Mixing
Probability of getting $n$ tails in a row
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Imagine we have an unusual coin. Probability of getting a tail is equal to $p in [0, 1]$. Of course that means that probability of getting a head is equal to $q = 1-p in [0, 1]$.
Let's define a random variable $X$
$$X = text{the amount of throws until getting n tails in a row}$$
I am to calculate $mathbb{E}(X)$, where $mathbb{E}$ is the expected value. How can it be found? I would appreciate any tips or hints.
probability
asked 22 hours ago
Hendrra
978413
|
show 3 more comments
up vote
0
down vote
favorite
Imagine we have an unusual coin. Probability of getting a tail is equal to $p in [0, 1]$. Of course that means that probability of getting a head is equal to $q = 1-p in [0, 1]$.
Let's define a random variable $X$
$$X = text{the amount of throws until getting n tails in a row}$$
I am to calculate $mathbb{E}(X)$, where $mathbb{E}$ is the expected value. How can it be found? I would appreciate any tips or hints.
probability
asked 22 hours ago
Hendrra
978413
@drhab I think your proposed solution was nearly correct. To get $k$ tails in a row I must first get $k-1$ tails in a row. So, we go out $mu_{k-1}$ turns and toss. either I win or I restart. If we restart I then expect $1+mu_{k-1}+mu_k$ tosses (including the $1+mu_{k-1}$ tosses I have already done). Thus $mu_k=p(1+mu_{k-1})+q(1+mu_{k-1}+mu_k)$. No?
– lulu
22 hours ago
@lulu That gives $mu_k=p^{-1}+mu_{k-1}$ and seems not to agree with what I found now after some struggling. On the other hand I have not the guts to say it is wrong, since it looks good. Good chance that my revised is answer is wrong also.
– drhab
20 hours ago
@drhab Does it? I get $mu_k=mu_{k-1}+1+qmu_kimplies pmu_k=mu_{k-1}+1$ Thus $mu_k=frac 1ptimes (1+mu_{k-1})$ Which seems to harmonize with your new answer.
– lulu
20 hours ago
@lulu Ah, yes. I misunderstood at first hand because in my new answer $mu_k$ is defined on a different way than in my first answer, $mu_n$ and $mu_0$ somehow "switched" of meaning.
– drhab
20 hours ago
@drhab I think your first answer (correctly modified) is the way to go. Comparatively clear.
– lulu
20 hours ago
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Imagine we have an unusual coin. Probability of getting a tail is equal to $p in [0, 1]$. Of course that means that probability of getting a head is equal to $q = 1-p in [0, 1]$.
Let's define a random variable $X$
$$X = text{the amount of throws until getting n tails in a row}$$
I am to calculate $mathbb{E}(X)$, where $mathbb{E}$ is the expected value. How can it be found? I would appreciate any tips or hints.
probability
asked 22 hours ago
Hendrra
978413
Imagine we have an unusual coin. Probability of getting a tail is equal to $p in [0, 1]$. Of course that means that probability of getting a head is equal to $q = 1-p in [0, 1]$.
Let's define a random variable $X$
$$X = text{the amount of throws until getting n tails in a row}$$
I am to calculate $mathbb{E}(X)$, where $mathbb{E}$ is the expected value. How can it be found? I would appreciate any tips or hints.
probability
probability
asked 22 hours ago
Hendrra
978413
asked 22 hours ago
Hendrra
978413
asked 22 hours ago
Hendrra
978413
asked 22 hours ago
Hendrra
978413
asked 22 hours ago
Hendrra
978413
978413
@drhab I think your proposed solution was nearly correct. To get $k$ tails in a row I must first get $k-1$ tails in a row. So, we go out $mu_{k-1}$ turns and toss. either I win or I restart. If we restart I then expect $1+mu_{k-1}+mu_k$ tosses (including the $1+mu_{k-1}$ tosses I have already done). Thus $mu_k=p(1+mu_{k-1})+q(1+mu_{k-1}+mu_k)$. No?
– lulu
22 hours ago
@lulu That gives $mu_k=p^{-1}+mu_{k-1}$ and seems not to agree with what I found now after some struggling. On the other hand I have not the guts to say it is wrong, since it looks good. Good chance that my revised is answer is wrong also.
