Let ( X n ) n ≥ 0 be a sequence of random variables and τ , t stopping times with respect to the sequence ( X n ) n ≥ 0 { τ + t = n } = { τ + t = n } ∩ { t ≤ n } = ⋃ k = 0 n { τ + t = n } ∩ { t = k } = ⋃ k = 0 n { τ = n − k } ∩ { t = k } . As { τ = n − k } ∈ F n − k ⊆ F n and { t = k } ∈ F k ⊆ F n for any k ≤ n, this implies that { τ + t = n } ∈ F n , and so τ + t is a stopping time.Now, my question is, let assume that I am considering τ − t I know that in general, τ − t is not a stopping time. However, if I were to consider my birthday this year (a stopping time), which is a deterministic stopping time. At any time, I know exactly when my birthday occurs. Also, I know two days before my birthday i.e, τ − 2. What kind of a formulated counterexample will show that τ − 2 is indeed a stopping time in this setting.

Question

Let   (      X          n            )          n      ≥      0       be a sequence of random variables and   τ  ,  t stopping times with respect to the sequence   (      X          n            )          n      ≥      0                          {        τ        +        t        =        n        }        =        {        τ        +        t        =        n        }        ∩        {        t        ≤        n        }                            =                  ⋃                      k            =            0                    n                {        τ        +        t        =        n        }        ∩        {        t        =        k        }                                          =                  ⋃                      k            =            0                    n                {        τ        =        n        −        k        }        ∩        {        t        =        k        }        .            As   {  τ  =  n  −  k  }  ∈            F              n      −      k        ⊆            F        n   and   {  t  =  k  }  ∈            F        k    ⊆            F        n   for any   k  ≤  n, this implies that   {  τ  +  t  =  n  }  ∈            F        n  , and so   τ  +  t is a stopping time.Now, my question is, let assume that I am considering   τ  −  t I know that in general,   τ  −  t is not a stopping time. However, if I were to consider my birthday this year (a stopping time), which is a deterministic stopping time. At any time, I know exactly when my birthday occurs. Also, I know two days before my birthday i.e,   τ  −  2. What kind of a formulated counterexample will show that   τ  −  2 is indeed a stopping time in this setting.

mnaonavl · Accepted Answer

If   τ is deterministic,   τ  −  2 is deterministic. As long as   τ  −  2  ≥  0, it is a stopping time because   {  τ  −  2  =  n  }  =  ∅ or   {  τ  −  2  =  n  }  =  Ω for all   n.

Let ( X <mrow class="MJX-TeXAtom-ORD"> n </mrow> </msub> )

Answered question

Answer & Explanation

New Questions in Pre-Algebra