Sum of $n$ coin flips, where the probability of heads at experiment $t$ is dependent on the outcomes of the...
$begingroup$
Suppose I play a game of $n$ coin flips, where heads is $1$ and tails is $0$. If each coin flip was independent, the expected sum of all $n$ coin flips is trivial. What if there is dependence? How can this be solved?
A general solution (if any) would be best, but let's construct a precise game just for simplicity. Let ${X_1, X_2, dots, X_n}$ be a sequence of $n$ coin flips, where:
$$
begin{cases}
P(X_t = 1) = (P(X_{t-1} = 1))^2text{ IF } X_{t-1} = 1; text{ ELSE } P(X_t = 1) = frac{1}{2}P(X_{t-1} = 1)\
P(X_t = 0) = 1 - P(X_t = 1)
end{cases}
$$
with base case:
$$
begin{cases}
P(X_1 = 1) = p\
P(X_1 = 0) = 1 - p
end{cases}
$$
In other words, the coin flip at time $t$ has two different probabilities of being heads. It is heads with half the probability of being heads at time $t-1$ if it was tails at $t-1$. On the other hand, if it was heads at $t-1$, then you take the squared value of the probability of heads at $t-1$.
How does one go about reasoning about this types of stochastic fields?
probability statistics stochastic-processes
asked Dec 4 '18 at 21:55
ux74bn1ux74bn1
263
$endgroup$
add a comment |
$begingroup$
Suppose I play a game of $n$ coin flips, where heads is $1$ and tails is $0$. If each coin flip was independent, the expected sum of all $n$ coin flips is trivial. What if there is dependence? How can this be solved?
A general solution (if any) would be best, but let's construct a precise game just for simplicity. Let ${X_1, X_2, dots, X_n}$ be a sequence of $n$ coin flips, where:
$$
begin{cases}
P(X_t = 1) = (P(X_{t-1} = 1))^2text{ IF } X_{t-1} = 1; text{ ELSE } P(X_t = 1) = frac{1}{2}P(X_{t-1} = 1)\
P(X_t = 0) = 1 - P(X_t = 1)
end{cases}
$$
with base case:
$$
begin{cases}
P(X_1 = 1) = p\
P(X_1 = 0) = 1 - p
end{cases}
$$
In other words, the coin flip at time $t$ has two different probabilities of being heads. It is heads with half the probability of being heads at time $t-1$ if it was tails at $t-1$. On the other hand, if it was heads at $t-1$, then you take the squared value of the probability of heads at $t-1$.
How does one go about reasoning about this types of stochastic fields?
probability statistics stochastic-processes
asked Dec 4 '18 at 21:55
ux74bn1ux74bn1
263
$endgroup$
$begingroup$
For every $p<1$, the expected sum of infinitely many flips is finite. The expected sum $s_n(p)$ of $n$ flips for parameter $p$ solves the recursion $$s_n(p)=(1-p)s_{n-1}left(tfrac12pright)+p(1+s_{n-1}(p^2))$$ with initialization $$s_1(p)=p$$ The way to solve these for the exact values $s_n(p)$, if possible, is not obvious.
$endgroup$
– Did
Dec 4 '18 at 22:35
$begingroup$
@Did these types of processes might also be described by something similar to Polya's urn process. Do you happen to have good references to them?
$endgroup$
– ux74bn1
Dec 5 '18 at 22:10
add a comment |
$begingroup$
Suppose I play a game of $n$ coin flips, where heads is $1$ and tails is $0$. If each coin flip was independent, the expected sum of all $n$ coin flips is trivial. What if there is dependence? How can this be solved?
A general solution (if any) would be best, but let's construct a precise game just for simplicity. Let ${X_1, X_2, dots, X_n}$ be a sequence of $n$ coin flips, where:
$$
begin{cases}
P(X_t = 1) = (P(X_{t-1} = 1))^2text{ IF } X_{t-1} = 1; text{ ELSE } P(X_t = 1) = frac{1}{2}P(X_{t-1} = 1)\
P(X_t = 0) = 1 - P(X_t = 1)
end{cases}
$$
with base case:
$$
begin{cases}
P(X_1 = 1) = p\
P(X_1 = 0) = 1 - p
end{cases}
$$
In other words, the coin flip at time $t$ has two different probabilities of being heads. It is heads with half the probability of being heads at time $t-1$ if it was tails at $t-1$. On the other hand, if it was heads at $t-1$, then you take the squared value of the probability of heads at $t-1$.
How does one go about reasoning about this types of stochastic fields?
probability statistics stochastic-processes
asked Dec 4 '18 at 21:55
ux74bn1ux74bn1
263
$endgroup$
Suppose I play a game of $n$ coin flips, where heads is $1$ and tails is $0$. If each coin flip was independent, the expected sum of all $n$ coin flips is trivial. What if there is dependence? How can this be solved?