– drhab
20 hours ago
@drhab Does it? I get $mu_k=mu_{k-1}+1+qmu_kimplies pmu_k=mu_{k-1}+1$ Thus $mu_k=frac 1ptimes (1+mu_{k-1})$ Which seems to harmonize with your new answer.
– lulu
20 hours ago
@lulu Ah, yes. I misunderstood at first hand because in my new answer $mu_k$ is defined on a different way than in my first answer, $mu_n$ and $mu_0$ somehow "switched" of meaning.
– drhab
20 hours ago
@drhab I think your first answer (correctly modified) is the way to go. Comparatively clear.
– lulu
20 hours ago
|
show 3 more comments
@drhab I think your proposed solution was nearly correct. To get $k$ tails in a row I must first get $k-1$ tails in a row. So, we go out $mu_{k-1}$ turns and toss. either I win or I restart. If we restart I then expect $1+mu_{k-1}+mu_k$ tosses (including the $1+mu_{k-1}$ tosses I have already done). Thus $mu_k=p(1+mu_{k-1})+q(1+mu_{k-1}+mu_k)$. No?
– lulu
22 hours ago
@lulu That gives $mu_k=p^{-1}+mu_{k-1}$ and seems not to agree with what I found now after some struggling. On the other hand I have not the guts to say it is wrong, since it looks good. Good chance that my revised is answer is wrong also.
– drhab
20 hours ago
@drhab Does it? I get $mu_k=mu_{k-1}+1+qmu_kimplies pmu_k=mu_{k-1}+1$ Thus $mu_k=frac 1ptimes (1+mu_{k-1})$ Which seems to harmonize with your new answer.
– lulu
20 hours ago
@lulu Ah, yes. I misunderstood at first hand because in my new answer $mu_k$ is defined on a different way than in my first answer, $mu_n$ and $mu_0$ somehow "switched" of meaning.
– drhab
20 hours ago
@drhab I think your first answer (correctly modified) is the way to go. Comparatively clear.
– lulu
20 hours ago
@drhab I think your proposed solution was nearly correct. To get $k$ tails in a row I must first get $k-1$ tails in a row. So, we go out $mu_{k-1}$ turns and toss. either I win or I restart. If we restart I then expect $1+mu_{k-1}+mu_k$ tosses (including the $1+mu_{k-1}$ tosses I have already done). Thus $mu_k=p(1+mu_{k-1})+q(1+mu_{k-1}+mu_k)$. No?
– lulu
22 hours ago
@drhab I think your proposed solution was nearly correct. To get $k$ tails in a row I must first get $k-1$ tails in a row. So, we go out $mu_{k-1}$ turns and toss. either I win or I restart. If we restart I then expect $1+mu_{k-1}+mu_k$ tosses (including the $1+mu_{k-1}$ tosses I have already done). Thus $mu_k=p(1+mu_{k-1})+q(1+mu_{k-1}+mu_k)$. No?
– lulu
22 hours ago
@lulu That gives $mu_k=p^{-1}+mu_{k-1}$ and seems not to agree with what I found now after some struggling. On the other hand I have not the guts to say it is wrong, since it looks good. Good chance that my revised is answer is wrong also.
– drhab
20 hours ago
@lulu That gives $mu_k=p^{-1}+mu_{k-1}$ and seems not to agree with what I found now after some struggling. On the other hand I have not the guts to say it is wrong, since it looks good. Good chance that my revised is answer is wrong also.
– drhab
20 hours ago
@drhab Does it? I get $mu_k=mu_{k-1}+1+qmu_kimplies pmu_k=mu_{k-1}+1$ Thus $mu_k=frac 1ptimes (1+mu_{k-1})$ Which seems to harmonize with your new answer.
– lulu
20 hours ago
@drhab Does it? I get $mu_k=mu_{k-1}+1+qmu_kimplies pmu_k=mu_{k-1}+1$ Thus $mu_k=frac 1ptimes (1+mu_{k-1})$ Which seems to harmonize with your new answer.
