{"id":5341,"date":"2012-12-08T15:44:08","date_gmt":"2012-12-08T14:44:08","guid":{"rendered":"http:\/\/djalil.chafai.net\/blog\/?p=5341"},"modified":"2012-12-10T10:39:58","modified_gmt":"2012-12-10T09:39:58","slug":"an-exercise-on-stopping-times","status":"publish","type":"post","link":"https:\/\/djalil.chafai.net\/blog\/2012\/12\/08\/an-exercise-on-stopping-times\/","title":{"rendered":"An exercise on stopping times"},"content":{"rendered":"

\"Dice\"<\/a><\/p>\n

This post provides the solution to a tiny exercise of probability theory, answering the question asked by a student during the MAP-432 class yesterday. Let \\( {(\\Omega,\\mathcal{F},\\mathbb{P})} \\) be a probability space equipped with a filtration \\( {{(\\mathcal{F}_n)}_{n\\geq0}} \\). Recall that a random variable \\( {\\tau} \\) taking values in \\( {\\mathbb{N}=\\{0,1,\\ldots\\}} \\) is a stopping time<\/em> when \\( {\\{\\tau=n\\}\\in\\mathcal{F}_n} \\) for any \\( {n\\in\\mathbb{N}} \\). We define the \\( {\\sigma} \\)-field \\( {\\mathcal{F}_\\tau:=\\{A\\in\\mathcal{F}:\\forall n,\\{\\tau=n\\}\\cap A\\in\\mathcal{F}_n\\}} \\). Remarkably, for any integrable random variable \\( {X} \\) and for any \\( {n\\in\\mathbb{N}} \\),<\/p>\n

\\[ \\mathbb{E}(X|\\mathcal{F}_\\tau)\\mathbf{1}_{\\{\\tau=n\\}} =\\mathbb{E}(X|\\mathcal{F}_n)\\mathbf{1}_{\\{\\tau=n\\}}. \\]<\/p>\n

To see it, for every \\( {A\\in\\mathcal{F}_n} \\), since \\( {A\\cap\\{\\tau=n\\}\\in\\mathcal{F}_\\tau} \\) and \\( {\\{\\tau=n\\}\\in\\mathcal{F}_\\tau} \\), we get<\/p>\n

\\[ \\mathbb{E}(\\mathbf{1}_AX\\mathbf{1}_{\\{\\tau=n\\}}) =\\mathbb{E}(\\mathbf{1}_{A\\cap\\{\\tau=n\\}}\\mathbb{E}(X|\\mathcal{F}_\\tau)) =\\mathbb{E}(\\mathbf{1}_A\\mathbb{E}(X\\mathbf{1}_{\\{\\tau=n\\}}|\\mathcal{F}_\\tau)), \\]<\/p>\n

and therefore \\( {\\mathbb{E}(X\\mathbf{1}_{\\{\\tau=n\\}}|\\mathcal{F}_n) =\\mathbb{E}(X\\mathbf{1}_{\\{\\tau=n\\}}|\\mathcal{F}_\\tau)} \\). Now \\( {\\{\\tau=n\\}\\in\\mathcal{F}_n\\cap\\mathcal{F}_\\tau} \\).<\/p>\n

A nice property.<\/b> If \\( {\\tau} \\) and \\( {\\theta} \\) are stopping times then \\( {\\tau\\wedge\\theta:=\\min(\\tau,\\theta)} \\) is a stopping time, and moreover for every integrable random variable \\( {X} \\),<\/p>\n

\\[ \\mathbb{E}(\\mathbb{E}(X|\\mathcal{F}_\\theta)|\\mathcal{F}_\\tau) =\\mathbb{E}(\\mathbb{E}(X|\\mathcal{F}_\\tau)|\\mathcal{F}_\\theta) =\\mathbb{E}(X|\\mathcal{F}_{\\tau\\wedge\\theta}). \\]<\/p>\n

