Find a function from a Recursive Formula

  • Context: Graduate 
  • Thread starter Thread starter loveinla
  • Start date Start date
  • Tags Tags
    Formula Function
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
loveinla
Messages
11
Reaction score
0
Hi,

I'd like to solve a function [tex]W_{t}^{x}[/tex] from a recursive formula below.

[tex]W_{t}^{x}=P_{t}^{x}*W_{t+1}^{x+1}+(1 - P_{t}^{x})*W_{t+1}^{x}[/tex],

where [tex]P_{t}^{x} = \frac{E[\theta^{x+1}*(1 - \theta)^{t-x}]}{E[\theta^{x}*(1-\theta)^{t-x}]}[/tex], [tex]{\underset{t\to\infty, x\to 0}{lim}}W_{t}^{x} =0[/tex]. Here [tex]E[/tex] stands for expectation.

Any suggestions? Thanks.
 
Physics news on Phys.org
Stephen Tashi said:
Expectation with respect to which variable? Over what range of integration?
I'd imagine on the only free variable in the expression for P, namely [itex]\theta[/itex], and over [itex](-\infty,\infty)[/itex]. But you do well in asking instead of guessing like yours truly. :)

What worries me most is that, in the recurrence rule for W, the "x" does not get any smaller. So what are supposed to be the initial values?
 
dodo said:
I'd imagine on the only free variable in the expression for P, namely [itex]\theta[/itex], and over [itex](-\infty,\infty)[/itex]. But you do well in asking instead of guessing like yours truly. :)

What worries me most is that, in the recurrence rule for W, the "x" does not get any smaller. So what are supposed to be the initial values?

Thanks, you are right, the expectation is with respect to [itex]\theta[/itex], which is on [0,1].

The initial value is indeed what I am trying to find.

In fact, it is a one-armed bandit problem with Bernoulli probability [itex]\theta[/itex], but [itex]\theta[/itex] is a random variable we are trying to figure out through the realization.

Any idea on how to determine the initial value [itex]W^{0}_{0}[/itex] or [itex]W^{x}_{t}[/itex]?