0

Let a probability space defined on $\Omega = [0 \ 1]$ and assume the probability of any interval be defined as the length of that interval. Let $\\$ $Y_i(w) = \begin{cases} 1 & \quad 0 \leq w \leq 1/i, \\ 0 & \quad otherwise \end{cases} $ and $X(w) = w$. I am trying to find $Z = E[X | Y_1, ..., Y_k]$ to show that the process Z is a martingale. To eliminate $w$, I write the probabilities as: $Y_i = \begin{cases} 1 & \quad w.p. \quad 1/i, \\ 0 & \quad w.p. \quad 1-1/i \end{cases} $ and $X \sim U[0,1]$. To show Z is a martingale we need to show $E[Z_k | Z_{k-1}, ..., Z_1] = Z_{k-1}$ but I'm struggling to find Z. Any ideas to approach this problem?

eet
  • 393
  • 1
    Do you mean the process is $Z_k = E[X|Y_1, \dots, Y_k]$? In this case, I believe conditional expectations of this form can be shown to be a martingale using the Tower property, for example. – JKL Jan 21 '21 at 04:47
  • Can you give more details? There are other terms in the conditioning such as $Z_1$, $Z_2$ etc. – eet Jan 21 '21 at 15:02

1 Answers1

0

Hints: You have to show that $\int_A Z_kdP=\int_A Z_{k-1}dP$ if $A \in \sigma (Y_1,Y_2,...,Y_{k-1})$. [This will automatically imply that the equation also holds for $A \in \sigma (Z_1,Z_2,...,Z_{k-1})$]. Verify that $\sigma (Y_1,Y_2,...,Y_{k-1})$ is simply the collection of all possible unions of the intervals $[0,\frac 1 {k-1}), [\frac 1 {k-1},\frac 1 {k-2}),....,[\frac 1 2,1)$. So it is enough to verify that $\int_A Z_kdP=\int_A Z_{k-1}dP$ for each of these $k-1$ intervals. This verification is straightforward.

Kavi Rama Murthy
  • 311,013
  • I'm not fully familiar with measure theory. Can you explain this in terms of probability theory, i.e. using the definition of martingales? Will the tower property useful in here? – eet Jan 21 '21 at 15:00