Note that if both \\( {\\tau} \\) and \\( {\\theta} \\) are deterministic (i.e. constant) then we recover the usual tower property of conditional expectations. Note also that the events \\( {\\{\\tau\\leq\\theta\\}} \\), \\( {\\{\\tau=\\theta\\}} \\), and \\( {\\{\\tau\\geq\\theta\\}} \\) all belong to \\( {\\mathcal{F}_\\tau\\cap\\mathcal{F}_\\theta} \\), and that \\( {\\mathcal{F}_{\\tau\\wedge\\theta}\\subset\\mathcal{F}_\\tau\\cap\\mathcal{F}_\\theta} \\).<\/p>\n

Proof.<\/b> The fact that \\( {\\tau\\wedge\\theta} \\) is a stopping time is left to the reader. To prove the property on conditional expectations, it suffices by symmetry to prove the second equality, i.e. that for every \\( {A\\in\\mathcal{F}_\\theta} \\), by denoting \\( {Y:=\\mathbb{E}(X|\\mathcal{F}_{\\tau\\wedge\\theta})} \\),<\/p>\n

\\[ \\mathbb{E}(Y\\mathbf{1}_A) = \\mathbb{E}(\\mathbb{E}(X|\\mathcal{F}_\\tau)\\mathbf{1}_A). \\]<\/p>\n

We start by observing that \\( {Y\\mathbf{1}_{\\{\\tau\\leq\\theta\\}}= \\mathbb{E}(X|\\mathcal{F}_\\tau)\\mathbf{1}_{\\{\\tau\\leq\\theta\\}}} \\) which gives immediately<\/p>\n

\\[ \\mathbb{E}(Y\\mathbf{1}_A\\mathbf{1}_{\\{\\tau\\leq\\theta\\}}) =\\mathbb{E}(\\mathbb{E}(X|\\mathcal{F}_\\tau)\\mathbf{1}_A\\mathbf{1}_{\\{\\tau\\leq\\theta\\}}). \\]<\/p>\n

On the other hand, we have<\/p>\n

\\[ \\mathbb{E}(Y\\mathbf{1}_A\\mathbf{1}_{\\{\\tau>\\theta\\}}) =\\sum_{k=0}^\\infty\\mathbb{E}(\\mathbb{E}(X|\\mathcal{F}_k)\\mathbf{1}_{A\\cap\\{\\tau>k\\}\\cap\\{\\theta=k\\}}). \\]<\/p>\n

Now, since \\( {B_k:=A\\cap\\{\\tau>k\\}\\cap\\{\\theta=k\\}\\in\\mathcal{F}_k\\cap\\mathcal{F}_\\tau} \\),<\/p>\n

\\[ \\mathbb{E}(\\mathbb{E}(X|\\mathcal{F}_k)\\mathbf{1}_{B_k}) =\\mathbb{E}(X\\mathbf{1}_{B_k}) =\\mathbb{E}(\\mathbb{E}(X|\\mathcal{F}_\\tau)\\mathbf{1}_{B_k}) \\]<\/p>\n

which gives, by summing over \\( {k} \\),<\/p>\n

\\[ \\mathbb{E}(Y\\mathbf{1}_A\\mathbf{1}_{\\{\\tau>\\theta\\}}) =\\mathbb{E}(\\mathbb{E}(X|\\mathcal{F}_\\tau)\\mathbf{1}_A\\mathbf{1}_{\\{\\tau>\\theta\\}}). \\]<\/p>\n","protected":false},"excerpt":{"rendered":"

This post provides the solution to a tiny exercise of probability theory, answering the question asked by a student during the MAP-432 class yesterday. Let…<\/p>\n

Continue readingAn exercise on stopping times<\/span><\/a><\/div>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":265},"categories":[1],"tags":[],"_links":{"self":[{"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/posts\/5341"}],"collection":[{"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/comments?post=5341"}],"version-history":[{"count":6,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/posts\/5341\/revisions"}],"predecessor-version":[{"id":5349,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/posts\/5341\/revisions\/5349"}],"wp:attachment":[{"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/media?parent=5341"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/categories?post=5341"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/tags?post=5341"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}