See my Github repo for full details.<\/p>\n\n\n\n
https:\/\/github.com\/eightsmile\/cqf<\/a><\/p>\n\n\n\n The transition probability density function, p(y,t;y’,t’)<\/span>, is defined by,<\/p>\n\n\n\n $$ Prob(a<y'<b, at\\ time \\ t’ | y \\ at \\ time\\ t) = \\int_a^b p(yet;y’,t’)dy’$$<\/p>\n\n\n\n In words this is \u201cthe probability that the random variable y \u2032 lies between a and b at time t \u2032 in the future, given that it started out with value y at time t.\u201d<\/p>\n\n\n\n Think of y and t as being current values with y \u2032 and t \u2032 being future values. The transition probability density function can be used to answer the question,<\/p>\n\n\n\n \u201cWhat is the probability of the variable y \u2032 being in a speci\ufb01ed range at time t \u2032 in the future given that it started out with value y at time t?\u201d<\/p>\n\n\n\n Our Goal is to find the transition probability p.d.f., and so we find the relationship between p(y,t;y’,t’)<\/span>, and p(y,t;y’,t’-\\delta t)<\/span>,<\/em><\/strong><\/p>\n\n\n\n The variable y can either rise, fall or take the same value after a time step \u03b4t. These movements have certain probabilities<\/strong> associated with them.<\/p>\n\n\n\n We are going to assume that the probability of a rise and a fall are both the same, \\alpha<\\frac{1}{2}<\/span> . (But, of course, this can be generalized. Why would we want to generalize this?)<\/p>\n\n\n\n Given {y,t}<\/span>, or says {y,t}<\/span> the current and previous. {y’,t’}<\/span> are variate in the future time.<\/p>\n\n\n\n The probability of being at y’<\/span> at time t’<\/span> is related to the probabilities of being at the previous three values and moving in the right direction:<\/p>\n\n\n\n $$ p(y,t;y’,t’) = \\alpha \\ p(y,t;y’+\\delta y,t’-\\delta t) + \\ (1-2\\alpha) \\ p(y,t;y’,t’-\\delta t) + \\alpha \\ p(y,t;y’-\\delta y,t’-\\delta t) $$<\/p>\n\n\n\n Given {y,t}<\/span>, we find relationship between {y’,t’}<\/span> and {y’\\pm \\delta y,t’-\\delta t}<\/span> that is y’ and t’ a bit time previously.<\/em><\/strong><\/p>\n\n\n\n Remember, our goal is to find a solution of p(.)<\/span>, we try to solve the above equation.<\/em><\/strong><\/p>\n\n\n\n We expand each term of the equation<\/strong>.<\/p>\n\n\n\n $$ p(y,t;y’,t’) = \\alpha \\ p(y,t;y’+\\delta y,t’-\\delta t) + \\ (1-2\\alpha) \\ p(y,t;y’,t’-\\delta t) + \\alpha \\ p(y,t;y’-\\delta y,t’-\\delta t) $$<\/p>\n\n\n\n Why we do that? Because there are too many variables in it, hard to solve it. We have to reduce the dimension.<\/em><\/strong><\/p>\n\n\n\n $$ p(y,t;y’+\\delta y,t’-\\delta t)\\approx \\ p(y,t;y’,t) – \\delta t \\frac{\\partial p}{\\partial t’} +\\delta y \\frac{\\partial p}{\\partial y’} + \\frac{1}{2}\\delta y^2 \\frac{\\partial^2 p}{\\partial y’^2} + O(\\frac{\\partial^2 p}{\\partial t’^2}) $$<\/p>\n\n\n\n $$ p(y,t;y’,t’-\\delta t)\\approx \\ p(y,t;y’,t) – \\delta t \\frac{\\partial p}{\\partial t’} + O(\\frac{\\partial^2 p}{\\partial t’^2}) $$<\/p>\n\n\n\n $$ p(y,t;y’-\\delta y,t’-\\delta t)\\approx \\ p(y,t;y’,t) – \\delta t \\frac{\\partial p}{\\partial t’} -\\delta y \\frac{\\partial