{"id":5562,"date":"2023-04-03T08:36:14","date_gmt":"2023-04-03T00:36:14","guid":{"rendered":"https:\/\/fanyuzhao.com\/?p=5562"},"modified":"2023-04-10T13:41:12","modified_gmt":"2023-04-10T05:41:12","slug":"girsanovs-theorem","status":"publish","type":"post","link":"https:\/\/fanyuzhao.com\/?p=5562","title":{"rendered":"Girsanov&#8217;s Theorem"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">Statement<\/h2>\n\n\n\n<p>We can change the probability measure, and then make a random variable follows a certain probability measure.<\/p>\n\n\n\n<ul><li>Radon-Nikodym Derivative:<\/li><\/ul>\n\n\n\n<p>$$Z(\\omega) = \\frac{\\tilde{P}(\\omega)}{P(\\omega)}$$<\/p>\n\n\n\n<ul><li>$\\tilde{P}(\\omega)$ is the risk-neutral probability measure.<\/li><li>${P}(\\omega)$ is the actual probability measure.<\/li><li>Properties:<ul><li>$Z(\\omega)&gt;0$<\/li><li>$\\mathbb{E}(Z)=1$<\/li><li>As <span class=\"katex math inline\">\\tilde{P}(\\omega) = Z(\\omega) P(\\omega)<\/span>, so if <span class=\"katex math inline\">Z(\\omega)<\/span>, then <span class=\"katex math inline\">\\tilde{P}(\\omega)>P(\\omega)<\/span>. <em>vice versa<\/em>.<\/li><\/ul><\/li><\/ul>\n\n\n\n<p>We can calculate that,<\/p>\n\n\n\n<p>$$ \\underbrace{\\tilde{\\mathbb{E}}(X)}_{\\text{Expectation under Risk-neutral Probability Measure}} = \\underbrace{\\mathbb{E}(ZX)}_{\\text{Expectation under Actual Probability Measure}} $$<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Proof &amp; Example<\/h2>\n\n\n\n<p>Under <span class=\"katex math inline\">(\\Omega,\\mathcal{F},P)<\/span>, <span class=\"katex math inline\">A\\in \\mathcal{F}<\/span>, let <span class=\"katex math inline\">X<\/span> be a random variable <span class=\"katex math inline\">X\\sim N(0,1)<\/span>. <span class=\"katex math inline\">\\mathbb{E}(X)=0<\/span>, and <span class=\"katex math inline\">\\mathbb{Var}(X)=1<\/span>.<\/p>\n\n\n\n<p>$Y=X+\\theta$, $\\mathbb{E}(Y)=\\theta$, and $\\mathbb{Var}(Y)=1$.<\/p>\n\n\n\n<p>$X$ here is s.d. normal under the actual probability measure.<\/p>\n\n\n\n<p>However, <span class=\"katex math inline\">Y<\/span> here is not standard normal under the current probability <span class=\"katex math inline\">P(.)<\/span>, because <span class=\"katex math inline\">\\mathbb{E}(Y)\\neq0<\/span>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">What do we do?<\/h3>\n\n\n\n<p><strong>We change the probability measure from <span class=\"katex math inline\">P(.)\\to\\tilde{P}(.)<\/span> to let <span class=\"katex math inline\">Y<\/span> be standard normal under the new probability measure!<\/strong><\/p>\n\n\n\n<p>We set the Radon-Nikodym Derivative,<\/p>\n\n\n\n<p>$$Z(\\omega) = exp\\{ -\\theta\\ X(\\omega) &#8211; \\frac{1}{2}\\theta^2 \\}$$<\/p>\n\n\n\n<p>Now, we can create the probability measure <span class=\"katex math inline\">\\tilde{P}(A)<\/span>, <span class=\"katex math inline\">A={ \\omega;Y(\\omega)\\leq b) }<\/span><\/p>\n\n\n\n<p>$$\\tilde{P}(A) = \\int_A Z(\\omega)\\ dP(\\omega)$$<\/p>\n\n\n\n<p>such that <span class=\"katex math inline\">Y=X+\\theta<\/span> would be standard normal distributed under the new probability measure <span class=\"katex math inline\">\\tilde{P}(A)<\/span>.