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I had asked a similar problem before and Didier Piau was very kind to help me with the answer. I have another question on the same problem setting.

Let $T$ be a random exponentially distributed time. $P(T>t)=e^{−t}$. Define $M$ via $M_t=1$ if $t−T∈Q^+$, $M_t=0$ otherwise. Where $Q^+$ being positive rationals. let $F_t$ be a filtration generated by the process $M$.

I can see that the process described above is a martingle but i am trying to prove it is not cadlag. i am trying to apply the definition $\lim_{h\rightarrow 0} M_{t+h} =M_t$ and $\lim_{h\rightarrow 0} M_{t-h} =exists$.

At $M_{t+h}$ i will generate a random variable $T$ and depending on whether $T>t+h$, $M_{t+h}$ will have 0 or 1. In the limit $h\rightarrow 0$ , $M_{t+h}$ does not necessarily go to $M_t$ as at $t+h$ no matter how small $h$ is a new random variable $T$ will decide the value of $M_{t+h}$ which can make $M_{t+h}$ different from $M_t$.

Is my thought process correct to arrive at the conclusion that $M_t$ is not cadlag? It would be very kind of someone to help as I am trying to learn stochastic process on my own reading from a book.

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up vote 4 down vote accepted

If I understand correct, you have $M_t = 0$ for $t<T$ and $M_t = D(t-T)$ for $t\geq T$ where $D(x)$ is a Dirichlet Function, which is continuous nowhere. It means that for $t\geq T$ the process $M_t$ has a trajectory which is continuous nowhere and has no right/left limits.

Edited: Since you have some questions on the construction of this process, I try to discuss it here. As you have described, the construction is following:

  1. you pick up a random variable $T\sim\mathcal E(1)$;

  2. you define $$ M_t = \begin{cases} 1, \text{ if }t-T\in\mathbb Q^+; \\ 0, \text{ otherwise}. \end{cases} $$

This means that whenever you know $T$ you can construct the process $M_t$. Vice versa, if you know $M_t$ then clearly $T = \min\{\tau\geq 0:M_t = 1\}$: we can take minimum since $M_T = 1$.

If $t-T\in \mathbb Q^+$ then $M_t = 1$ but there is a sequence $t_k\geq T$ such that $M_{t_k} = 0$ but $t_k\to t$ with $k\to\infty$. You can also just change notation a bit and write $$ M_t = \begin{cases} D(t), \text{ if }t\geq T; \\ 0, \text{ if }t<T. \end{cases} $$ where $D(t)$ is a Dirichlet function. This immediately implies that $M_t$ is not cadlag.

Note that here you simulate $T$ only once, if I understand your question correct. That's why you don't need to simulate two different random variables $T_1,T_2$ at times $t,t+h$. There is one-to-one correspondence between $T\in\mathbb R_{\geq0}$ and $M_t(T)$.

Let me show you an example which is more easy to visualize. Let us take $$ M'_t = \begin{cases}1+\log{t}, \text{ if }t\geq T; \\ 0, \text{ if }t<T. \end{cases} $$

Here is the plot, for $T\approx 4.32$. In your case situation is almost the same - just you have $D(t)$ instead of $1+\log{t}$, but the former function is hard to visualize.


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most probably you did not answer the correct question. You introduced an additional function $D(x)$. It is not there in my question. all I say is $M_t=1$ if $t>T$ and trying to see if $M_t$ is cadlag. – babu Jul 12 '11 at 16:16
You wrote that $M_t = 1$ iff $t-T\in \mathbb Q^+$, didn't you? Now you need to verify the cadlag property. I wrote that $M_t = D(t-T)$ where $D(x) = 1$ if $x\in \mathbb Q$ and $0$ otherwise. It's easy to prove that this function does not have any limits in any point: neither left nor right - hence the process $M_t$ is not cadlag. If you say that $M_t =1$ iff $t>T$ then the process is not cadlag since $M_t$ is not right-continuous at $T$. – Ilya Jul 12 '11 at 16:21
thanks you very much.please help me to understand one more issue.the way i am trying to visualize right continuity is given a $t$ and $t+h$ i generate a random number $T_1$ at $t$,and $T_2$ at $t+h$ and then check the right continuity.the reason i try to visualize it this way is because $M_t$ is defined as a process which takes in different values at different times.following this reasoning i fail to come to a conclusion.But if i am given $T$,and i try to see changes in $M_t$ as $t$ changes then i see the right continuity easily.what is my mistake in visualizing the first way i described? – babu Jul 12 '11 at 17:28
@babu: I've edited the answer, hope it's clear now. – Ilya Jul 13 '11 at 9:48
Thank you very very much Gortaur. This explanation was of extreme help. I have struggled with this concept for a long long time and now I understand the fallacy of my reasoning. The understanding of this will also help me tremendously in other things that I had problem with. You see this is a common problem for someone who does not have a teacher and is trying to learn by picking up a book. – babu Jul 13 '11 at 15:45

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