A general solution (if any) would be best, but let's construct a precise game just for simplicity. Let ${X_1, X_2, dots, X_n}$ be a sequence of $n$ coin flips, where:
$$
begin{cases}
P(X_t = 1) = (P(X_{t-1} = 1))^2text{ IF } X_{t-1} = 1; text{ ELSE } P(X_t = 1) = frac{1}{2}P(X_{t-1} = 1)\
P(X_t = 0) = 1 - P(X_t = 1)
end{cases}
$$
with base case:
$$
begin{cases}
P(X_1 = 1) = p\
P(X_1 = 0) = 1 - p
end{cases}
$$
In other words, the coin flip at time $t$ has two different probabilities of being heads. It is heads with half the probability of being heads at time $t-1$ if it was tails at $t-1$. On the other hand, if it was heads at $t-1$, then you take the squared value of the probability of heads at $t-1$.
How does one go about reasoning about this types of stochastic fields?
probability statistics stochastic-processes
probability statistics stochastic-processes
asked Dec 4 '18 at 21:55
ux74bn1ux74bn1
263
asked Dec 4 '18 at 21:55
ux74bn1ux74bn1
263
asked Dec 4 '18 at 21:55
ux74bn1ux74bn1
263
asked Dec 4 '18 at 21:55
ux74bn1ux74bn1
263
asked Dec 4 '18 at 21:55
ux74bn1ux74bn1
263
263
$begingroup$
For every $p<1$, the expected sum of infinitely many flips is finite. The expected sum $s_n(p)$ of $n$ flips for parameter $p$ solves the recursion $$s_n(p)=(1-p)s_{n-1}left(tfrac12pright)+p(1+s_{n-1}(p^2))$$ with initialization $$s_1(p)=p$$ The way to solve these for the exact values $s_n(p)$, if possible, is not obvious.
$endgroup$
– Did
Dec 4 '18 at 22:35
$begingroup$
@Did these types of processes might also be described by something similar to Polya's urn process. Do you happen to have good references to them?
$endgroup$
– ux74bn1
Dec 5 '18 at 22:10
add a comment |
$begingroup$
For every $p<1$, the expected sum of infinitely many flips is finite. The expected sum $s_n(p)$ of $n$ flips for parameter $p$ solves the recursion $$s_n(p)=(1-p)s_{n-1}left(tfrac12pright)+p(1+s_{n-1}(p^2))$$ with initialization $$s_1(p)=p$$ The way to solve these for the exact values $s_n(p)$, if possible, is not obvious.
$endgroup$
– Did
Dec 4 '18 at 22:35
$begingroup$
@Did these types of processes might also be described by something similar to Polya's urn process. Do you happen to have good references to them?
$endgroup$
– ux74bn1
Dec 5 '18 at 22:10
$begingroup$
For every $p<1$, the expected sum of infinitely many flips is finite. The expected sum $s_n(p)$ of $n$ flips for parameter $p$ solves the recursion $$s_n(p)=(1-p)s_{n-1}left(tfrac12pright)+p(1+s_{n-1}(p^2))$$ with initialization $$s_1(p)=p$$ The way to solve these for the exact values $s_n(p)$, if possible, is not obvious.
$endgroup$
– Did
Dec 4 '18 at 22:35
$begingroup$
For every $p<1$, the expected sum of infinitely many flips is finite. The expected sum $s_n(p)$ of $n$ flips for parameter $p$ solves the recursion $$s_n(p)=(1-p)s_{n-1}left(tfrac12pright)+p(1+s_{n-1}(p^2))$$ with initialization $$s_1(p)=p$$ The way to solve these for the exact values $s_n(p)$, if possible, is not obvious.
$endgroup$
– Did
Dec 4 '18 at 22:35
$begingroup$
@Did these types of processes might also be described by something similar to Polya's urn process. Do you happen to have good references to them?
$endgroup$
– ux74bn1
Dec 5 '18 at 22:10
$begingroup$
@Did these types of processes might also be described by something similar to Polya's urn process. Do you happen to have good references to them?
$endgroup$
– ux74bn1
Dec 5 '18 at 22:10
add a comment |
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$begingroup$
For every $p<1$, the expected sum of infinitely many flips is finite. The expected sum $s_n(p)$ of $n$ flips for parameter $p$ solves the recursion $$s_n(p)=(1-p)s_{n-1}left(tfrac12pright)+p(1+s_{n-1}(p^2))$$ with initialization $$s_1(p)=p$$ The way to solve these for the exact values $s_n(p)$, if possible, is not obvious.
$endgroup$
– Did
Dec 4 '18 at 22:35
$begingroup$
@Did these types of processes might also be described by something similar to Polya's urn process. Do you happen to have good references to them?
$endgroup$
– ux74bn1
Dec 5 '18 at 22:10