p}{\\partial y’} + \\frac{1}{2}\\delta y^2 \\frac{\\partial^2 p}{\\partial y’^2} + O(\\frac{\\partial^2 p}{\\partial t’^2}) $$<\/p>\n\n\n\n Plug them back into that equation<\/strong>, and after cancel out terms repeated we would left with,<\/p>\n\n\n\n $$ \\frac{\\partial p}{\\partial t’} =\\alpha \\frac{\\delta y^2}{\\delta t} \\frac{\\partial^2 p}{\\partial y’^2} + O(\\frac{\\partial^2 p}{\\partial t’^2})$$<\/p>\n\n\n\n We drop those derivative terms with order greater and equal than O(\\frac{\\partial^2 p}{\\partial t’^2})<\/span>.<\/p>\n\n\n\n $$ \\frac{\\partial p}{\\partial t’} =\\alpha \\frac{\\delta y^2}{\\delta t} \\frac{\\partial^2 p}{\\partial y’^2} $$<\/p>\n\n\n\n In the RHS, we focus on \\alpha \\frac{\\delta y^2}{\\delta t}<\/span>, firstly. The denominator and numerator have to be in the same order to make that term definite. Or, say \\delta y \\sim O(\\sqrt{\\delta t})<\/span>.<\/p>\n\n\n\n We thus let c^2 = \\alpha \\frac{\\delta y^2}{\\delta t}<\/span><\/p>\n\n\n\n $$ \\frac{\\partial p}{\\partial t’} =c^2 \\frac{\\partial^2 p}{\\partial y’^2} $$<\/p>\n\n\n\n The above equation is also named Fokker\u2013Planck<\/strong> or forward Kolmogorov equation.<\/strong><\/p>\n\n\n\n Now, we have a partial differential equation. Solve it, we can get the form of p<\/span>.<\/em><\/strong><\/p>\n\n\n\n $$ p(y,t;y’,t’) = \\alpha \\ p(y+\\delta y,t+\\delta t;y’,t’) + \\ (1-2\\alpha) \\ p(y,t+\\delta t;y’,t’) + \\alpha \\ p(y-\\delta y,t+\\delta t;y’,t’) $$<\/p>\n\n\n\n the dimension-reduced result is the blow, and it is called the backward Kolmogorov equation<\/strong>.<\/p>\n\n\n\n $$ \\frac{\\partial p}{\\partial t’} + c^2 \\frac{\\partial^2 p}{\\partial y’^2} =0 $$<\/p>\n\n\n\n We will solve for p<\/span> right now! However, we will solve it by assuming similarity solution.<\/p>\n\n\n\n $$ \\frac{\\partial p}{\\partial t’} =c^2 \\frac{\\partial^2 p}{\\partial y’^2} $$<\/p>\n\n\n\n This equation has an in\ufb01nite number of solutions. It has di\ufb00erent solutions for di\ufb00erent initial conditions and di\ufb00erent boundary conditions. We need only a special solution here.<\/u><\/em><\/strong> The detailed process of finding that solution is showing as the following,<\/p>\n\n\n\n $$ p=t’^a f(\\frac{y’}{t’^b}) = t’^a f(\\xi)$$<\/p>\n\n\n\n $$ \\xi = \\frac{y’}{t’^b}$$<\/p>\n\n\n\n , where a<\/span>, and b<\/span> are indefinite variables.<\/p>\n\n\n\n Again, don’t ask why it is in this form, because it is a special<\/strong> solution!<\/p>\n\n\n\n $$\\frac{\\partial p}{\\partial y’}=t’^{a-b}\\frac{df}{d\\xi}$$<\/p>\n\n\n\n $$\\frac{\\partial^2 p}{\\partial y’^2}=t’^{a-2b}\\frac{d^2f}{d\\xi^2}$$<\/p>\n\n\n\n $$\\frac{\\partial p}{\\partial t’}=at’^{a-1}f(\\xi)+by’t’^{a-b-1}\\frac{df}{d\\xi}$$<\/p>\n\n\n\n Substitue back into the forward Kolmogorov equation (remember y’ = t’^b \\xi<\/span>), we get,<\/p>\n\n\n\n $$ af(\\xi) – b\\xi \\frac{df}{d\\xi} = c^2 t’^{-2b+1} \\frac{d^2f}{d\\xi^2}$$<\/p>\n\n\n\n As we need the RHS to be independent of t’<\/span>, we could choose the value of b=\\frac{1}{2}<\/span>, to let the t’<\/span> has a power of 0. Why we do that? Because we aim to reduce the partial differential equation to be a ordinary differential equation, in which the only variable is \\xi<\/span>, and t’<\/span> disappear.