I'm asked to show that given $\tau$ is a stopping time and if $$I_n = \begin{cases} 1 & \text{if}\ n\leq\tau \\ 0 & \text{if}\ n>\tau \end{cases}$$ then $(I_n)_{n\geq1}$ is a predictable process.

I know that $I_n$ is predictable if it's $\mathbb{F}_{n-1}$-measurable but I could really need some help getting through this. I was wondering if I should look at three different possibilities for the time $n$.

If $n-1\leq n\leq \tau$ it holds that $I_{n-1}=I_n=1$ and if $n>n-1>\tau$ it holds that $I_{n-1}=I_n=0$ and lastly if $n-1\leq \tau$ and $n>\tau$ it holds that $0=I_n<I_{n-1}=1$. But is this enough?


1 Answer 1


Hint: Show that $I_n = 1-1\{\tau \leq n-1\}$, and show this is $\mathbb{F}_{n-1}$-measurable.

Comparing the value of $I_n$ to $I_{n-1}$ is not enough to show that $I_n$ is $\mathbb{F}_{n-1}$-measurable (after all, how do we even know $I_{n-1}$ is $\mathbb{F}_{n-1}$-measurable?). Instead you have to show that at time $n-1$ (given the information in the sigma algebra $\mathbb{F}_{n-1}$) you can determine the value of $I_n$.

  • $\begingroup$ Just to make sure; when you're writing $1\{\tau\leq n-1\}$ is that the indicator function $1_{(\tau \leq n-1)}$? $\endgroup$
    – mas2
    Dec 11, 2019 at 21:02
  • $\begingroup$ Yes, that is what I mean. $\endgroup$
    – kccu
    Dec 12, 2019 at 15:14

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