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Construct a continuous local martingale $(M(t))_{t \in [0,1]}$ with $M(0)=0$, $M(t)$ is not always equal to $0$, and $M(t)$ is constant with positive probability.

Is the above construction possible?

I am thinking that the $M(t)$ can divide as constant part and non-constant part, then I can show the positive part prob higher than 0.

However, I don't know how to find such.

Please provide me a example.

saz
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Alex Wu
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  • What exactly do you mean by "$M(t)$ is constant with positive probability"...? Should the process satisfy $$\mathbb{P}(\forall t: M_t =M_0)>0$$ ...? – saz Jun 01 '18 at 10:58
  • what alternative understanding？ – Alex Wu Jun 01 '18 at 12:16
  • Well, for instance that with probability $>0$ the sample path $M_t(\omega)$ is constant on some time interval ... or that there exists $c=c(t)>0$ such that $\mathbb{P}(M_t=c)>0$.. whatsoever, there are plenty of possibilities. Obviously @Stefan didn't understand it the way you meant it. – saz Jun 01 '18 at 13:01
  • sorry let me be clear.the word constant means that M(t,w)=M(0,w),t∈[0,1] – Alex Wu Jun 01 '18 at 13:58
  • ... but then you can take any (local) martingale $(X_t){t \geq 0}$ starting at $X_0=0$ (e.g. a Brownian motion) and $$M_t := \begin{cases} 0, & t \leq 1, \ X{t-1}, & t>1 \end{cases}$$will do the job. – saz Jun 01 '18 at 14:42
  • t>1 is not the range to be considered，so i dont understand the purpose.can you explain the above more in detail？ – Alex Wu Jun 01 '18 at 14:53
  • You didn't mention any range so I was assuming that you are talking about martingales $(M_t)_{t \geq 0}$ with time index $t \in [0,\infty)$. – saz Jun 01 '18 at 15:06
  • oh come on，i wrote clearly t∈[0,1] – Alex Wu Jun 01 '18 at 15:55
  • Everything would be much clearer if you would format your question properly (=use MathJax). – saz Jun 01 '18 at 16:54

3 Answers3

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Let $(B_t)_{t \geq 0}$ be a martingale with continuous sample paths (e.g. a Brownian motion), and let $X$ be a random variable which is independent from $(B_t)_{t \geq 0}$ and which satisfies $$\mathbb{P}(X= 1) = \frac{1}{2} \qquad \mathbb{P}(X=0) = \frac{1}{2}.$$

If we consider

$$M_t := X \cdot B_t$$

with the filtration

$$\mathcal{F}_t := \sigma(X, B_s; s \leq t),$$

then $(M_t)_{t \geq 0}$ has all desired properties. Clearly, $(M_t)_{t \geq 0}$ has continuous sample paths, it satisfies $$\mathbb{P}(\forall t \geq 0: \, \, M_t=0)= \frac{1}{2},$$

and $(M_t,\mathcal{F}_t)_{t \geq 0}$ is a martingale since

$$\mathbb{E}(M_t \mid \mathcal{F}_s) = X \cdot \mathbb{E}(B_t \mid \mathcal{F}_s) = X \cdot B_s = M_s$$

for all $s \leq t$.

saz
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Take $N\in exp(1)$ exponential distributed and consider the Brownian motion $W_t$ independant of $N$, and construct

$$ X_t = \int_N^t dW_{\tau} $$

Hence $$ P(\{X_s\}_{s=0}^t \text{ is constant }) = P(N \geq t) > 0 $$

Also $$ E(X_{t+s}|X_{t}) = P(N < t)E(X_{t+s}|X_{t},N<t)+ P(N \geq t)E(X_{t+s}|X_{t},N\geq) $$ Now, $$ E(X_{t+s}|X_{t},N\geq t) = \int E(\int_N^{t+s} dW_\tau|X_t=0,N > t,N=n)f_{N|N>t}(n) =\int 0 f_{N|N>t} = 0 = (X_t|N\geq t) $$

and $$ E(X_{t+s}|X_{t},N < t) = E(X_t+\int_t^{s+t}dW\tau|X_t,N<t) = X_t|N<t + 0 $$

hence we have $$ E(X_{t+s}|X_{t}) = X_t $$ becaus conditoin an both the events $N<t,N\geq t$ the expectation is $X_t$.

So this is a martingale and hence also a local martingale.

Stefan
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An example without extra randomization would be $X_t=\int_0^t 1_{\{|W_s|\ge k\}}\,dW_s$, with $k$ a large constant. This is a local martingale (for $t\ge 0$, not just $t\in [0,1]$), and $X_t=0$ for $0\le t\le 1$ on the event $\{\max_{0\le s\le 1}|W(s)|<k\}$, which has probability close to 1 for $k$ large enough.

John Dawkins
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