Generalized Polya's Urn
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In Polya's urn, we have $b$ black balls and $w$ white balls at time $t$. At time $t+1$, we have $b+1$ black balls with probability $frac{b}{w+b}$ and $w+1$ white balls with probability $frac{w}{w+b}$.
Is there a way to generalize the probability function? Can you construct a generalized Polya's urn such that the probability of $b+1$ black and $w+1$ white balls at $t+1$ is $f(b, w)$, where $f: {mathbb{N}^+}^2 rightarrow {0, 1}^2$?
probability statistics stochastic-processes random-variables random
asked Dec 5 '18 at 23:57
ux74bn1ux74bn1
163
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add a comment |
$begingroup$
In Polya's urn, we have $b$ black balls and $w$ white balls at time $t$. At time $t+1$, we have $b+1$ black balls with probability $frac{b}{w+b}$ and $w+1$ white balls with probability $frac{w}{w+b}$.
Is there a way to generalize the probability function? Can you construct a generalized Polya's urn such that the probability of $b+1$ black and $w+1$ white balls at $t+1$ is $f(b, w)$, where $f: {mathbb{N}^+}^2 rightarrow {0, 1}^2$?
probability statistics stochastic-processes random-variables random
asked Dec 5 '18 at 23:57
ux74bn1ux74bn1
163
$endgroup$
add a comment |
$begingroup$
In Polya's urn, we have $b$ black balls and $w$ white balls at time $t$. At time $t+1$, we have $b+1$ black balls with probability $frac{b}{w+b}$ and $w+1$ white balls with probability $frac{w}{w+b}$.
Is there a way to generalize the probability function? Can you construct a generalized Polya's urn such that the probability of $b+1$ black and $w+1$ white balls at $t+1$ is $f(b, w)$, where $f: {mathbb{N}^+}^2 rightarrow {0, 1}^2$?
probability statistics stochastic-processes random-variables random
asked Dec 5 '18 at 23:57
ux74bn1ux74bn1
163
$endgroup$
In Polya's urn, we have $b$ black balls and $w$ white balls at time $t$. At time $t+1$, we have $b+1$ black balls with probability $frac{b}{w+b}$ and $w+1$ white balls with probability $frac{w}{w+b}$.
Is there a way to generalize the probability function? Can you construct a generalized Polya's urn such that the probability of $b+1$ black and $w+1$ white balls at $t+1$ is $f(b, w)$, where $f: {mathbb{N}^+}^2 rightarrow {0, 1}^2$?
probability statistics stochastic-processes random-variables random
probability statistics stochastic-processes random-variables random
asked Dec 5 '18 at 23:57
ux74bn1ux74bn1
163
asked Dec 5 '18 at 23:57
ux74bn1ux74bn1
163
asked Dec 5 '18 at 23:57
ux74bn1ux74bn1
163
asked Dec 5 '18 at 23:57
ux74bn1ux74bn1
163
asked Dec 5 '18 at 23:57
ux74bn1ux74bn1
163
163
add a comment |
add a comment |
1 Answer
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It is possible to use any function $f:mathbb N^+ times mathbb N^+ to [0,1]$ to represent
$f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b+1$ black balls and $w$ white balls
$1-f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b$ black balls and $w+1$ white balls
and given the starting position, then to find the distribution of the number or proportion of black balls or of white balls a given number of steps later. With the original Polya urn you have $f(b,w)=frac{b}{b+w}$ and $1-f(b,w)=frac{w}{b+w}$, but any other function can be used instead
A further generalisation could be to allow transitions to other states, not just adding a single ball with each step, or to have more than two colours, though these might require a more complicated function or a switch to a Markov-chain type of analysis
answered Dec 6 '18 at 12:02
HenryHenry
99.5k479165
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$begingroup$
What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
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– ux74bn1
Dec 6 '18 at 16:10
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@ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
$endgroup$
– Henry
Dec 6 '18 at 19:34
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
It is possible to use any function $f:mathbb N^+ times mathbb N^+ to [0,1]$ to represent
$f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b+1$ black balls and $w$ white balls
$1-f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b$ black balls and $w+1$ white balls
and given the starting position, then to find the distribution of the number or proportion of black balls or of white balls a given number of steps later. With the original Polya urn you have $f(b,w)=frac{b}{b+w}$ and $1-f(b,w)=frac{w}{b+w}$, but any other function can be used instead