<\/p>\n\n\n\n By assuming the special form of p<\/span>, and letting b= 1\/2<\/span>, our forward Kolmogorov becomes,<\/p>\n\n\n\n $$ af(\\xi) – \\frac{1}{2}\\xi \\frac{df}{d\\xi} = c^2 \\frac{d^2f}{d\\xi^2}$$<\/p>\n\n\n\n $$ p=t’^a f(\\frac{y’}{\\sqrt{t’}}) = t’^a f(\\xi)$$<\/p>\n\n\n\n $$ \\xi = \\frac{y’}{\\sqrt{t’}}$$<\/p>\n\n\n\n $$p=t’^a f(\\frac{y’}{\\sqrt{t’}}) $$<\/p>\n\n\n\n We know that p<\/span> is the transition p.d.f., its integral must be equal to ‘1’. t’<\/span> is independent by the definition of random walk behaviour, so we do only integrate p<\/span>, w.r.t. y’<\/span>.<\/p>\n\n\n\n $$\\int_{\\mathbb{R}}p\\ dy’ = \\int_{\\mathbb{R}} t’^a f(\\frac{y’}{\\sqrt{t’}})\\ dy’ = 1$$<\/p>\n\n\n\n $$ \\int_{\\mathbb{R}} t’^a f(\\frac{y’}{\\sqrt{t’}})\\ dy’ = 1 $$<\/p>\n\n\n\n , by replace x = \\frac{y’}{\\sqrt{t’}}<\/span>,<\/p>\n\n\n\n $$ \\int_{\\mathbb{R}} t’^{a+1\/2} f(x)\\ dx =t’^{a+1\/2} \\int_{\\mathbb{R}} f(x)\\ dx= 1 $$<\/p>\n\n\n\n $t’$ is independent, so the above equation would be equal to ‘1’ regardless the power of $t’$. Thus, $a = -\\frac{1}{2}$ for sure.<\/p>\n\n\n\n Also, we get \\int_{\\mathbb{R}} f(x)\\ dx= 1<\/span>.<\/p>\n\n\n\n By assuming the special form of p<\/span>, and letting a=-1\/2<\/span>, b=1\/2<\/span>, we get,<\/p>\n\n\n\n $$ -\\frac{1}{2}f(\\xi) – \\frac{1}{2}\\xi \\frac{df}{d\\xi} = c^2 \\frac{d^2f}{d\\xi^2}$$<\/p>\n\n\n\n $$ p=\\frac{1}{\\sqrt{t’}} f(\\frac{y’}{\\sqrt{t’}}) = \\frac{1}{\\sqrt{t’}}f(\\xi)$$<\/p>\n\n\n\n $$ \\xi = \\frac{y’}{\\sqrt{t’}}$$<\/p>\n\n\n\n The forward Kolmogorov equation becomes,<\/p>\n\n\n\n $$ -\\frac{1}{2}\\bigg(f(\\xi) – \\xi \\frac{df}{d\\xi} \\bigg)= c^2 \\frac{d^2f}{d\\xi^2}$$<\/p>\n\n\n\n $$ -\\frac{1}{2}\\bigg( \\frac{d \\xi f(\\xi)}{d \\xi} \\bigg)= c^2 \\frac{d^2f}{d\\xi^2}$$<\/p>\n\n\n\n , as f(\\xi) – \\xi \\frac{df}{d\\xi} = \\frac{d \\xi f(\\xi)}{d \\xi}<\/span>.<\/p>\n\n\n\n $$ -\\frac{1}{2}\\xi f(\\xi)= c^2 \\frac{df}{d\\xi} + constant$$<\/p>\n\n\n\n There\u2019s an arbitrary constant of integration that could go in here but for the answer we want this is zero. We need only a special solution, so we can set that arbitrary constant term be zero.<\/p>\n\n\n\n So, the eq could be rewritten as,<\/p>\n\n\n\n $$ -\\frac{1}{2c^2}\\xi d\\xi = \\frac{1}{f(\\xi)}df $$<\/p>\n\n\n\n $$ ln\\ f(\\xi) = -\\frac{\\xi^2}{4c^2} + C$$<\/p>\n\n\n\n Take exponential, f(\\xi) = e^C e^{-\\frac{\\xi^2}{4c^2}} = A e^{-\\frac{\\xi^2}{4c^2}}<\/span> .<\/p>\n\n\n\n The Last Step here is to find the exact value of A<\/span>. A<\/span> is chosen such that the integral of f<\/span> is one.<\/p>\n\n\n\n $$\\int_{\\mathbb{R}}f(\\xi)\\ d\\xi =1$$<\/p>\n\n\n\n $$ \\int_{\\mathbb{R}}A e^{-\\frac{\\xi^2}{4c^2}} \\ d\\xi = 2cA\\int_{\\mathbb{R}} e^{-\\frac{\\xi^2}{4c^2}} \\ d\\big(\\frac{\\xi}{2c}\\big) =1 $$<\/p>\n\n\n\n $$ 2cA \\sqrt{\\pi} = 1 $$<\/p>\n\n\n\n , so we get A = \\frac{1}{2c\\sqrt{\\pi}}<\/span><\/p>\n\n\n\n Plug f(\\xi), a, b, A<\/span> back into p = t^a f(\\xi)<\/span>.