Question
$N$ Daleks are invading Earth, where $N$ is a Poisson random variable with mean $\lambda$. Assume each Dalek dies independently. Let $X_i$ be the lifetime of the $i^{th}$ Dalek in the battlefield, which is a positive random variable with PDF $f$ and CDF $F$ and $N(t)$ be the number of Daleks who have died by time $t$.
$(a)\quad$ Find the mean of $N(t)$ for each $t > 0$.
$(b)\quad$ Identify the distribution of $N(t)$ for each $t > 0$.
$(c)\quad$ Show that, for any $0 \leq t_1 < t_2 \leq t_3 < t_4$, $N(t_2) - N(t_1)$ and $N(t_4) - N(t_3)$ are independent. What type of process is $N(t)$?
I have thought long and hard about this problem, but as my professor has just covered Poisson processes, I probably do not have enough exposure to know what the problem requires, so I would like to know how to approach each part. In particular, how do I show that $N(t_2) - N(t_1)$ and $N(t_4) - N(t_3)$ are independent? Any intuitive solutions for the three parts will be greatly appreciated :)
Edit
I have made some progress following some help from an answer. In particular, I know that $N(t)$ is a thinned Poisson random variable with parameter $\lambda F(t)$. However, I am still unsure of my working for part $(c)$, which I replicate below:
Consider the case when $t_3 = t_2$.
$$\begin{aligned} \implies & \mathbb{P}[N(t_4) - N(t_3) = x, N(t_2) - N(t_1) = y)]\\[1 mm] & = \mathbb{P}[N(t_4) - N(t_2) = x, N(t_2) - N(t_1) = y)]\\[1 mm] & = \sum^{\infty}_{z = 0} \mathbb{P}[N(t_1) = z, N(t_2) = z + y, N(t_4) = z + y + x]\\[1 mm] & = \sum^{\infty}_{z = 0} \mathbb{P}[N(t_1) = z]P_{z, z + y}(t_2 - t_1)P_{z + y, z + y + x}(t_4 - t_2)\\[1 mm] & = \sum^{\infty}_{z = 0} \mathbb{P}[N(t_1) = z]P_{0y}(t_2 - t_1)P_{0x}(t_4 - t_2)\\[1 mm] & = P_{0y}(t_2 - t_1)P_{0x}(t_4 - t_2)\\[1 mm] & = \mathbb{P}[N(t_4) - N(t_2) = x]\mathbb{P}[N(t_2) - N(t_1) = y)]\\[1 mm] & = \mathbb{P}[N(t_4) - N(t_3) = x]\mathbb{P}[N(t_2) - N(t_1) = y)] \end{aligned}$$
Now, consider the case when $t_3 > t_2$.
$$\begin{aligned} \implies & \mathbb{P}[N(t_4) - N(t_3) = w, N(t_3) - N(t_2) = x, N(t_2) - N(t_1) = y)]\\[1 mm] & = \sum^{\infty}_{z = 0} \mathbb{P}[N(t_1) = z, N(t_2) = z + y, N(t_3) = z + y + x, N(t_4) = z + y + x + w]\\[1 mm] & = \sum^{\infty}_{z = 0} \mathbb{P}[N(t_1) = z]P_{z, z + y}(t_2 - t_1)P_{z + y, z + y + x}(t_3 - t_2)P_{z + y + x, z + y + x + w}(t_4 - t_3)\\[1 mm] & = \sum^{\infty}_{z = 0} \mathbb{P}[N(t_1) = z]P_{0y}(t_2 - t_1)P_{0x}(t_3 - t_2)P_{0w}(t_4 - t_3)\\[1 mm] & = P_{0y}(t_2 - t_1)P_{0x}(t_3 - t_2)P_{0w}(t_4 - t_3)\\[1 mm] & = \mathbb{P}[N(t_4) - N(t_3) = w]\mathbb{P}[N(t_3) - N(t_2) = x)]\mathbb{P}[N(t_2) - N(t_1) = y) \end{aligned}$$
Thus, in both cases, we have independence and can conclude that $N(t_2) - N(t_1)$ and $N(t_4) - N(t_3)$ are independent $\forall\ 0 \leq t_1 < t_2 \leq t_3 < t_4$. In particular, $N(t)$ is a homogenous Poisson process.
Is my reasoning sound and in particular, would I be able to actually combine both cases?
Conditioned on $N=n$, the distribution of $N(t)$ is $\text{Binomial}(n, F(t))$.
(a) So, $E[N(t) \mid N=n] = nF(t)$. Thus, $$E[N(t)] = E[E[N(t) \mid N]] = E[N\cdot F(t)] = \lambda \cdot F(t).$$
(b) For the unconditional distribution of $N(t)$, again go through conditioning on $N=n$. \begin{align}P(N(t)=k) &= \sum_{n = k}^\infty P(N(t)=k \mid N=n) P(N=n) \\ &= \sum_{n=k}^\infty \binom{n}{k} F(t)^k (1-F(t))^{n-k} e^{-\lambda} \frac{\lambda^n}{n!} \\ &= \sum_{n=k}^\infty e^{-\lambda F(t)}\frac{(\lambda F(t))^k}{k!} \cdot e^{-\lambda (1-F(t))}\frac{(\lambda (1-F(t)))^{n-k}}{(n-k)!} \\ &= e^{-\lambda F(t)} \frac{(\lambda F(t))^k}{k!}. \end{align}
(c)
$$P(N(t_2)-N(t_1) = k, N(t_4)-N(t_3)=k') = P(N(t_2)-N(t_1) = k) \cdot P(N(t_4)-N(t_3)=k')$$ for any $k,k'$. I suspect if you start from the left-hand side and do a similar conditioning on $N=n$ you can arrive at the final result with a lot of tedious accounting. Not sure if there is a slicker way. I may revisit this later.