Probability of Random walk with 2 absorbing walls hitting one wall during N steps
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Assume a i.i.d. one dimensional random walk S with symmetrical probabilities of $frac{1}{2}$ for a unit plus or minus step. Start is at $S = 0$ and absorbing walls are at $S = - B$ and $S = + A$.
What is the probability $P_a$ that during $N$ steps the particle will be absorbed at $S = A$ before reaching $S = - B$ or escaping unabsorbed. Assume $N > 2A + B$ and assume if necessary $A > B$.
What is the probability $P_u$ that the particle hits neither wall in $N$ steps?
The solution for $N rightarrow infty$ or until particle is absorbed is $P_a = frac{B}{A + B}$. But for finite $N$ this is solved tediously by counting outcomes for specific examples. Can anyone find a closed solution for $P_a$ in terms of $A, B$ and $N$?
This is one of the well-known Gambler's Ruin problems.
probability
asked Dec 22 '18 at 1:23
artbenisartbenis
112
$endgroup$
add a comment |
$begingroup$
Assume a i.i.d. one dimensional random walk S with symmetrical probabilities of $frac{1}{2}$ for a unit plus or minus step. Start is at $S = 0$ and absorbing walls are at $S = - B$ and $S = + A$.
What is the probability $P_a$ that during $N$ steps the particle will be absorbed at $S = A$ before reaching $S = - B$ or escaping unabsorbed. Assume $N > 2A + B$ and assume if necessary $A > B$.
What is the probability $P_u$ that the particle hits neither wall in $N$ steps?
The solution for $N rightarrow infty$ or until particle is absorbed is $P_a = frac{B}{A + B}$. But for finite $N$ this is solved tediously by counting outcomes for specific examples. Can anyone find a closed solution for $P_a$ in terms of $A, B$ and $N$?
This is one of the well-known Gambler's Ruin problems.
probability
asked Dec 22 '18 at 1:23
artbenisartbenis
112
$endgroup$
1
$begingroup$
$P_a=(0; 0; ..., 0; 1; 0; ...;0)^TP^N(0;0;...;0;1)$ for $P$ the transition probability matrix, where $x^T$ denotes the transpose of the column vector $x$.
$endgroup$
– Michael
Dec 22 '18 at 2:34
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@Michael Could you please suggest another method? I am not familiar with transition matrix method you suggest. Thanks.
$endgroup$
– artbenis
Jan 2 at 14:30
$begingroup$
Let $S(t)$ be the Markov chain in the state space $mathcal{S}={-b,...,a}$. By the law of total probability $$ underbrace{P[S(t+1)=j]}_{p_j(t+1)}=sum_{iin mathcal{S}} underbrace{P[S(t+1)=j|S(t)=i]}_{P_{ij}}underbrace{P[S(t)=i]}_{p_i(t)}$$ which reduces to $p_j(t+1)=sum_{i in mathcal{S}}p_i(t)P_{ij}$ for all $jin mathcal{S}$. Define row vector $p(t)^T = (p_{-b}(t), ..., p_a(t))$. The matrix version of this is $p(t+1)^T = p(t)^T P$ for all $t in {0, 1, 2, ...}$. So start with $p(0)^T=(0,...,0,1,0,...,0)$ and iteratively multiply by $P$.
$endgroup$
– Michael
Jan 2 at 18:17
add a comment |
$begingroup$
Assume a i.i.d. one dimensional random walk S with symmetrical probabilities of $frac{1}{2}$ for a unit plus or minus step. Start is at $S = 0$ and absorbing walls are at $S = - B$ and $S = + A$.
What is the probability $P_a$ that during $N$ steps the particle will be absorbed at $S = A$ before reaching $S = - B$ or escaping unabsorbed. Assume $N > 2A + B$ and assume if necessary $A > B$.
What is the probability $P_u$ that the particle hits neither wall in $N$ steps?
The solution for $N rightarrow infty$ or until particle is absorbed is $P_a = frac{B}{A + B}$. But for finite $N$ this is solved tediously by counting outcomes for specific examples. Can anyone find a closed solution for $P_a$ in terms of $A, B$ and $N$?