<\/p>\n\n\n\n $$ p(y’)=\\frac{1}{2c\\sqrt{\\pi \\ t’}}e^{-\\frac{\\xi^2}{4c^2}} =\\frac{1}{2c\\sqrt{\\pi \\ t’}}e^{-\\frac{y’^2}{4c^2t’}} $$<\/p>\n\n\n\n $p(.)$ now is normal like distributed.<\/p>\n\n\n\n $$N(x) = \\frac{1}{\\sqrt{2\\pi\\sigma^2}}e^{-\\frac{(x-\\mu)^2}{2\\sigma^2}}$$<\/p>\n\n\n\n So, we may say \\mu_{y’}=0<\/span>, and \\sigma^2_{y’}=2c^2t’<\/span>. Or, y’ \\sim N(0, 2c^2t’)<\/span>.<\/p>\n\n\n\n $$p(y’)=\\frac{1}{2c\\sqrt{\\pi \\ t’}}e^{-\\frac{\\xi^2}{4c^2}} =\\frac{1}{2c\\sqrt{\\pi \\ t’}}e^{-\\frac{y’^2}{4c^2t’}} $$<\/p>\n\n\n\n Finally, we solved the transition probability density function p(.)<\/span><\/strong>. By assuming the forward or backward type of trinomial model, we find a partial differential relationship. Then, assuming a special form of p(.)<\/span> by similarity method, we solve it.<\/p>\n\n\n\n The meaning is that p(y’)=\\frac{1}{2c\\sqrt{\\pi \\ t’}}e^{-\\frac{\\xi^2}{4c^2}} =\\frac{1}{2c\\sqrt{\\pi \\ t’}}e^{-\\frac{y’^2}{4c^2t’}}<\/span> is one of the transition probability density function that can satisfy the trinomial random walk.<\/p>\n\n\n\n Also, we find that p(.)<\/span> is normally liked distributed.<\/p>\n","protected":false},"excerpt":{"rendered":" See my Github repo for full details. https:\/\/github.com\/eightsmile\/cqf 1. Trinomial Random Walk 2. Transition Probability Density Function The transition probability density function, p(y,t;y’,t’), is defined by, $$ Prob(a<y'<b, at\\ time \\ t’ | y \\ at \\ time\\ t) = \\int_a^b p(yet;y’,t’)dy’$$ In words this is \u201cthe probability that the random variable y \u2032 lies … Continue reading Taylor Series and Transition Density Functions<\/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":[6,18,26],"tags":[],"_links":{"self":[{"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=\/wp\/v2\/posts\/5524"}],"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=5524"}],"version-history":[{"count":8,"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=\/wp\/v2\/posts\/5524\/revisions"}],"predecessor-version":[{"id":5533,"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=\/wp\/v2\/posts\/5524\/revisions\/5533"}],"wp:attachment":[{"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5524"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5524"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/fanyuzhao.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5524"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Trinomial Random Walk<\/h2>\n\n\n\n
2. Transition Probability Density Function<\/h2>\n\n\n\n
3. From the Trinomial model to the Transition Probability Density function<\/h2>\n\n\n\n
3.1 The Forward Equation<\/h3>\n\n\n\n
3.2 Taylor Series Expansion<\/h3>\n\n\n\n
3.3 Backward Equation works similar.<\/h3>\n\n\n\n
4. Solve the Forward Kolmogorov Equation<\/h2>\n\n\n\n
4. 1 Assume a Solution Form<\/h3>\n\n\n\n
4.2 Derivation<\/h3>\n\n\n\n
4.3 Choose b<\/span><\/h3>\n\n\n\n
4.4 Choose a<\/span><\/h3>\n\n\n\n
4.5 Integrate! Solve it!<\/h3>\n\n\n\n
Integrate 1st Time<\/h4>\n\n\n\n
Integrate 2nd Time<\/h4>\n\n\n\n
Find A<\/span><\/h4>\n\n\n\n
5. Summary<\/h2>\n\n\n\n