Rain Probability from a Markov Chain

A simple weather model: given that it rains today, it will rain tomorrow with probability 0.7, and given that it does not rain today, it will rain tomorrow with probability 0.4. Given that it is raining today, represent this model with a discrete Markov process.

A representation of the weather as a discrete Markov process:

In[1]:=

weather=DiscreteMarkovProcess[1,{{0.7,0.3},{0.4,0.6}}];

Draw the model as a directed graph based on its transition matrix:

In[2]:=

Graph[weather,VertexLabels{1"rain",2"no rain"},VertexSizeSmall,EdgeLabelsWith[{sm=MarkovProcessProperties[weather,"TransitionMatrix"]},Flatten@Table[DirectedEdge[i,j]sm〚i,j〛,{i,2},{j,2}]],ImageSizeMedium]

Out[2]=

The probability of rain in four days:

In[3]:=

Probability[rain[4]1,rainweather]

Out[3]=

0.5749

Now, build a weather model based on whether it rained on the two previous days. Suppose it will rain tomorrow with probability 0.7 if it rained today and yesterday, with probability 0.5 if it rained today but not yesterday, with probability 0.4 if it rained yesterday but not today, and with probability 0.2 if it did not rain yesterday or today. Use a product state space

{x,y}

, where

indicates whether it rained yesterday and

whether it rained today.

Define the transition probabilities:

In[4]:=

rain[{"R","R"},{"R","R"}]=0.7;rain[{"R","R"},{"R","N"}]=1-rain[{"R","R"},{"R","R"}];rain[{"N","R"},{"R","R"}]=0.5;rain[{"N","R"},{"R","N"}]=1-rain[{"N","R"},{"R","R"}];rain[{"R","N"},{"N","R"}]=0.4;rain[{"R","N"},{"N","N"}]=1-rain[{"R","N"},{"N","R"}];rain[{"N","N"},{"N","R"}]=0.2;rain[{"N","N"},{"N","N"}]=1-rain[{"N","N"},{"N","R"}];

The remaining cases cannot happen:

In[5]:=

rain[s_,ns_]:=0

The resulting transition probability matrix:

In[6]:=

m=ArrayFlatten@Table[rain[{x,y},{z,w}],{x,{"R","N"}},{y,{"R","N"}},{z,{"R","N"}},{w,{"R","N"}}]

Out[6]=

{{0.7,0.3,0,0},{0,0,0.4,0.6},{0.5,0.5,0,0},{0,0,0.2,0.8}}

Create a table of the transition matrix:

In[7]:=

TableForm[m,TableHeadings{StringJoin/@Tuples[{"R","N"},2],StringJoin/@Tuples[{"R","N"},2]}]

Out[7]//TableForm=

	RR	RN	NR	NN
RR	0.7	0.3	0	0
RN	0	0	0.4	0.6
NR	0.5	0.5	0	0
NN	0	0	0.2	0.8

The resulting discrete Markov process:

In[8]:=

weather=DiscreteMarkovProcess[1,m]

Out[8]=

DiscreteMarkovProcess[1,{{0.7,0.3,0,0},{0,0,0.4,0.6},{0.5,0.5,0,0},{0,0,0.2,0.8}}]

In[9]:=

labels={"RR","RN","NR","NN"};Graph[Flatten@Table[DirectedEdge[labels〚i〛,labels〚j〛,m〚i,j〛],{i,4},{j,4}],EdgeLabels"EdgeTag",VertexLabels"Name",VertexSizeSmall,ImageSizeMedium]

Out[9]=

Create a map to match the states to numbers from 1 to 4:

In[10]:=

{state["RR"],state["RN"],state["NR"],state["NN"]}=Range[4];

Use the map to calculate the probability of rain in 4 days:

In[11]:=

Probability[rain[4]state["RR"]||rain[4]state["NR"],rainweather]

Out[11]=

0.4789

Find the stationary distribution of the weather chain (the long-run probabilities):

In[12]:=

longtermWeather=StationaryDistribution[weather]

Out[12]=

ProbabilityDistribution[0.25Boole[x.1]+0.15Boole[x.2]+0.15Boole[x.3]+0.45Boole[x.4],{x.,1,4,1}]

Using that long-run distribution, return the probability that the chain is in state RR (two rainy days in a row):

In[13]:=

Probability[rainstate["RR"],rainlongtermWeather]

Out[13]=

0.25

Compare to a simulation over a long time period:

In[14]:=

Count[RandomFunction[weather,{1,10^4}]["Values"],1]/10^4//N

Out[14]=

0.2557

External Links

https://reference.wolfram.com/language/guide/MarkovProcesses.html

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Rain Probability from a Markov Chain

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Rain Probability from a Markov Chain

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