"Mathematics Standard level Internal Assessment TitleHow can Markov Chains be utilized to determine the probability of weatherswitching between rainy, cloudy or sunny in Italy?RationalDuring the first term of doing the International Baccalaureate I noticed the topicof Math surface in general conversations between students. My peers wereconstantly arguing over if they should stay in their current mathematic intensityor move to either a higher or lower level of mathematics. After some time passedI realized the vast amount of students that drifted from level to level. Ieffortlessly attempted to analyse the phenomenon to which I failed. As monthspassed my Standard level maths teacher finally announced the internalassessment. This internal assessment incited thoughts about the circulation ofstudents between mathematical intensities. I wanted to analyse the probabilitydistribution associated with this trend but after conducting a lot of research Icame to the conclusion that this phenomenon is stochastic in nature and Markovchains would provide a suitable exploration. Unfortunately this data of studentschanging classes was not available, but I still was lucky that I had been given theidea of Markov chains through these observations. This gave me the possibilityto expand on the concept of Markov chains with the subject of weather in myhome country, Italy.IntroductionMarkov chains are suitable for this situation because Markov chains describe thebehaviour of stochastic systems. A stochastic system obtains an unsystematicprobability distribution or pattern that may be analyzed statistically but may notbe predicted precisely. A stochastic system may be described as a Markov chainif the next state does not depend on any precedingstates but only depends on thecurrent state. This can be applied to weatherchangingstates in a given timeperiod. There are three states that will be taken into consideration: rainyweather, cloudy weather and sunny weather. Between these three statesatmospheric conditions tend to transfer randomly across different time periodsin a year. Atmospheric conditions are directly and only affected by their previousstate.AimTo use the theory surrounding Markov Chains to calculate the long runprobability of the climatic conditions in Italy Mathematical procedure Deriving the transition matrix In a Markov Chain there are a set of states, S = {s1, s2, . . . ,sr}.The process startsin one of these states and moves successively from one state to another. Eachmove is called a step.There are three states: sunny, cloudy and rainy. K= {S, C,R}.In order to find out the probabilities of steps between our three states atransition matrix is required. The probabilities of the climatic conditions inRome, Italy 2015 can be calculated across 3 months of the following seasons:summer, winter and autumn In order to calculate the probability of the likelyhood of the states (K= {S, C, R}) during a specific month the following formulacan be utilized.Let K denote any one of the states belonging to K x Let K M denote the total amount of the occurrences of K where K M1 is the totalx x x amount of the occurrences of K in month 1 (specific to season, eg for summer,x K M1 is total amount of occurences of K for June)x x ? ? ? P(K )=x ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Summer Data10 June Sunny = 30 17 June Cloudy = 30 3 June Rainy = 30 12 July Sunny = 31 17 July Cloudy = 31 2 July Rainy = 31 16 August Sunny = 31 10 August Cloudy = 31 5 August Rainy = 31Now to calculate the probability of the probability of K across 3 months, we addx up the previously calculated values of P(K ) of all three months and divide thex whole equation by the number of days for each month.Let Mx Denote the total days of a month specific to a season where (M1 is month1, M2 is month 2 … MN) 3 K? M? ? =1× 1003M? ? =1 Applying the formula for the probability of a sunny day during summer? