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...knowing that $X_t$ has independent increments and is adapted to its natural filtration, $u \in \mathrm{R}$

My problem is in particular how to show this process has finite mean...(can I use the fact that $e^{iux}$ is bounded $\forall x\in\mathrm{R}$?)

EDIT $E[e^{iuX_t}]\neq 0$

mg91
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    Hint: for u sufficiently close to zero, the expectation in the denominator is bounded below and for all u, the expression in the numerator is bounded by $1$ in modulus. Notice also that it is possible that there exist u for which the expectation in the denominator is zero. – Chris Janjigian Jul 23 '14 at 23:58
  • "Notice also that it is possible that there exist u for which the expectation in the denominator is zero."

    how is this possible, since the exponential never reaches zero...? can you provide me an example to better understand?

     – mg91 Jul 24 '14 at 00:07
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    Take a uniform random variable on the interval $[0,2\pi]$ or on the discrete set ${0, \pi}$ and set $u = 1$ for some simple examples. The exponential can have cancellations which sum to zero. – Chris Janjigian Jul 24 '14 at 00:16
  • What remains of the question since the result does not hold in general? Supplementary hypothesis or deletion? – Did Jul 26 '14 at 14:59
  • sorry for being so late, adding the condition that $E[e^{iuX_t}]\neq 0$ is sufficient (other than necessary) to show that this process is a martingale? using increments I arrive till $\frac {e^{iuX_s} E[e^{iu(X_t-X_s)}]}{E[e^{iuX_t}]}$

    I'm not sure if I am allowed to simplify to arrive to $\frac{e^{iuX_s}}{E[e^{iuX_s}]}$.

    ps I added the extra-condition to the original post

     – mg91 Aug 15 '14 at 16:59

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