Question

Let \(T, Y_{0}, Y_{1}, \ldots\) be random variables. Suppose the possible values for \(T\) are \(\\{0,1, \ldots\\}\) and, for every \(n \geq 0\), the event \(\\{T \geq n\\}\) is determined by \(\left(Y_{0}, \ldots, Y_{n}\right)\). Is \(T\) necessarily a Markov time with respect to \(\left\\{Y_{n}\right\\}\) ? Provide a proof or counterexample to support your claim.

Short answer

No, \(T\) is not necessarily a Markov time with respect to \(\{Y_n\}\). A counterexample is given by having independent and identically distributed \(Y_n\)s, where \(T = \min \{n \geq 0: Y_n = Y_0\}\). This choice of \(T\) does not satisfy the Markov property because there exists an event \(A = \{Y_{n-1} = Y_0\}\) such that \(P(A \cap \{T \geq n\} | Y_n) \neq P(A | Y_n) P(\{T \geq n\} | Y_n)\), which contradicts the Markov property.

Step-by-step solution

Counterexample

Consider the case where \(Y_n\) are independent random variables and have identical distributions, i.e., for each \(n \geq 0\), \(Y_n\) has the same distribution as \(Y_0\). Let us choose \(T\) as: \[T = \min \{n \geq 0: Y_n = Y_0\}\] In this case, the possible values for \(T\) are \(\{0, 1, \ldots\}\). The event \(\{T \geq n\}\) is determined by \(\left(Y_{0}, \ldots, Y_{n}\right)\) since the condition means there is no \(k\) such that \(0 \leq k < n\) and \(Y_k = Y_0\). Now we want to show that this does not satisfy the Markov property, by proving that there exists an event \(A\) such that: \[P\left(A \cap \{T \geq n\} | Y_n\right) \neq P\left(A | Y_n\right) P\left(\{T \geq n\} | Y_n\right)\] Let \(A= \{Y_{n-1} = Y_0\}\), then: \[P\left(A \cap \{T \geq n\} | Y_n\right) = 0\] since if \(T \geq n\), then there is no \(k\) such that \(0 \leq k < n\) and \(Y_k = Y_0\). However, we have: \[P\left(A | Y_n\right) = P\left(\{Y_{n-1} = Y_0\} | Y_n\right) > 0\] (we know that this probability is greater than 0 because \(Y_n\) are identically distributed) and \[P\left(\{T \geq n\} | Y_n\right) = P\left(\{T \geq n\} | Y_n\right) > 0\] (we know that this probability is also greater than 0 because there is no guarantee that \(Y_n\) has appeared among \(\left(Y_{0}, \ldots, Y_{n}\right))\). Thus, \(P\left(A | Y_n\right) P\left(\{T \geq n\} | Y_n\right) > 0\) and \[P\left(A \cap \{T \geq n\} | Y_n\right) \neq P\left(A | Y_n\right) P\left(\{T \geq n\} | Y_n\right)\] which is a contradiction to the Markov property. So, in this case, \(T\) is not a Markov time with respect to \(\{Y_n\}\), and the claim is false.