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 $$ \varphi_n(\alpha) \to \varphi(\alpha)\quad \text{for} \quad n \to \infty? $$

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

  • $\begingroup$ 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? $\endgroup$ – Christina Mar 13 at 14:21
  • $\begingroup$ Apart of permuting the sums I just used Taylor on: $\frac{\sigma^2}{\varepsilon_n^2}(e^{-\alpha \varepsilon_n} -1 +\alpha\varepsilon_n) = \frac{1}{2} \alpha^2 \sigma^2 + o(\varepsilon_n)$. $\endgroup$ – Kore-N Mar 13 at 14:55

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