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Let $X_t$ be a Levy process, and consider $$Y_t=X_t-\sum_{0<s\le t}\Delta X_s1_{\{|\Delta X_s|\ge 1\}},$$ which is a Levy process with jumps bounded by 1. Then $Y_t\in L^p$ for all $p\ge 1$ and $t>0$.

I aim to show $E[Y_t]=\alpha t$ with $\alpha=E[Y_1]$.


Let $f(t)=E[Y_t]$. By the stationary increments, we have $f(t+s)=f(t)+f(s)$. Hence it is easy to show for $\alpha=E[Y_1]$, we have $$f(q)=q\alpha, \quad q\in \mathbb{Q}^+.$$ Then given $t>0$, consider $q_n$ decreases to $t$, then $$q_n\alpha=f(q_n)=f(t)+f(q_n-t),$$ hence it suffices to show $\lim_{s_n\to 0} E[Y_{s_n}]=E[Y_0]=0$. I know for each $\omega$, the right continuity of sample paths gives that $Y_{s_n}(\omega)\to Y_0(\omega)=0$, then if I can obtain a integrable bounded of $Y_{s_n}$ for all $n$ and $\omega$, the Dominated Convergence Theorem gives me the desired result.

However, I don't know how to obtain the integrable bound.

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2 Answers 2

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Claim: as $t\to 0$ we have $f(t):=\mathbb{E}(Y_t)\to 0.$

Suppose that this is not the case, i.e. there exists $t_n \to 0$ such that for all $n, f(t_n)>\epsilon >0$ say. Then $f(t_n \times[1/t_n])\ge \epsilon \times[1/t_n]\to \infty$ which contradicts the fact that $f$ is bounded. (for that you can use the fact that $Y_t-\alpha t $ is abounded $L^p$ martingale and use Doob maximal inequality. )

There should be a simpler way to say this.

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  • $\begingroup$ Thanks for your answer. But I still hope to get a more direct proof since in fact, the result in OP is one of the steps to show $Y_t-\alpha t$ is a martingale. However, inspired by your arguments, even though I don't have the explicit expression of $E[Y_t]$, I can instead use the fact $Y_t-E[Y_t]$ is a $L^p$ martingale and then Doob maximal inequality. $\endgroup$
    – John
    Mar 28, 2017 at 16:31
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I just found an elementary proof.

Since $X$ is a Levy process with bounded jumps, it has finite moments of all orders. By stationary and independent increments, we have additive property of $E[X_t]$ and $Var[X_t]$, which leads to $$E[X_q]=bq, \quad Var[X_q]=\sigma^2 q,\quad q\in \mathbb{Q}^+,$$ for some $b\in \mathbb{R}$ and $\sigma\ge 0$.

For any given $t>0$, we find $q_n\in \mathbb{Q}$ decreases to $t$. Let $n\to \infty$, it suffices to show $\lim_{q_n\to 0, q_n\in \mathbb{Q}}E[X_{q_n}]=0$.

To demonstrate the above statement, we show $X_{q_n}$ is Cauchy in $L^2$. In fact, for $q_1<q_2$. \begin{align} E[|X_{q_1}-X_{q_2}|^2]&=E[|X_{q_2-q_1}|^2]=Var[X_{q_2-q_1}]+E[X_{q_2-q_1}]^2\\ &=\sigma^2(q_2-q_1)+b^2(q_2-q_1)^2\to 0, \end{align} as $q_1,q_2\to 0$. Thus $X_{q_n}\to Y$ in $L^2$, and in probability. Since $X$ is Levy, we know it is continuous in probability, hence $Y=X_0=0$. Thus $X_{q_n}\to 0$ in $L^2$, which proves the desired result.

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