Hawkes Process Basics

/(1 - n)$ is the self-excitation multiplier. As $\alpha \to \beta$, the process becomes critical and the mean intensity diverges.

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(iii) Log-Likelihood

For a point process on $[0, T]$ with conditional intensity $\lambda(t)$, the log-likelihood is:

$\ell = \sum_{i=1}^{n} \ln \lambda(t_i) - \int_0^T \lambda(t) \, dt$

First, the conditional intensity at each event time:

$\lambda(t_i) = \mu + \sum_{j < i} \alpha e^{-\beta(t_i - t_j)}$

Second, the compensator (integrated intensity). Split the integral:

$\int_0^T \lambda(t) \, dt = \mu T + \sum_{i=1}^{n} \int_{t_i}^{T} \alpha e^{-\beta(t - t_i)} \, dt$

Each kernel integral evaluates to:

$\int_{t_i}^{T} \alpha e^{-\beta(t - t_i)} \, dt = \frac{\alpha}{\beta}\left(1 - e^{-\beta(T - t_i)}\right)$

Putting it together:

$\boxed{\ell = \sum_{i=1}^{n} \ln\!\left(\mu + \sum_{j < i} \alpha e^{-\beta(t_i - t_j)}\right) - \mu T - \frac{\alpha}{\beta} \sum_{i=1}^{n} \left(1 - e^{-\beta(T - t_i)}\right)}$

Note: the double sum in the first term can be computed in $O(n)$ time using a recursive update. Define $R_i = \sum_{j < i} e^{-\beta(t_i - t_j)}$. Then $R_1 = 0$ and $R_i = e^{-\beta(t_i - t_{i-1})}(R_{i-1} + 1)$ for $i \geq 2$. This makes MLE fitting practical even for large datasets.

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(iv) Expected Descendants by Time $T$

Let $D(T)$ be the expected total number of descendants (all generations) of an immigrant at time $0$, counted up to time $T$.

First generation (direct children): The immigrant triggers offspring at rate $\alpha e^{-\beta t}$. The expected number of direct children by time $T$ is:

$m_1(T) = \int_0^{T} \alpha e^{-\beta t} \, dt = \frac{\alpha}{\beta}\left(1 - e^{-\beta T}\right) = n\left(1 - e^{-\beta T}\right)$

All generations: A child born at time $s$ itself triggers a sub-cascade with the same statistics, but the remaining observation window is $T - s$. By the branching property, $D(T)$ satisfies the renewal equation:

$D(T) = m_1(T) + \int_0^{T} \alpha e^{-\beta s} \, D(T - s) \, ds$

This is solved via Laplace transforms. Let $\hat{D}(z) = \int_0^{\infty} e^{-zt} D(t) \, dt$ and $\hat{\phi}(z) = \alpha/(z + \beta)$. The renewal equation gives:

$\hat{D}(z) = \frac{\hat{\phi}(z)}{z(1 - \hat{\phi}(z))} = \frac{\alpha}{z(z + \beta - \alpha)}$

Partial fractions:

$\hat{D}(z) = \frac{\alpha}{\beta - \alpha}\left(\frac{1}{z} - \frac{1}{z + \beta - \alpha}\right)$

Inverting:

$\boxed{D(T) = \frac{\alpha}{\beta - \alpha}\left(1 - e^{-(\beta - \alpha)T}\right) = \frac{n}{1 - n}\left(1 - e^{-(\beta - \alpha)T}\right)}$

Sanity checks: - As $T \to \infty$: $D(\infty) = n/(1 - n)$, the total progeny of a sub-critical branching process. - As $T \to 0$: $D(T) \approx \alpha T$, just the first-generation children in a short window. - As $\alpha \to 0$ (no excitation): $D(T) \to 0$. - The rate of convergence is governed by $\beta - \alpha$: the closer to criticality, the longer the cascade takes to die out.

Answer: The stability condition is $\alpha / \beta < 1$. The branching ratio is $n = \alpha / \beta$. The stationary mean intensity is $\bar{\lambda} = \mu\beta/(\beta - \alpha)$. The log-likelihood is the standard point-process formula with the exponential kernel admitting an $O(n)$ recursive computation. The expected descendants by time $T$ are $D(T) = \frac{n}{1-n}(1 - e^{-(\beta - \alpha)T})$.