– lulu
20 hours ago
@lulu Ah, yes. I misunderstood at first hand because in my new answer $mu_k$ is defined on a different way than in my first answer, $mu_n$ and $mu_0$ somehow "switched" of meaning.
– drhab
20 hours ago
@lulu Ah, yes. I misunderstood at first hand because in my new answer $mu_k$ is defined on a different way than in my first answer, $mu_n$ and $mu_0$ somehow "switched" of meaning.
– drhab
20 hours ago
@drhab I think your first answer (correctly modified) is the way to go. Comparatively clear.
– lulu
20 hours ago
@drhab I think your first answer (correctly modified) is the way to go. Comparatively clear.
– lulu
20 hours ago
|
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
Let's say that we are in status $k$ if our last $k$ throws ended up in tail and the (eventual) throw before this sequence of tails is not a tail.
Let $mu_k$ denote the expectation of the number of throws needed to arrive at $n$ tails in a row if we are in status $k$.
Then $mu_n=0$ and to be found is $mu_0$.
For $k=0,1,2,dots n-1$ we have:$$mu_{k}=p(1+mu_{k+1})+q(1+mu_0)=1+pmu_{k+1}+qmu_0$$
So we have the following equalities:
$mu_0=1+pmu_1+qmu_0$ so that $mu_0=frac1p+mu_1$
$mu_1=1+pmu_2+qmu_0=1+pmu_2+frac{q}p+qmu_1$ so that $mu_1=frac1p+frac q{p^2}+mu_2=frac1{p^2}+mu_2$
This makes us suspect that $mu_k=sum_{i=k+1}^np^{-i}$ for $k=0,1,dots,n$ and substitution in $(1)$ confirms that conjecture.
The the outcome is:$$mathbb EX=sum_{i=1}^np^{-i}$$
addendum:
Another way to look at it is this:
Let $nu_k$ denote the expectation of the number of throws needed to achieve exactly $k$ tails in a row.
Then $nu_0=0$ and to be found is $nu_n$.
Now let it be that after $X$ steps we have a consecutive row of exactly $k-1$ tails. Then by throwing tails we need $X+1$ throws to get a consecutive row of exactly $k$ tails. By throwing heads we come back in start position and from there $X+1+Y$ steps will be needed to get a consecutive row of exactly $k$ tails. This with $mathbb EX=nu_{k-1}$ and $mathbb EY=nu_{k}$ hence giving the equality:$$nu_k=p(1+nu_{k-1})+q(1+nu_{k-1}+nu_k)$$or shorter:$$pnu_k=1+nu_{k-1}$$
This leads easily to:$$nu_k=sum_{i=1}^kp^{-i}text{ for }k=1,2,dots$$so that: $$nu_n=sum_{i=1}^np^{-i}$$
Credit for the second solution (definitely the most elegant one) goes to @lulu.
I was on a path that was nice but not completely okay.
Someone attended me on that (thank you @saulspatz).
Then someone said to me: take the original path, but with an adaption (thank you @lulu).
answered 22 hours ago
drhab
94.2k543125
1
I did this too, but on reflection, I don't think it's right. If we toss a tail, we can't say we just need $n-1$ tails in a row. If the next toss is heads, we're back to needing $n$ tails in a row.
– saulspatz
22 hours ago
@saulspatz You are correct. I will delete this within some minutes. Thank you.
– drhab
22 hours ago
add a comment |
up vote
0
down vote
This is a finite-state absorbing Markov chain. The state is the number of consecutive tails we have tossed so far. We start in state $0$ and the absorbing state is $n$. If we are in state $k<n$ then we transition to state $k+1$ with probability $p$ and we transition to state $0$ with probability $q,$ so that the first column on the matrix $Q$ described in the wiki has $q$ in every position, and the entries on the superdiagonal are all $p.$
You need to figure out $(I-Q)^{-1}$
answered 21 hours ago
saulspatz
12.8k21327
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let's say that we are in status $k$ if our last $k$ throws ended up in tail and the (eventual) throw before this sequence of tails is not a tail.
Let $mu_k$ denote the expectation of the number of throws needed to arrive at $n$ tails in a row if we are in status $k$.
Then $mu_n=0$ and to be found is $mu_0$.
