1
$\begingroup$

Let $X$ be a one dimensional Ito diffusion given by

$$X_t = b \,W_t$$

where $b$ is a constant, and $W$ is a standard Brownian motion.

Let $B$ be another Brownian motion independent of $W$, and define the observation process $Y$ by

$$Y_t = X_t + B_t.$$

Fix $T > 0$. A choice of sampling times is simply a choice of real numbers $0 \leq t_1 \leq \dots \leq t_n \leq T$.

Question: For fixed $n > 1$, what choice of sampling times $t_1, \dots, t_n$ minimises the expression

$$\mathbb E\left [ |\mathbb E[X_T| \sigma(Y_{t_1}, \dots, Y_{t_n})] - X_T | \right ]?$$

Where $\sigma(Y_{t_1}, \dots, Y_{t_n})$ denotes the sigma algebra generated by the $Y_{t_i}$.

Remark: It is not certain that there exists a minimiser - to prove existence first, it would suffice to show that the given function is continuous in the $t_i$ and apply compactness.

$\endgroup$

1 Answer 1

3
$\begingroup$

Write $$Z_t = W_t - b B_t,$$ so that $Y_t$ and $Z_t$ are independent Brownian motions, $$X_t = b W_t = b \cdot \frac{b Y_t + Z_t}{1 + b^2} \, ,$$ and the question asks for the distance between $X_T$ and $$ \mathbb E[X_T | \sigma(Y_{t_1},\ldots,Y_{t_n})] = b \cdot \frac{b Y_{t_n} + 0}{1 + b^2} \, . $$ This distance is of course $$ \frac{b}{1 + b^2} \mathbb E[|b (Y_{t_n} - Y_T) - Z_T|] , $$ which is minimised when $t_n = T$.

$\endgroup$
6
  • $\begingroup$ Great answer! I think you meant $X/b=\frac{bY+Z}{1+b^2}$? $\endgroup$
    – Pierre PC
    Jan 30 at 11:57
  • $\begingroup$ Huh, so it doesn’t depend on the earlier times.. did you mean $Z_t = bW_t - B_t$ instead though? I get that they are independent if $Z_t$ is defined to be that. $\endgroup$
    – Nate River
    Jan 30 at 12:50
  • $\begingroup$ @PierrePC: Right, thanks! Too many processes for a Sunday morning. :-) $\endgroup$ Jan 30 at 13:20
  • $\begingroup$ @NateRiver: I think it is correct with $Z_t=W_t-bB_t$: then $\mathbb EY_tZ_t=\mathbb E(bW_t^2+(1-b^2)B_tW_t-bB_t^2)=bt+0-bt=0$. But I just get my second coffee today, sorry if that makes no sense. $\endgroup$ Jan 30 at 13:24
  • $\begingroup$ Oh, by the way: what is $\mu$ for? $\endgroup$ Jan 30 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy