I am trying to derive the mean and variance of a non-homogeneous Poisson process. For a homogeneous Poisson process with parameter $\lambda$, our class notes show the following derivation for the mean, $\mathbb{E}[N(s)]$, and variance, $var[N(s)]$, of the number of events over a duration of $s$:
$ \begin{align} \quad \mathbb{E}[N(s)] &= \sum_{n=0}^{\infty} n \cdot P(N(s) = n) \\ &= 0 + \sum_{n=1}^{\infty} n \cdot P(N(s) = n) \\ &= \sum_{n=1}^{\infty} n e^{-\lambda s} \frac{(\lambda s)^n}{n!} \\ &= \sum_{n=1}^{\infty} e^{-\lambda s} \frac{(\lambda s)^n}{(n-1)!} \\ &= (\lambda s) (e^{-\lambda s}) \sum_{n=1}^{\infty} \frac{(\lambda s)^{n-1}}{(n-1)!} \\ &= (\lambda s) (e^{-\lambda s}) (e^{\lambda s}) \\ &= \lambda s \end{align} $
$ \begin{align} \quad \mathbb{E}[\left ( N(s) \right ) \left ( N(s) - 1 \right )] &= \sum_{n=0}^{\infty} n(n-1) \cdot P(N(s) = n) \\ &= 0 + 0 + \sum_{n=2}^{\infty} n(n-1) e^{-\lambda s} \frac{(\lambda s)^n}{n!} \\ &= \sum_{n=2}^{\infty} e^{-\lambda s} \frac{(\lambda s)^n}{(n-2)!} \\ &= (\lambda s)^2 (e^{-\lambda s}) \sum_{n=2}^{\infty} \frac{(\lambda s)^{n-2}}{(n-2)!} \\ &= (\lambda s)^2 (e^{-\lambda s}) (e^{\lambda s}) \\ &= (\lambda s)^2 \end{align} $
$ \begin{align} \quad var[N(s)] &= \mathbb{E}[(N(s))^2] - \left ( \mathbb{E}[N(s)] \right )^2 \\ &= \mathbb{E}[(N(s))^2] - \mathbb{E}[N(s)] + \mathbb{E}[N(s)] - \left ( \mathbb{E}[N(s)] \right )^2 \\ &= \mathbb{E}[(N(s))^2 - N(s)] + \mathbb{E}[N(s)] - \left ( \mathbb{E}[N(s)] \right )^2 \\ &= \mathbb{E}[(N(s))(N(s) - 1)] + \mathbb{E}[N(s)] - \left ( \mathbb{E}[N(s)] \right )^2 \\ &= (\lambda s)^2 + \lambda s - (\lambda s)^2 \\ &= \lambda s \end{align} $
where we used the Taylor expansion $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ in the infinite sums.
When I try to apply this to a non-homogeneous process with parameter $\lambda(t)$ that is dependent on $t$, I get the mean:
$ \begin{align} \quad \mathbb{E}[N(s)] &= \lim_{\delta r \to 0} \sum_{r=0}^s \left ( \sum_{n=0}^{\infty} n \cdot P(N_r(\delta r) = n) \right ) \\ &= \int_0^s \left ( \sum_{n=0}^{\infty} n \cdot P(N_r(\delta r) = n) \right ) \\ &= \int_0^s \left ( \sum_{n=1}^{\infty} n e^{-\lambda(r) dr} \frac{(\lambda(r) dr)^n}{n!} \right ) \\ &= \int_0^s \left ( \lambda(r) dr \sum_{n=1}^{\infty} e^{-\lambda(r) dr} \frac{(\lambda(r) dr)^{n-1}}{(n-1)!} \right ) \\ &= \int_0^s \left ( \lambda(r) dr \right ) \left ( e^{-\lambda(r) dr} \right ) \left ( e^{\lambda(r) dr} \right ) \\ &= \int_0^s \lambda(r) dr \\ &= \Lambda(t, s) \end{align} $
But when I try to derive the $\mathbb{E}[\left ( N(s) \right ) \left ( N(s) - 1 \right )]$ term, I get stuck:
$ \begin{align} \quad \mathbb{E}[\left ( N(s) \right ) \left ( N(s) - 1 \right )] &= \int_0^s \left ( \sum_{n=0}^{\infty} n(n-1) \cdot P(N_r(dr) = n) \right ) \\ &= \int_0^s \left ( \sum_{n=2}^{\infty} n(n-1) e^{-\lambda(r) dr} \frac{(\lambda(r) dr)^n}{n!} \right ) \\ &= \int_0^s \left ( \left ( \lambda(r) dr \right )^2 e^{-\lambda(r) dr} \sum_{n=2}^{\infty} \frac{(\lambda(r) dr)^{n-2}}{(n-2)!} \right ) \\ &= \int_0^s \left ( \left ( \lambda(r) dr \right )^2 \left ( e^{-\lambda(r) dr} \right ) \left ( e^{-\lambda(r) dr} \right ) \right ) \\ &= \int_0^s \left ( \lambda(r) dr \right )^2 \\ &=^? \left ( \int_0^s \lambda(r) dr \right )^2 \end{align} $
If the last line is true, then the variance becomes $\Lambda(t, s)$ like I would expect. However, I feel like this is generally not true, and to be honest I'm not even sure how to interpret the square inside of the integral, which is making me question my derivation altogether...
Let $\Lambda(t) = \int_0^t\lambda(s)\ \mathsf ds$ be the cumulative rate function. Then \begin{align} \mathbb E[N(t)] &= \sum_{n=1}^\infty n e^{-\Lambda(t)}\frac{\Lambda(t)^n}{n!}\\ &= e^{-\Lambda(t)}\Lambda(t)\sum_{n=0}^\infty\frac{\Lambda(t)^n}{n!}\\ &= e^{-\Lambda(t)}\Lambda(t)e^{\Lambda(t)}\\ &= \Lambda(t). \end{align} Similarly, \begin{align} \mathbb E[N(t)(N(t)-1)] &= \sum_{n=2}^\infty n(n-1) e^{-\Lambda(t)}\frac{\Lambda(t)^n}{n!}\\ &= e^{-\Lambda(t)}\Lambda(t)^2\sum_{n=0}^\infty\frac{\Lambda(t)^n}{n!}\\\\ &= e^{-\Lambda(t)}\Lambda(t)^2 e^{\Lambda(t)}\\ &= \Lambda(t)^2. \end{align} It follows that \begin{align} \mathrm{Var}(N(t)) &= \mathbb E[N(t)(N(t)-1)] + \mathbb E[N(t)] -\mathbb E[N(t)]^2\\ &= \Lambda(t)^2 + \Lambda(t) - \Lambda(t)^2\\ &= \Lambda(t). \end{align}