For $k=0,1,2,dots n-1$ we have:$$mu_{k}=p(1+mu_{k+1})+q(1+mu_0)=1+pmu_{k+1}+qmu_0$$
So we have the following equalities:
$mu_0=1+pmu_1+qmu_0$ so that $mu_0=frac1p+mu_1$
$mu_1=1+pmu_2+qmu_0=1+pmu_2+frac{q}p+qmu_1$ so that $mu_1=frac1p+frac q{p^2}+mu_2=frac1{p^2}+mu_2$
This makes us suspect that $mu_k=sum_{i=k+1}^np^{-i}$ for $k=0,1,dots,n$ and substitution in $(1)$ confirms that conjecture.
The the outcome is:$$mathbb EX=sum_{i=1}^np^{-i}$$
addendum:
Another way to look at it is this:
Let $nu_k$ denote the expectation of the number of throws needed to achieve exactly $k$ tails in a row.
Then $nu_0=0$ and to be found is $nu_n$.
Now let it be that after $X$ steps we have a consecutive row of exactly $k-1$ tails. Then by throwing tails we need $X+1$ throws to get a consecutive row of exactly $k$ tails. By throwing heads we come back in start position and from there $X+1+Y$ steps will be needed to get a consecutive row of exactly $k$ tails. This with $mathbb EX=nu_{k-1}$ and $mathbb EY=nu_{k}$ hence giving the equality:$$nu_k=p(1+nu_{k-1})+q(1+nu_{k-1}+nu_k)$$or shorter:$$pnu_k=1+nu_{k-1}$$
This leads easily to:$$nu_k=sum_{i=1}^kp^{-i}text{ for }k=1,2,dots$$so that: $$nu_n=sum_{i=1}^np^{-i}$$
Credit for the second solution (definitely the most elegant one) goes to @lulu.
I was on a path that was nice but not completely okay.
Someone attended me on that (thank you @saulspatz).
Then someone said to me: take the original path, but with an adaption (thank you @lulu).
answered 22 hours ago
drhab
94.2k543125
1
I did this too, but on reflection, I don't think it's right. If we toss a tail, we can't say we just need $n-1$ tails in a row. If the next toss is heads, we're back to needing $n$ tails in a row.
– saulspatz
22 hours ago
@saulspatz You are correct. I will delete this within some minutes. Thank you.
– drhab
22 hours ago
add a comment |
up vote
2
down vote
Let's say that we are in status $k$ if our last $k$ throws ended up in tail and the (eventual) throw before this sequence of tails is not a tail.
Let $mu_k$ denote the expectation of the number of throws needed to arrive at $n$ tails in a row if we are in status $k$.
Then $mu_n=0$ and to be found is $mu_0$.
For $k=0,1,2,dots n-1$ we have:$$mu_{k}=p(1+mu_{k+1})+q(1+mu_0)=1+pmu_{k+1}+qmu_0$$
So we have the following equalities:
$mu_0=1+pmu_1+qmu_0$ so that $mu_0=frac1p+mu_1$
$mu_1=1+pmu_2+qmu_0=1+pmu_2+frac{q}p+qmu_1$ so that $mu_1=frac1p+frac q{p^2}+mu_2=frac1{p^2}+mu_2$
This makes us suspect that $mu_k=sum_{i=k+1}^np^{-i}$ for $k=0,1,dots,n$ and substitution in $(1)$ confirms that conjecture.
The the outcome is:$$mathbb EX=sum_{i=1}^np^{-i}$$
addendum:
Another way to look at it is this:
Let $nu_k$ denote the expectation of the number of throws needed to achieve exactly $k$ tails in a row.
Then $nu_0=0$ and to be found is $nu_n$.
Now let it be that after $X$ steps we have a consecutive row of exactly $k-1$ tails. Then by throwing tails we need $X+1$ throws to get a consecutive row of exactly $k$ tails. By throwing heads we come back in start position and from there $X+1+Y$ steps will be needed to get a consecutive row of exactly $k$ tails. This with $mathbb EX=nu_{k-1}$ and $mathbb EY=nu_{k}$ hence giving the equality:$$nu_k=p(1+nu_{k-1})+q(1+nu_{k-1}+nu_k)$$or shorter:$$pnu_k=1+nu_{k-1}$$
This leads easily to:$$nu_k=sum_{i=1}^kp^{-i}text{ for }k=1,2,dots$$so that: $$nu_n=sum_{i=1}^np^{-i}$$
Credit for the second solution (definitely the most elegant one) goes to @lulu.
