Let $X(t)$ denote the number of claims received by an insurance company in the time interval $[0,t]$. We will assume that ${X(t) : t ≥ 0}$ can be modelled as a Poisson process, where $t$ is measured in days since January 1st at 00:00.00. Assume that the rate of the Poisson process is given by $λ(t) = 1.5, t ≥ 0$. Further, Assume that the monetary claims are independent, and are independent of the claim arrival times. Each claim amount (in mill. kr.) has an exponential distribution with rate parameter $\gamma = 10$. This means that claim $C_i ∼ Exp(\gamma)$, i = 1,2,....
The policy of the insurance company is to investigate a claim if and only if it exceeds 250000 kr.. For $t ≥ 0$, let $Y_t$ denote the number of claims, which need to be investigated, received in the time interval $[0, t]$. Prove that ${Y (t) : t ≥ 0}$ is a Poisson process and compute its rate.
So we can write that $Y(t) = \sum_{i=1}^{X(t)}I(C_i > 0.25)$ where $I$ is the indicator function. Then clearly, $Y(0) = 0$ since $X(0)$ is $0$ because no claims have been made yet at time $0$. Further, $Y(t)$ must have independent increments, since the monetary claims are independent, and are independent of the claim arrival times.
Now this is where I struggle to continue. Im not really sure how to go about computing the rate of $Y(t)$. I believe that $$E[Y(t)] = P(C_i > 0.25) X(t).$$ Perhaps this can be used to find the rate somehow? I know the expected value of a Poisson distributed random variable is $\lambda t$, however the expected value I found is not directly proportional to $t$...