There are two equally general methods for computing the long term probability of any given day being sunny or rainy but they differ in convenience and scope. The first is the analytic method.<\/p>\n\n\n\n
In the analytic method, we express the state of the system as a vector ${\\bar S}$ with two components stating the probability of getting a sunny or rainy day. The transition rules express themselves nicely in matrix form according to<\/p>\n\n\n\n
\\[ {\\bar S}_{tomorrow} = M {\\bar S}_{today} \\; ,\\]<\/p>\n\n\n\n
which has the explicit form<\/p>\n\n\n\n
\\[ \\left[ \\begin{array}{c} S \\\\ R \\end{array} \\right]_{tomorrow} = \\left[ \\begin{array}{cc} 0.9 & 0.5 \\\\ 0.1 & 0.5 \\end{array} \\right] \\left[ \\begin{array}{c} S \\\\ R \\end{array} \\right]_{today} \\; . \\]<\/p>\n\n\n\n
The explicit form of $M$ can be checked against an initial, known state of either ${\\bar S}^{(0)} = [1,0]^{T}$ for sunny or ${\\bar S}^{(0)} = [0,1]^{T}$ for rainy.<\/p>\n\n\n\n
The matrix $M$ links the weather state today to that of tomorrow and we can think of it as an adjacency matrix for the following graph<\/p>\n\n\n\n
![\"\"](\"http:\/\/aristotle2digital.blogwyrm.com\/wp-content\/uploads\/2022\/04\/Sunny_days_transitions.png\"){kind=spacer}
<\/figure><\/div>\n\n\n\nwith the interpretation that the edge weightings are the conditional transition probabilities.<\/p>\n\n\n\n
Now lets place an index on any given day just to keep track. The state of the nth<\/sup> day is related to the initial state by<\/p>\n\n\n\n \\[ {\\bar S}^{(n)} = M {\\bar S}^{(n-1)} = M^n {\\bar S}^{0} \\; .\\]<\/p>\n\n\n\n Since we would like to know the probability of picking a sunny day from a collection of days, we are looking for a steady state solution, one which is immune to random fluctuations associated with a string of good or bad luck. This special state ${\\bar S}^{*}$ must correspond to $n \\rightarrow \\infty$ and so it must remain unchanged by successive applications of $M$ (otherwise it doesn\u2019t represent a steady state). Mathematically, this condition can be expressed as<\/p>\n\n\n\n \\[ {\\bar S}^{*} = M {\\bar S}^{*} \\; , \\]<\/p>\n\n\n\n or, in words, ${\\bar S}^{*}$ is an eigenvector of $M$. Computing the eigenvectors of $M$ is easy with numpy and the results are<\/p>\n\n\n\n \\[ {\\bar S}^{*}_{\\lambda_1} = [0.98058068, 0.19611614]^{T} \\; \\]<\/p>\n\n\n\n and<\/p>\n\n\n\n \\[ {\\bar S}^{*}_{\\lambda_2} = [-0.70710678, 0.70710678]^{T} \\; \\]<\/p>\n\n\n\n for $\\lambda_1 = 1$ and $\\lambda_2 = 0.4$. We reject the vector corresponding to $\\lambda_2$ because it has negative entries which can\u2019t be interpreted as probabilities. All that is left is to scale ${\\bar S}^{*}_{\\lambda_1}$ so that the sum of the entries is 1. Doing so gives,<\/p>\n\n\n\n \\[ {\\bar S}^{*} = [0.8333{\\bar 3},0.16666{\\bar 6}]^{T} \\; ,\\]<\/p>\n\n\n\n where the subscript has been dropped as it serves no purpose.<\/p>\n\n\n\n While mathematically correct, one might feel queasy about $\\lim_{n\\rightarrow \\infty} M^n$ converging to something meaningful. But a few numerical results will help support this argument. Consider the following 3 powers of $M$<\/p>\n\n\n\n \\[ M^2 = \\left[ \\begin{array}{cc} 0.86 & 0.70 \\\\ 0.14 & 0.3 \\end{array} \\right] \\; ,\\]<\/p>\n\n\n\n \\[ M^{10} = \\left[ \\begin{array}{cc} 0.83335081 & 0.83324595 \\\\ 0.16664919 & 0.16675405 \\end{array} \\right] \\; ,\\]<\/p>\n\n\n\n and<\/p>\n\n\n\n \\[ M^{20} = \\left[ \\begin{array}{cc} 0.83333334 & 0.83333332 \\\\ 0.16666666 & 0.16666668 \\end{array} \\right] \\; .\\]<\/p>\n\n\n\n Clearly, repeated powers are causing the matrix to converge and the elements that obtain in the limit are the very probabilities calculated above and, by the time $n=20$ we have essentially reached steady state.<\/p>\n\n\n\n The second method is numerical simulation where we \u2018propagate\u2019 an initial state forward in time according to the following rules:<\/p>\n\n\n\n An alternative way of writing the same thing reads<\/p>\n\n\n\n Of course, hybridizations of these rule sets where block 2 in the first set if matched with block 3 in the second and vice versa also give the same results. It is a matter of taste.<\/p>\n\n\n\n In any event, these rules are easy to implement in python and the following plot shows the fraction of $S$\u2019s as a function of the number of times the rule has been implemented<\/p>\n\n\n\n The dotted line shows the value for the probability of a sunny day obtained from the first method and clearly the simulations shows that it converges to that value. This sunny\/rainy day simulation is an example of Markov Chain Monte Carlo, which will be the subject of upcoming blogs.<\/p>\n","protected":false},"excerpt":{"rendered":" In the last post, we demonstrated that not all conditional probabilities are conceptually equal even if they are linguistically equal. The specific case discussed at the end \u2013 the sunny… Read more ><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-1455","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/aristotle2digital.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts\/1455","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/aristotle2digital.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/aristotle2digital.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/aristotle2digital.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/aristotle2digital.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1455"}],"version-history":[{"count":0,"href":"https:\/\/aristotle2digital.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts\/1455\/revisions"}],"wp:attachment":[{"href":"https:\/\/aristotle2digital.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1455"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/aristotle2digital.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1455"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/aristotle2digital.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1455"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"http:\/\/aristotle2digital.blogwyrm.com\/wp-content\/uploads\/2022\/04\/Sunny_day_probs.png\")
<\/figure><\/div>\n\n\n\n