Determining probability of a rainy day
up vote
3
down vote
favorite
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.
probability probability-theory
edited Dec 12 at 23:32
JYelton
1226
asked Dec 12 at 18:24
smiljanic997
1818
add a comment |
up vote
3
down vote
favorite
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.
probability probability-theory
edited Dec 12 at 23:32
JYelton
1226
asked Dec 12 at 18:24
smiljanic997
1818
2
You can diagonalize the transition matrix and raise it to the desired power.
– SmileyCraft
Dec 12 at 18:28
1
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.
– lulu
Dec 12 at 18:49
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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.
probability probability-theory
edited Dec 12 at 23:32
JYelton
1226
asked Dec 12 at 18:24
smiljanic997
1818
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.
probability probability-theory
probability probability-theory
edited Dec 12 at 23:32
JYelton
1226
asked Dec 12 at 18:24
smiljanic997
1818
edited Dec 12 at 23:32
JYelton
1226
asked Dec 12 at 18:24
smiljanic997
1818
edited Dec 12 at 23:32
JYelton
1226
edited Dec 12 at 23:32
JYelton
1226
edited Dec 12 at 23:32
JYelton
1226
1226
asked Dec 12 at 18:24
smiljanic997
1818
asked Dec 12 at 18:24
smiljanic997
1818
asked Dec 12 at 18:24
smiljanic997
1818
1818
2
You can diagonalize the transition matrix and raise it to the desired power.
– SmileyCraft
Dec 12 at 18:28
1
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.
– lulu
Dec 12 at 18:49
add a comment |
2
You can diagonalize the transition matrix and raise it to the desired power.
– SmileyCraft
Dec 12 at 18:28
1
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.
– lulu
Dec 12 at 18:49
2
2
You can diagonalize the transition matrix and raise it to the desired power.
– SmileyCraft
Dec 12 at 18:28
You can diagonalize the transition matrix and raise it to the desired power.
– SmileyCraft
Dec 12 at 18:28
1
1
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.
– lulu
Dec 12 at 18:49
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.
– lulu
Dec 12 at 18:49
add a comment |
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
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: $begin{pmatrix} s \ rend{pmatrix}$ 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 $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.
More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
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.
answered Dec 12 at 19:26
8910
1562
New contributor
8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
This is hands down the most elegant solution ever on this site. Wow.
– smiljanic997
Dec 13 at 18:19
add a comment |
up vote
6
down vote
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$
$ds{P_{n}}$ 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}
answered Dec 12 at 19:45
Felix Marin
66.8k7107139
I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
– Henry
Dec 12 at 23:49
@Henry That's true.
– Felix Marin
Dec 13 at 1:21
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
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: $begin{pmatrix} s \ rend{pmatrix}$ 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 $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.
More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
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.
answered Dec 12 at 19:26
8910
1562
New contributor
8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
This is hands down the most elegant solution ever on this site. Wow.
– smiljanic997
Dec 13 at 18:19
add a comment |
up vote
8
down vote
accepted
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: $begin{pmatrix} s \ rend{pmatrix}$ 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 $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.
More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
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.
answered Dec 12 at 19:26
8910
1562
New contributor
8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
This is hands down the most elegant solution ever on this site. Wow.
– smiljanic997
Dec 13 at 18:19
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
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: $begin{pmatrix} s \ rend{pmatrix}$ 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 $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.
More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
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.
answered Dec 12 at 19:26
8910
1562
New contributor
8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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: $begin{pmatrix} s \ rend{pmatrix}$ 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 $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.
More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
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.
answered Dec 12 at 19:26
8910
1562
New contributor
8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 12 at 19:26
8910
1562
New contributor
8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 12 at 19:26
8910
1562
answered Dec 12 at 19:26
8910
1562
1562
New contributor
8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
This is hands down the most elegant solution ever on this site. Wow.
– smiljanic997
Dec 13 at 18:19
add a comment |
1
This is hands down the most elegant solution ever on this site. Wow.
– smiljanic997
Dec 13 at 18:19
1
1
This is hands down the most elegant solution ever on this site. Wow.
– smiljanic997
Dec 13 at 18:19
This is hands down the most elegant solution ever on this site. Wow.
– smiljanic997
Dec 13 at 18:19
add a comment |
up vote
6
down vote
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$
$ds{P_{n}}$ 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}
answered Dec 12 at 19:45
Felix Marin
66.8k7107139
I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
– Henry
Dec 12 at 23:49
@Henry That's true.
– Felix Marin
Dec 13 at 1:21
add a comment |
up vote
6
down vote
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$
$ds{P_{n}}$ 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}
answered Dec 12 at 19:45
Felix Marin
66.8k7107139
I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
– Henry
Dec 12 at 23:49
@Henry That's true.
– Felix Marin
Dec 13 at 1:21
add a comment |
up vote
6
down vote
up vote
6
down vote
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$
$ds{P_{n}}$ 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}
answered Dec 12 at 19:45
Felix Marin
66.8k7107139
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$
$ds{P_{n}}$ 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}
answered Dec 12 at 19:45
Felix Marin
66.8k7107139
edited Dec 12 at 20:56
answered Dec 12 at 19:45
Felix Marin
66.8k7107139
answered Dec 12 at 19:45
Felix Marin
66.8k7107139
answered Dec 12 at 19:45
Felix Marin
66.8k7107139
66.8k7107139
I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
– Henry
Dec 12 at 23:49
@Henry That's true.
– Felix Marin
Dec 13 at 1:21
add a comment |
I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
– Henry
Dec 12 at 23:49
@Henry That's true.
– Felix Marin
Dec 13 at 1:21
I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
– Henry
Dec 12 at 23:49
I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
– Henry
Dec 12 at 23:49
@Henry That's true.
– Felix Marin
Dec 13 at 1:21
@Henry That's true.
– Felix Marin
Dec 13 at 1:21
add a comment |
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2
You can diagonalize the transition matrix and raise it to the desired power.
– SmileyCraft
Dec 12 at 18:28
1
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.
– lulu
Dec 12 at 18:49