A further generalisation could be to allow transitions to other states, not just adding a single ball with each step, or to have more than two colours, though these might require a more complicated function or a switch to a Markov-chain type of analysis
answered Dec 6 '18 at 12:02
HenryHenry
99.5k479165
$endgroup$
$begingroup$
What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
$endgroup$
– ux74bn1
Dec 6 '18 at 16:10
$begingroup$
@ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
$endgroup$
– Henry
Dec 6 '18 at 19:34
add a comment |
$begingroup$
It is possible to use any function $f:mathbb N^+ times mathbb N^+ to [0,1]$ to represent
$f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b+1$ black balls and $w$ white balls
$1-f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b$ black balls and $w+1$ white balls
and given the starting position, then to find the distribution of the number or proportion of black balls or of white balls a given number of steps later. With the original Polya urn you have $f(b,w)=frac{b}{b+w}$ and $1-f(b,w)=frac{w}{b+w}$, but any other function can be used instead
A further generalisation could be to allow transitions to other states, not just adding a single ball with each step, or to have more than two colours, though these might require a more complicated function or a switch to a Markov-chain type of analysis
answered Dec 6 '18 at 12:02
HenryHenry
99.5k479165
$endgroup$
$begingroup$
What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
$endgroup$
– ux74bn1
Dec 6 '18 at 16:10
$begingroup$
@ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
$endgroup$
– Henry
Dec 6 '18 at 19:34
add a comment |
$begingroup$
It is possible to use any function $f:mathbb N^+ times mathbb N^+ to [0,1]$ to represent
$f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b+1$ black balls and $w$ white balls
$1-f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b$ black balls and $w+1$ white balls
and given the starting position, then to find the distribution of the number or proportion of black balls or of white balls a given number of steps later. With the original Polya urn you have $f(b,w)=frac{b}{b+w}$ and $1-f(b,w)=frac{w}{b+w}$, but any other function can be used instead
A further generalisation could be to allow transitions to other states, not just adding a single ball with each step, or to have more than two colours, though these might require a more complicated function or a switch to a Markov-chain type of analysis
answered Dec 6 '18 at 12:02
HenryHenry
99.5k479165
$endgroup$
It is possible to use any function $f:mathbb N^+ times mathbb N^+ to [0,1]$ to represent
$f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b+1$ black balls and $w$ white balls
$1-f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b$ black balls and $w+1$ white balls
and given the starting position, then to find the distribution of the number or proportion of black balls or of white balls a given number of steps later. With the original Polya urn you have $f(b,w)=frac{b}{b+w}$ and $1-f(b,w)=frac{w}{b+w}$, but any other function can be used instead
A further generalisation could be to allow transitions to other states, not just adding a single ball with each step, or to have more than two colours, though these might require a more complicated function or a switch to a Markov-chain type of analysis
answered Dec 6 '18 at 12:02
HenryHenry
99.5k479165
answered Dec 6 '18 at 12:02
HenryHenry
99.5k479165
answered Dec 6 '18 at 12:02
HenryHenry
99.5k479165
answered Dec 6 '18 at 12:02
HenryHenry
99.5k479165
99.5k479165
$begingroup$
What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
$endgroup$
– ux74bn1
Dec 6 '18 at 16:10
$begingroup$
@ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
$endgroup$
– Henry
Dec 6 '18 at 19:34
add a comment |
$begingroup$
What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
$endgroup$
– ux74bn1
Dec 6 '18 at 16:10
$begingroup$
@ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
$endgroup$
– Henry
Dec 6 '18 at 19:34
$begingroup$
What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
$endgroup$
– ux74bn1
Dec 6 '18 at 16:10
$begingroup$
What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
$endgroup$
– ux74bn1
Dec 6 '18 at 16:10
$begingroup$
@ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
$endgroup$
– Henry
Dec 6 '18 at 19:34
$begingroup$
@ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
$endgroup$
– Henry
Dec 6 '18 at 19:34
add a comment |
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