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For a given Process $K_{t}=\exp(B_{t}+\theta t)$ with $\theta\in\mathbb{R}$ and $B_{t}$ a Wiener process i want to show, that $K_{t}$ does have dependent increments. My idea is: $$ \begin{split} K_{t+s} &= \exp(B_{t+s}+\theta(t+s)) \\ &= \exp(B_{t}+\theta t)\exp(B_{t+s}-B_{t}+\theta s)\\ &= K_{t}\exp(B_{t+s}-B_{t}+\theta s) \end{split} $$ With $B_{0}=0$ We have $K_{0}=1$ and we define the two increments \begin{align} Z_{1}:=K_{1}-K_{0}=K_{1}-1\,,\,Z_{2}:=K_{2}-K_{1}=K_{1}\left[\exp(\theta)\exp(B_{2}-B_{1})-1\right] \end{align} Since $\exp(\theta)\exp(B_{2}-B_{1})$ is independent of $K_{1}$ and both increments depend on $K_{1}$ it follows, that $K_{t}$ doesn't have independent increments.

An argumentation i found is, that $K_{t}=\exp(B_{t}+\theta t)$ solves the SDG \begin{align} dK_{t}=(\theta+1/2 )K_{t}dt+K_{t}dB_{t} \end{align} with initial value $K_{0}=1$. It is said, that by the form of the SDG, the dependence of the increments follow. How can you argue this way?

gt6989b
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ziT
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  • "How can you argue this way?" Hint: by the uniqueness of the solutions of this SDE, if $K$ solves it then $K_2=K_1K'_1$ where $K'_1$ is independent of $K_1$ and follows the same distribution. Can you conclude? – Did Sep 05 '15 at 21:24
  • @Did $K_{2}=\exp(B_{2}+2\theta)=\exp(B_{1}+\theta)\cdot\exp(B_{2}-B_{1}+\theta)=K_{1}K_{1}'$. Due to the properties of the Brownian motion we have that $B_{1}\sim B_{2}-B_{1}\sim \mathcal{N}(0,1)$ so they have the same distribution. So far so good. I am stuck with that independence, because i would have argued like in the mid-section of my ownpost: "Since $K_{1}'$ is independen of $K_{1}$ and the increments $K_{1}-K_{0}$ and $K_{2}-K_{1}$ depend on $K_{1}$ they doesn't have independent increments." But in the nex post Ian mentoined, that this argumentation is wrong. Can you give a little hint? – ziT Sep 05 '15 at 22:33
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    If $K_2-K_1=K_1(K'_1-1)$ and $K_1-K_0=K_1-1$ are independent, then $E(K_1(K'_1-1)(K_1-1))=E(K_1(K'_1-1))E(K_1-1)$, that is, $E(K_1(K_1-1))E(K_1-1)=E(K_1)E(K_1-1)^2$. Can you continue massaging this? – Did Sep 06 '15 at 00:26
  • @Did Assuming the independence the right side of your last eqauation is erqual to $E[K_{1}]E^{2}[K_{1}-1]=EK_{1}$. The left side is equal to: $E[K_{1}^{2}K_{1}']-E[K_{1}^{2}]-E[K_{1}']E[K_{1}]+E[K_{1}]$. It holds that $K_{1}^{2}=\exp(2B_{1}+2\theta)]=\exp(B_{4}+4\theta)\exp(-2\theta)$ (because $2B_{1} \sim\mathcal{N}(0,2^{2})$ is independent of $K_{1}'$ since the Brownian motion has independent increments. So we have this expression for the left side of the equation: $E[K_{1}^{2}]E[K_{1}']-E[K_{1}^{2}]-E^{2}[K_{1}]+E[K_{1}]$ . Continuing in the next post – ziT Sep 06 '15 at 12:49
  • now in combination with the right side: $E[K_{1}^{2}]E[K_{1}']-E[K_{1}^{2}]-E^{2}[K_{1}]+E[K_{1}]=EK_{1}$. Shortening : $E[K_{1}^{2}]E[K_{1}']-E[K_{1}^{2}]=E[K_{1}]\cdot E^{2}[K_{1}]-E^{2}[K_{1}]$. This leads to $EK_{1}^{2}=E^{2}K_{1}$. But for $K_{1}: E[K_{1}^{2}]\not=E^{2}[K_{1}]$. Thus the assumption is wrong – ziT Sep 06 '15 at 12:56
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    ...And $E(K_1)\ne1$ (if ever $\theta$ is such that $E(K_1)=1$, then $E(K_2)\ne1$ hence repeat the argument with $K_2$ and $K_4$ instead of $K_1$ and $K_2$). Well done. – Did Sep 06 '15 at 13:30

1 Answers1

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We consider a diffusion process satisfying

$$dX_t = b(X_t) dt + \sigma(X_t) dB_t.$$

A process with independent increments with finite mean is a martingale and therefore has constant expectation. A process of the form above with a nontrivial drift term $b$ is "generically" not a martingale, because

$$\int_s^t b(X_u) du = E[X_t]-E[X_s]$$

provided $\sigma$ is sufficiently nice that the stochastic integral term forms a martingale. (For example this is true if $\sigma$ is bounded, as well as under much milder assumptions.) Now if $b$ is not identically zero then the integral on the left side "generically" can't be zero for all $s,t$.

Assuming your work is correct, you must be more careful in the case $\theta=-1/2$, for then the drift is in fact zero. In this case you basically need to argue that when $K$ is growing, the noise intensity is strengthening.

Edit: as @Did pointed out, there is a flaw here: a process with independent mean zero increments is a martingale. So one should use something like the second argument even when $\theta \neq -1/2$.

Ian
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  • This example is from : "Uwe Küchler, Michael Sorensen Exponential Families of Stochastic Processes 1997" Example 4.2.2. The authors describe the problem like it is trivial. But not for me. What do you mean by noise intensity? and stregnthening? My easy approach should work or? – ziT Sep 04 '15 at 16:43
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    @ziT I don't really see how you're going to show that $Z_1,Z_2$ are independent. Anyway, for nonzero drift, you don't even have a martingale, so you can't have independent increments. With zero drift you do have a martingale, so you should be more careful. But if, say, $s<t<u$, $K_s>0$, then $K_u-K_t$ will have more variance if $K_t-K_s>0$ than if $K_t-K_s<0$, since the effect of the noise on $[t,u]$ will be larger. – Ian Sep 04 '15 at 16:51
  • They should be dependend since, they both depend on $K_{1}$ by my argumentation. Ah ok thanks for the explanation. – ziT Sep 04 '15 at 16:53
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    @ziT To see why that argument doesn't really work, notice that if $X,Y$ are iid standard normals, then $X+Y$ and $X-Y$ are independent, even though they both "depend on $X$". – Ian Sep 04 '15 at 22:29
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    "A process with independent increments with finite mean is a martingale and therefore has constant expectation." Well, no. Example: $X_t=t$. – Did Sep 05 '15 at 21:13
  • @Did Woops, I needed to say "independent mean zero increments". But that does actually throw a wrench into the rest of the argument. Hmm... – Ian Sep 05 '15 at 21:21
  • Indeed (sorry). You might want to complete the argument delineated in my two comments on main and post it as an answer. – Did Sep 06 '15 at 08:18