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Let $(X_t)_{t\ge 0}$ be a stochastic process. Let $$M_n(a_1,\dotsc, a_n; t_1, \dotsc, t_n) = \mathbb{E}e^{\sum_{i=1}^n a_i X_{t_i}},\,$$ where $t_i,\, 1\le i \le n$ are distinct.

  1. Is it essential that $$\lim_{h_1\to 0,\dotsc, h_n\to 0}M_n(a_1, \dotsc, a_n; t+h_1,\dotsc, t+h_n) = M_1\left(\sum_{i=1} a_i; t\right)$$ Almost every process that I can think of including Levy processes, Gaussian process satisfy this condition.
  2. If the answer to the above question is no, I was wondering about the implication of a process not satisfying this condition.

Example

Let $(X_t)_{t\ge 0}$ be a Gaussian process with mean function $\mu$ and covariance function $\Sigma$. Then $$M_n(a_1,\dotsc, a_n; t_1,\dotsc, t_n) = \exp\left(\sum_{i=1}^n a_i \mu(t_i) + \frac{1}{2}\sum_{i=1}^n \sum_{j=1}^n a_ia_j \Sigma(t_i, t_j) \right)$$

Moreover, $$\lim_{h_1\to 0,\dotsc, h_n\to 0}M_n(a_1, \dotsc, a_n; t+h_1,\dotsc, t+h_n) = \exp\left(\sum_{i=1}^n a_i \mu(t) + \frac{1}{2}\sum_{i=1}^n \sum_{j=1}^n a_ia_j \Sigma(t, t) \right) = M_1(\sum_{i=1}^n a_i; t)$$

user2808118
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1 Answers1

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Here is a counterexample. Let $X_t$ be the process that is $0$ for all $t \leq 1$ and $1$ for all $t > 1$. Then $$M_1(a;t) = E[e^{aX_t}]$$ This function is $1$ for $t \leq 1$ and $e^a$ for $t>1$. Now your condition is violated with $n = 1$ and $a \neq 0$ as $t$ approaches $1$ from the right. $$e^a = \lim_{h \to 0} M_1(a; 1+h) \neq M_1(a;1) = 1$$

The key to this example is that there is a discontinuous jump in the "density" of the process at time $t = 1$.

muaddib
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