I have the following problem:

  • If today is a sunny day, a probability that it will rain tomorrow is $0.2$.
  • If today is a rainy day, a probability that it will be sunny tomorrow is $0.4$.

I need to find the probability that if it's rainy on the third of May, it will also rain on the third of June.

My initial idea was to write a program that will create the binary tree with all possible combinations and then I just traverse through all of them and sum the probabilities accordingly, but unfortunately, I have to do this by hand, so any help is very welcome.

  • 2
    $\begingroup$ You can diagonalize the transition matrix and raise it to the desired power. $\endgroup$ – SmileyCraft Dec 12 '18 at 18:28
  • 1
    $\begingroup$ It is easy to compute the limiting probabilities...you might imagine that $30$ days (or whatever) is long enough for the probabilities to be in their limiting state. Running it quickly on a spreadsheet, it looks like they get to the limit much quicker than that. $\endgroup$ – lulu Dec 12 '18 at 18:49

A binary tree is definitely a possible way to solve this problem.

Another way to think about it though is maybe in the language or linear algebra.

We can represent day as the vector: $\begin{pmatrix} s \\ r\end{pmatrix}$ where $s$ is the probability of sun on that day and $r$ represents the chance of rain, and then the matrix: $$\begin{pmatrix} 0.8 & 0.4 \\ 0.2 & 0.6 \end{pmatrix}$$ would represent the transition function from one day to another.

So if we have rain on the 3rd of May, the probability vector for the 4th of May will be $\begin{pmatrix} 0.8 & 0.4 \\ 0.2 & 0.6 \end{pmatrix} \begin{pmatrix} 0 \\ 1\end{pmatrix}$.

More generally, $$\begin{pmatrix} 0.8 & 0.4 \\ 0.2 & 0.6 \end{pmatrix}^n \begin{pmatrix} 0 \\ 1\end{pmatrix}$$ the probability vector for the nth day after the 3rd of May. For your problem, I think $n = 31$.

edit I notice now that SmileyCraft makes a good point to diagonize this transition matrix and this makes the power easier to work with.

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    $\begingroup$ This is hands down the most elegant solution ever on this site. Wow. $\endgroup$ – smiljanic997 Dec 13 '18 at 18:19

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \left\{\begin{array}{rcl} \ds{P_{n}} & \ds{\equiv} & Rainy\ Probability\ \mbox{at the}\ n_{th}\ \mbox{Day} \\ \ds{P_{1}} & \ds{=} & \ds{1} \\ \ds{P_{31}} & \ds{=} & \ds{\large ?} \end{array}\right. $$ $\ds{P_{n}}$ is given by \begin{align} &P_{n} = \pars{1 - P_{n - 1}}0.2 + P_{n - 1}\pars{1 - 0.4}\,,\qquad P_{1} = 1 \\[5mm] \implies &\pars{P_{n} - {1 \over 3}} - {2 \over 5}\pars{P_{n - 1} - {1 \over 3}} = 0 \\[5mm] \implies & \bbx{P_{n} = {1 \over 3} + {5 \over 3}\pars{2 \over 5}^{n}} \\[5mm] \implies & P_{31} = {310440858205875333091 \over 931322574615478515625} \approx \bbox[#ffd,10px,border:1px groove navy]{0.3333} \end{align}

  • $\begingroup$ I think there are more than four $3$s at the start of the final decimal. Perhaps $11$? $\endgroup$ – Henry Dec 12 '18 at 23:49
  • $\begingroup$ @Henry That's true. $\endgroup$ – Felix Marin Dec 13 '18 at 1:21

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