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? +? ? ? ? ? ? ? ? ? ? ? ? ? ? ? +? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 10 + 12 + 16 38 =30 + 31 + 31 92 38× 100 = 41.30 %92 This is one of the 9 values that will be used on the transition matrix. Using theformulas generated the remaining 8 values can be calculated for the transitionmatrix.Probability of a cloudy day during summer17 + 17 + 10 44 =30 + 31 + 31 92 44 × 100 = 47.83%92 Probability of a rainy day during summer3 + 2 + 5 10 =30 + 31 + 31 92 10 × 100 = 10.87%92 Winter Data15 December Sunny = 3112 December Cloudy = 31 4 December rainy = 31 8 January Sunny = 31 14 January Cloudy = 31 9 January Rainy = 31 3 February Sunny = 29 15 February Cloudy = 29 11 February Rainy =29 Probability of a Sunny day during winter15 + 8 + 3 26 =31 + 31 + 29 91 26× 100 = 28.57%91 Probability of a Cloudy day during winter12 + 14 + 15 41 =31 + 31 + 29 91 41× 100 = 45.06%91 Probability of a Rainy day during winter4 + 9 + 11 24 =31 + 31 + 29 91 24× 100 = 26.37%91Autumn data7 September Sunny = 30 17 September Cloudy = 30 6 September rainy = 30 7 October Sunny = 31 14 October Cloudy = 31 10 October Rainy =31 10 November Sunny = 30 10 November Cloudy = 30 10 November Rainy = 30 Probability of a sunny day during autumn 7 + 7 + 10 24 = 30 + 31 + 30 91 24× 100 = 26.37%91 Probability of a cloudy day during autumn 17 + 14 + 10 41 = 30 + 31 + 30 91 41× 100 = 45.06%91Probability of a rainy day during autumn 6 + 10 + 10 26 = 30 + 31 + 30 91 26× 100 = 28.57%91 State transition diagramThe calculated probabilities can be utilized in order to create a visual depictionof the transition probabilities.0.4506Cloudy0.2637 0.28570.4783 0.45060.1087Sunny0.4130 Rainy 0.2637 0.2857Transition probability matrixWe can use these values to produce a transition probability matrix SCRS 0.4130 0.4783 0.1087 0.2637 0.4506 0.2857 C0.2857 0.4506 0.2637 R We can denote this transition probability matrix as PTransition probability matrix validity check To test the validity of the transition probability matrix the row values can beadded. The addition of the rows on any transition diagram should add up to 1. Ifall three rows add up to exactly 1, then the transition diagram has correct values,if it does not add up to exactly 1, errors have been made in the previouscalculations.0.4130 + 0.4783 + 0.1087 = 10.2637 + 0.4506 + 0.2857 = 10.2857 + 0.4506 + 0.2647 = 1N-step transition matrixLet P be the transition matrix of a Markov chain. The ijth en- (n) ntry p of the matrix P gives the probability that the Markov chain, starting inij stateK , will be in stateK after n steps.i j 0.4130 0.4783 0.1087 0.4130 0.4783 0.1087 (2) P = x 0.2637 0.4506 0.2857 0.2637 0.4506 0.2857 0.2857 0.4506 0.2637 0.2857 0.4506 0.2637 0.3278 0.4620 0.2102 (2) P =0.3094 0.4579 0.2374 0.3122 0.0.4585 0.2293 0.3278 0.4620 0.2104 (5) P =0.3093 0.4579 0.2330 0.3127 0.4594 0.2304 0.3418 0.4667 0.2109 (10) P = 0.3141 0.4567 0.2327 0.3173 0.4572 0.2293 0.3158 0.4593 0.2248 (20) P = 0.3158 0.4593 0.2248 0.3158 0.4593 0.22480.3158 0.4593 0.2248 (50) P = 0.3158 0.4593 0.2248 0.3158 0.4593 0.2248 In the long run the n-step transition probability matrix converge tothe followingmatrix:0.32 0.46 0.22 (x) P = where x= 200.32 0.46 0.22 0.32 0.46 0.22 We can conclude that in the long run the probability of sunny days converge to32%, the probability for cloudy days are 46% and the probability rainy days is22%Application Dartmouth. 2016. Markov Chains. [ONLINE] Available at:https://www.dartmouth.edu/~chance/teaching_aids/books_articles/probabilit y_book/Chapter11.pdf. [Accessed 21 November 2016].Statslab. 2016. Markov chains. [ONLINE] Available at:http://www.statslab.cam.ac.uk/~rrw1/markov/M.pdf. [Accessed 22 November2016].Mathspace. 2016. Mathspace. [ONLINE] Available at:https://mathspace.co/learn/world-of-maths/binomial-distribution/markov- chains-and-transition-matrices-34075/markov-chains-and-transition-matrices- 1672/. [Accessed 24 November 2016]. "