# Convergence of the Laplace Exponent of a Compound Poisson Process, Lèvy Fluctuation Theory

The question is about spectrally positive Lévy processes.

For certain $$d, \sigma^{2} \geq 0$$ and measure $$\Pi_{\varphi}(\cdot)$$ such that $$\int_{(0,\infty)} \min \{1, x^2 \} \Pi_{\varphi}(\cdot) (dx) < \infty$$, the Laplace exponent reads $$\varphi(\alpha)= \alpha d + \frac{1}{2} \alpha^2 \sigma^2 + \int_{(0,\infty)} \big(e^{-\alpha x} - 1 + \alpha x \quad \mathbb{1}_{x \in (0,1) } \big) \Pi_{\varphi}(dx).$$ For a sequence $$\epsilon_n$$ such that $$\epsilon_n \to 0$$ as $$n \to \infty$$, we define: $$\varphi_n(\alpha)= \bigg(d+\int_{\epsilon_n}^{1} x \Pi_{\varphi}(dx) + \frac{\sigma^2}{\epsilon_n} \bigg) \alpha + \frac{\sigma^2}{\epsilon_n^2} \big(e^{-\alpha \epsilon_n} - 1 \big) + \int_{\epsilon_n}^{\infty} \big(e^{-\alpha x} - 1 \big)\Pi_{\varphi}(dx).$$

How can I prove that for $$\alpha \geq 0$$, it holds thats $$\varphi_n(\alpha) \to \varphi(\alpha)\quad \text{for} \quad n \to \infty?$$

• Welcome to MSE. Shouldn't there be a $1$ instead of an $a$ in the integral, between the exponential and the liniear term? – Kore-N Mar 13 at 13:59
• Thank you, you are right. I corrected my mistake! – Christina Mar 13 at 14:18

I think only two ingredients are necessary: Taylor expansion and dominated convergence theorem. Let us rewrite $$\varphi_n(\alpha)$$:
$$\varphi_n(\alpha) = \alpha d {+} \frac{1}{2} \alpha^2 \sigma^2 + o(\varepsilon_n) + \int_{\varepsilon_n}^{{+}\infty} \big(e^{-\alpha x} - 1 + \alpha x 1_{(0,1)}(x) \big) \Pi_{\varphi}(dx).$$
Now, since $$|e^{-\alpha x} - 1 + \alpha x | \leq C|x|^2$$ for $$x \in [0,1]$$ we can pass the limit under integral sign in the last integral and deduce the convergence.
• Thank you for your answer! Can you give the steps how you rewrote $\varphi_n(\alpha)$? The first and last term is clear for me (I ended up with the same when I rewrote it), but how do you arrive at the second term? – Christina Mar 13 at 14:21
• Apart of permuting the sums I just used Taylor on: $\frac{\sigma^2}{\varepsilon_n^2}(e^{-\alpha \varepsilon_n} -1 +\alpha\varepsilon_n) = \frac{1}{2} \alpha^2 \sigma^2 + o(\varepsilon_n)$. – Kore-N Mar 13 at 14:55