# Interchanging the order of limit and integration

I'm trying to interchange the order of limit and integration: $$\displaystyle\lim_{t\to0}\int_0^\infty{(1-e^{-v})\frac{P(X_t\geq v)}{t}dv}$$ where $X_t$ is a compound Poisson process so $\displaystyle X_t = \alpha t + \sum_{i=0}^{N_t}{\Delta X_i}$ where $N_t\sim Poi(\lambda t)$

To begin with I assumed $\alpha=0$ because I felt that term did not matter, and I was able to move the limit inside the integral by conditioning on the number of jumps. However, if $\alpha>0$ and I condition on the number of jumps, the case of no jump (which vanishes if $\alpha=0$) gives: \begin{aligned} \displaystyle\lim_{t\to0}\int_0^\infty{(1-e^{-v})\frac{P(\alpha t\geq v)}{t}dv} =\displaystyle\lim_{t\to0}\int_0^\infty{(1-e^{-v})\frac{{\bf 1}\{v\leq\alpha t\}}{t}dv} \end{aligned} and the dominated convergence theorem does not apply. However, $$\int_0^\infty{(1-e^{-v})\frac{{\bf 1}\{v\leq\alpha t\}}{t}dv}\leq C\frac{\alpha t}{t}=C\alpha,\forall t\leq t_0$$ where $C$ is a constant. Does this uniform boundedness somehow enable me to move the limit inside the integral?

• Got something from the answer below? – Did Apr 2 '13 at 7:59

First note that, for every random variable $X$ and every real number $v$, $\mathbb P(X\gt v)=\mathbb E(\mathbf 1_{X\gt v})$. Hence, when $X$ is almost surely nonnegative, Fubini-Tonelli theorem yields $$\int_0^\infty(1-\mathrm e^{-v})\mathbb P(X\gt v)\ \mathrm dv=\mathbb E\left(\int_0^X(1-\mathrm e^{-v})\ \mathrm dv\right)=\mathbb E\left(X-1+\mathrm e^{-X}\right).$$ Thus, the integral whose limit when $t\to0$ is to be computed is $$t^{-1}\mathbb E(u(X_t))\quad\text{with}\quad u(x)=x-1+\mathrm e^{-x}.$$ When $t\to0$, $\mathbb P(N_t=0)=1-\lambda t+o(t)$, $\mathbb P(N_t=1)=\lambda t+o(t)$, and $\mathbb P(N_t\geqslant2)=o(t)$. On the event $[N_t=0]$, $X_t=\alpha t$. On the event $[N_t=1]$, $X_t=\alpha t+Z$, where the jumps of the compound Poisson process are distributed like $Z$. Thus, $$\mathbb E(u(X_t))=u(\alpha t)(1-\lambda t+o(t))+\mathbb E(u(\alpha t+Z))(\lambda t+o(t))+\mathbb E(u(X_t);N_t\geqslant2).$$ Since $u(0)=u'(0)=0$, the RHS is $$o(t)+\big[\mathbb E(u(Z))\lambda t+o(t)\big]+o(t),$$ and finally, $$\lim_{t\to0}t^{-1}\mathbb E(u(X_t))=\lambda\,\mathbb E(u(Z))=\lambda\,\mathbb E(Z-1+\mathrm e^{-Z}).$$