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A simple question:

Let $H$ be a cadlag, adapted process and $A$ a process of finite variation. Then also

$\int_t^T HdA_t$ is a finite variation process (see "Limit Theorems... "Jacod&Shiryaev Prop. 3.5 p. 28). Is $\mathbb{E}\left[\int_t^T HdA_t\right]$ also a finite variation process?

I think one can refute that:

Suppose $H\equiv1$. Then $\mathbb{E}\left[\int_t^T HdA_t\right]=\mathbb{E}\left[A_T-A_t\right]=\mathbb{E}\left[A_T\right]-A_t$

$\mathbb{E}[A_T|\mathcal{F}_t]$ is a true martingale, if $A_T$ is integrable. Now if $\mathbb{E}[A_T|\mathcal{F}_t]$ is predictable (esp. if it is continuous) then it can be of finite variation only, if it is constant. In other words, $A$ must be deterministic, if $\mathbb{E}[A_T|\mathcal{F}_t]$ is of finite variation.

Do you agree?

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1 Answer 1

up vote 0 down vote accepted

Your example works fine for me. You can specify it a little further if you like by taking any non integrable random variable $X$ (e.g. a Cauchy r.v.), and make it a process $X_t$ defined by $X_t=0$ for all $t\in [0,1[$ and $X_1=X$ . Your $H_t=1$ $\forall t$ process is still fine for your purpose and as the integral $Z_t=\int_t^1H_s dX_s=X$ is well defined, has finite variation almost surely but as it has no finite expectation for all t, it is not a finite variation process.

Regarding the second part of your question, I don't really get what you want to claim exactly could be a little more precise so that it makes more sense to me ?

Best regards

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Sorry, there were some typos. What I used in the second part is the fact, that the only predictable, finite variation martingales are constants. This is usually used to show that naive integration (Riemann) is impossible for Brownian motion. –  user13655 Sep 27 '12 at 6:13
    
Sorry I still don't get what your point is. –  TheBridge Sep 27 '12 at 20:51
    
You may want to check Cor. 3.16 in the same book as above. Note that the process $M_t:=\mathbb{E}[A_T|\mathcal{F}_t]$ is a martingale and can thus either be constant or it is not of finite variation. Sorry for creating confusion. –  user13655 Sep 28 '12 at 6:52

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