This is one of the well-known Gambler's Ruin problems.
probability
asked Dec 22 '18 at 1:23
artbenisartbenis
112
$endgroup$
Assume a i.i.d. one dimensional random walk S with symmetrical probabilities of $frac{1}{2}$ for a unit plus or minus step. Start is at $S = 0$ and absorbing walls are at $S = - B$ and $S = + A$.
What is the probability $P_a$ that during $N$ steps the particle will be absorbed at $S = A$ before reaching $S = - B$ or escaping unabsorbed. Assume $N > 2A + B$ and assume if necessary $A > B$.
What is the probability $P_u$ that the particle hits neither wall in $N$ steps?
The solution for $N rightarrow infty$ or until particle is absorbed is $P_a = frac{B}{A + B}$. But for finite $N$ this is solved tediously by counting outcomes for specific examples. Can anyone find a closed solution for $P_a$ in terms of $A, B$ and $N$?
This is one of the well-known Gambler's Ruin problems.
probability
probability
asked Dec 22 '18 at 1:23
artbenisartbenis
112
asked Dec 22 '18 at 1:23
artbenisartbenis
112
edited Jan 2 at 14:28
artbenis
asked Dec 22 '18 at 1:23
artbenisartbenis
112
asked Dec 22 '18 at 1:23
artbenisartbenis
112
asked Dec 22 '18 at 1:23
artbenisartbenis
112
112
1
$begingroup$
$P_a=(0; 0; ..., 0; 1; 0; ...;0)^TP^N(0;0;...;0;1)$ for $P$ the transition probability matrix, where $x^T$ denotes the transpose of the column vector $x$.
$endgroup$
– Michael
Dec 22 '18 at 2:34
$begingroup$
@Michael Could you please suggest another method? I am not familiar with transition matrix method you suggest. Thanks.
$endgroup$
– artbenis
Jan 2 at 14:30
$begingroup$
Let $S(t)$ be the Markov chain in the state space $mathcal{S}={-b,...,a}$. By the law of total probability $$ underbrace{P[S(t+1)=j]}_{p_j(t+1)}=sum_{iin mathcal{S}} underbrace{P[S(t+1)=j|S(t)=i]}_{P_{ij}}underbrace{P[S(t)=i]}_{p_i(t)}$$ which reduces to $p_j(t+1)=sum_{i in mathcal{S}}p_i(t)P_{ij}$ for all $jin mathcal{S}$. Define row vector $p(t)^T = (p_{-b}(t), ..., p_a(t))$. The matrix version of this is $p(t+1)^T = p(t)^T P$ for all $t in {0, 1, 2, ...}$. So start with $p(0)^T=(0,...,0,1,0,...,0)$ and iteratively multiply by $P$.
$endgroup$
– Michael
Jan 2 at 18:17
add a comment |
1
$begingroup$
$P_a=(0; 0; ..., 0; 1; 0; ...;0)^TP^N(0;0;...;0;1)$ for $P$ the transition probability matrix, where $x^T$ denotes the transpose of the column vector $x$.
$endgroup$
– Michael
Dec 22 '18 at 2:34
$begingroup$
@Michael Could you please suggest another method? I am not familiar with transition matrix method you suggest. Thanks.
$endgroup$
– artbenis
Jan 2 at 14:30
$begingroup$
Let $S(t)$ be the Markov chain in the state space $mathcal{S}={-b,...,a}$. By the law of total probability $$ underbrace{P[S(t+1)=j]}_{p_j(t+1)}=sum_{iin mathcal{S}} underbrace{P[S(t+1)=j|S(t)=i]}_{P_{ij}}underbrace{P[S(t)=i]}_{p_i(t)}$$ which reduces to $p_j(t+1)=sum_{i in mathcal{S}}p_i(t)P_{ij}$ for all $jin mathcal{S}$. Define row vector $p(t)^T = (p_{-b}(t), ..., p_a(t))$. The matrix version of this is $p(t+1)^T = p(t)^T P$ for all $t in {0, 1, 2, ...}$. So start with $p(0)^T=(0,...,0,1,0,...,0)$ and iteratively multiply by $P$.
