those that are not over $10000$. So the claims of policy $A$ that are over $10000$ and claims of policy $B$ that are over $10000$ are
$$N_{A1}(t) \sim \mathrm{Poisson}(3t \cdot \frac{2}{3})=\mathrm{Poisson}(2t) \\ N_{B1}\sim\mathrm{Poisson}(6t \cdot \frac{2}{9}) = \mathrm{Poisson}(\frac{4t}{3}).$$
Hence, if we define $X(t):= N_{A1}(t) + N_{B1}(t)$ to be the number of claims in total that are over $10000$ for $t$ time units, then
$$X(t) \sim \mathrm{Poisson}(2t+\frac{4t}{3}) = \mathrm{Poisson}(\frac{10t}{3})$$
The question asks for "a given day", so set $t=1$ and so
$$\mathbb{E}(X(1)) = \frac{10}{3}$$
Is this correct? If so, did I make any awkward defining or working out? I felt awkward when trying to explain what $t$ was supposed to mean... so i feel like I did it incorrectly.
$P(Claim > 10000) = P(Policy = A)P(A > 10000) + P(Policy = B)P(B > 10000)$
The expected number of claims per day is always $9$ of which the expected number of claims for policy A is always $3$ and the rest is for policy B.