I'm trying to prove the next:

If $X$ is cád (right continuous) and adapted process, then $\displaystyle\lim_{n\rightarrow\infty}X_{Z_{n}}=X_{T}$ and $X_{T}$ is random variable.

Here $T$ is a stopping time such that $\displaystyle\lim_{n\rightarrow\infty}Z_{n}(\omega)=T(\omega),$ where $$Z_{n}(\omega)=\left\{ \begin{array}{lcc} \frac{X}{2^{n}} & if & \frac{k-1}{2^{n}}\leq T(\omega)<\frac{k}{2^{n}},\space n=1,2,\ldots \\ \\ \infty & if & T(\omega)=\infty. \end{array} \right.$$

I cannot see how the right continuos of the process works on this sequence to prove the desire limit. Also, $X_{T}$ will be random variable because is limit of random variables, isn't?

Any kind of help is thanked in advanced.

  • $\begingroup$ The question seems ill-posed. $T$ is defined as a (presumably almost sure) limit of $Z_n$, yet each $Z_n$ is determined by the value of $T$. $\endgroup$ – Math1000 Apr 2 '18 at 6:48
  • $\begingroup$ There is a typo in the definition of $Z_n$, I suppose. $\endgroup$ – saz Apr 7 '18 at 18:40
  • $\begingroup$ What typo @zas? $\endgroup$ – Squird37 Apr 9 '18 at 4:45

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