$endgroup$
– Michael
Jan 2 at 18:17
1
1
$begingroup$
$P_a=(0; 0; ..., 0; 1; 0; ...;0)^TP^N(0;0;...;0;1)$ for $P$ the transition probability matrix, where $x^T$ denotes the transpose of the column vector $x$.
$endgroup$
– Michael
Dec 22 '18 at 2:34
$begingroup$
$P_a=(0; 0; ..., 0; 1; 0; ...;0)^TP^N(0;0;...;0;1)$ for $P$ the transition probability matrix, where $x^T$ denotes the transpose of the column vector $x$.
$endgroup$
– Michael
Dec 22 '18 at 2:34
$begingroup$
@Michael Could you please suggest another method? I am not familiar with transition matrix method you suggest. Thanks.
$endgroup$
– artbenis
Jan 2 at 14:30
$begingroup$
@Michael Could you please suggest another method? I am not familiar with transition matrix method you suggest. Thanks.
$endgroup$
– artbenis
Jan 2 at 14:30
$begingroup$
Let $S(t)$ be the Markov chain in the state space $mathcal{S}={-b,...,a}$. By the law of total probability $$ underbrace{P[S(t+1)=j]}_{p_j(t+1)}=sum_{iin mathcal{S}} underbrace{P[S(t+1)=j|S(t)=i]}_{P_{ij}}underbrace{P[S(t)=i]}_{p_i(t)}$$ which reduces to $p_j(t+1)=sum_{i in mathcal{S}}p_i(t)P_{ij}$ for all $jin mathcal{S}$. Define row vector $p(t)^T = (p_{-b}(t), ..., p_a(t))$. The matrix version of this is $p(t+1)^T = p(t)^T P$ for all $t in {0, 1, 2, ...}$. So start with $p(0)^T=(0,...,0,1,0,...,0)$ and iteratively multiply by $P$.
$endgroup$
– Michael
Jan 2 at 18:17
$begingroup$
Let $S(t)$ be the Markov chain in the state space $mathcal{S}={-b,...,a}$. By the law of total probability $$ underbrace{P[S(t+1)=j]}_{p_j(t+1)}=sum_{iin mathcal{S}} underbrace{P[S(t+1)=j|S(t)=i]}_{P_{ij}}underbrace{P[S(t)=i]}_{p_i(t)}$$ which reduces to $p_j(t+1)=sum_{i in mathcal{S}}p_i(t)P_{ij}$ for all $jin mathcal{S}$. Define row vector $p(t)^T = (p_{-b}(t), ..., p_a(t))$. The matrix version of this is $p(t+1)^T = p(t)^T P$ for all $t in {0, 1, 2, ...}$. So start with $p(0)^T=(0,...,0,1,0,...,0)$ and iteratively multiply by $P$.
$endgroup$
– Michael
Jan 2 at 18:17
add a comment |
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1
$begingroup$
$P_a=(0; 0; ..., 0; 1; 0; ...;0)^TP^N(0;0;...;0;1)$ for $P$ the transition probability matrix, where $x^T$ denotes the transpose of the column vector $x$.
$endgroup$
– Michael
Dec 22 '18 at 2:34
$begingroup$
@Michael Could you please suggest another method? I am not familiar with transition matrix method you suggest. Thanks.
$endgroup$
– artbenis
Jan 2 at 14:30
$begingroup$
Let $S(t)$ be the Markov chain in the state space $mathcal{S}={-b,...,a}$. By the law of total probability $$ underbrace{P[S(t+1)=j]}_{p_j(t+1)}=sum_{iin mathcal{S}} underbrace{P[S(t+1)=j|S(t)=i]}_{P_{ij}}underbrace{P[S(t)=i]}_{p_i(t)}$$ which reduces to $p_j(t+1)=sum_{i in mathcal{S}}p_i(t)P_{ij}$ for all $jin mathcal{S}$. Define row vector $p(t)^T = (p_{-b}(t), ..., p_a(t))$. The matrix version of this is $p(t+1)^T = p(t)^T P$ for all $t in {0, 1, 2, ...}$. So start with $p(0)^T=(0,...,0,1,0,...,0)$ and iteratively multiply by $P$.
$endgroup$
– Michael
Jan 2 at 18:17