<\/p>\n\n\n\n<p>$$\\tilde{P}(A) = \\tilde{P}(Y(\\omega \\leq b)$$<\/p>\n\n\n\n<p>$$ = \\int_{{ Y(\\omega)\\leq b } } exp{ -\\theta\\ X(\\omega) &#8211; \\frac{1}{2}\\theta^2 } \\ dP(\\omega)$$<\/p>\n\n\n\n<p>, then change the integral range from the set <span class=\"katex math inline\">A<\/span> to <span class=\"katex math inline\">\\Omega<\/span> by multiplying that indicator.<\/p>\n\n\n\n<p>$$ = \\int_{\\Omega }\\mathbb{1}_{ Y(\\omega)\\leq b }\\ exp{ -\\theta\\ X(\\omega) &#8211; \\frac{1}{2}\\theta^2 } \\ dP(\\omega)$$<\/p>\n\n\n\n<p>, change from <span class=\"katex math inline\">dP<\/span> to <span class=\"katex math inline\">dX<\/span>,<\/p>\n\n\n\n<p>$$ = \\int_{-\\infty }^{\\infty }\\mathbb{1}_{ b-\\theta}\\ exp{ -\\theta\\ X(\\omega) &#8211; \\frac{1}{2}\\theta^2 } \\ \\frac{1}{\\sqrt{2\\pi}}e^{-\\frac{1}{2}X^2(\\omega)} \\ dX(\\omega)$$<\/p>\n\n\n\n<p>$$ =\\frac{1}{\\sqrt{2\\pi}} \\int_{-\\infty }^{b-\\theta}\\ exp{ -\\theta\\ X(\\omega) &#8211; \\frac{1}{2}\\theta^2- \\frac{1}{2}X^2(\\omega)} \\ dX(\\omega)$$<\/p>\n\n\n\n<p>$$ =\\frac{1}{\\sqrt{2\\pi}} \\int_{-\\infty }^{b-\\theta}\\ exp\\Bigg\\{ -\\frac{1}{2}\\bigg(\\theta+ X(\\omega)\\bigg)^2\\Bigg\\} \\ dX(\\omega)$$<\/p>\n\n\n\n<p>, as <span class=\"katex math inline\">Y=X+\\theta<\/span>, <span class=\"katex math inline\">dY = dX<\/span>, we now change <span class=\"katex math inline\">dX<\/span> to <span class=\"katex math inline\">dY<\/span>,<\/p>\n\n\n\n<p>$$ =\\frac{1}{\\sqrt{2\\pi}} \\int_{-\\infty }^{b}\\ exp\\big\\{ -\\frac{1}{2}Y(\\omega)^2\\big\\} \\ dY(\\omega)$$<\/p>\n\n\n\n<p>, the above is now a standard normal distribution for <span class=\"katex math inline\">Y(\\omega)<\/span>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Statement We can change the probability measure, and then make a random variable follows a certain probability measure. Radon-Nikodym Derivative: $$Z(\\omega) = \\frac{\\tilde{P}(\\omega)}{P(\\omega)}$$ $\\tilde{P}(\\omega)$ is the risk-neutral probability measure. ${P}(\\omega)$ is the actual probability measure. Properties: $Z(\\omega)&gt;0$ $\\mathbb{E}(Z)=1$ As \\tilde{P}(\\omega) = Z(\\omega) P(\\omega), so if Z(\\omega), then \\tilde{P}(\\omega)>P(\\omega). vice versa. We can calculate that, $$ &hellip; <a href=\"https:\/\/fanyuzhao.com\/?p=5562\" class=\"more-link\">Continue reading <span class=\"screen-reader-text\">Girsanov&#8217;s Theorem<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6,26],"tags":[],"_links":{"self":[{"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=\/wp\/v2\/posts\/5562"}],"collection":[{"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5562"}],"version-history":[{"count":7,"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=\/wp\/v2\/posts\/5562\/revisions"}],"predecessor-version":[{"id":5575,"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=\/wp\/v2\/posts\/5562\/revisions\/5575"}],"wp:attachment":[{"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5562"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5562"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5562"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}