I was on a path that was nice but not completely okay.
Someone attended me on that (thank you @saulspatz).
Then someone said to me: take the original path, but with an adaption (thank you @lulu).
answered 22 hours ago
drhab
94.2k543125
1
I did this too, but on reflection, I don't think it's right. If we toss a tail, we can't say we just need $n-1$ tails in a row. If the next toss is heads, we're back to needing $n$ tails in a row.
– saulspatz
22 hours ago
@saulspatz You are correct. I will delete this within some minutes. Thank you.
– drhab
22 hours ago
add a comment |
up vote
2
down vote
up vote
2
down vote
Let's say that we are in status $k$ if our last $k$ throws ended up in tail and the (eventual) throw before this sequence of tails is not a tail.
Let $mu_k$ denote the expectation of the number of throws needed to arrive at $n$ tails in a row if we are in status $k$.
Then $mu_n=0$ and to be found is $mu_0$.
For $k=0,1,2,dots n-1$ we have:$$mu_{k}=p(1+mu_{k+1})+q(1+mu_0)=1+pmu_{k+1}+qmu_0$$
So we have the following equalities:
$mu_0=1+pmu_1+qmu_0$ so that $mu_0=frac1p+mu_1$
$mu_1=1+pmu_2+qmu_0=1+pmu_2+frac{q}p+qmu_1$ so that $mu_1=frac1p+frac q{p^2}+mu_2=frac1{p^2}+mu_2$
This makes us suspect that $mu_k=sum_{i=k+1}^np^{-i}$ for $k=0,1,dots,n$ and substitution in $(1)$ confirms that conjecture.
The the outcome is:$$mathbb EX=sum_{i=1}^np^{-i}$$
addendum:
Another way to look at it is this:
Let $nu_k$ denote the expectation of the number of throws needed to achieve exactly $k$ tails in a row.
Then $nu_0=0$ and to be found is $nu_n$.
Now let it be that after $X$ steps we have a consecutive row of exactly $k-1$ tails. Then by throwing tails we need $X+1$ throws to get a consecutive row of exactly $k$ tails. By throwing heads we come back in start position and from there $X+1+Y$ steps will be needed to get a consecutive row of exactly $k$ tails. This with $mathbb EX=nu_{k-1}$ and $mathbb EY=nu_{k}$ hence giving the equality:$$nu_k=p(1+nu_{k-1})+q(1+nu_{k-1}+nu_k)$$or shorter:$$pnu_k=1+nu_{k-1}$$
This leads easily to:$$nu_k=sum_{i=1}^kp^{-i}text{ for }k=1,2,dots$$so that: $$nu_n=sum_{i=1}^np^{-i}$$
Credit for the second solution (definitely the most elegant one) goes to @lulu.
I was on a path that was nice but not completely okay.
Someone attended me on that (thank you @saulspatz).
Then someone said to me: take the original path, but with an adaption (thank you @lulu).
answered 22 hours ago
drhab
94.2k543125
Let's say that we are in status $k$ if our last $k$ throws ended up in tail and the (eventual) throw before this sequence of tails is not a tail.
Let $mu_k$ denote the expectation of the number of throws needed to arrive at $n$ tails in a row if we are in status $k$.
Then $mu_n=0$ and to be found is $mu_0$.
For $k=0,1,2,dots n-1$ we have:$$mu_{k}=p(1+mu_{k+1})+q(1+mu_0)=1+pmu_{k+1}+qmu_0$$
So we have the following equalities:
$mu_0=1+pmu_1+qmu_0$ so that $mu_0=frac1p+mu_1$
$mu_1=1+pmu_2+qmu_0=1+pmu_2+frac{q}p+qmu_1$ so that $mu_1=frac1p+frac q{p^2}+mu_2=frac1{p^2}+mu_2$
This makes us suspect that $mu_k=sum_{i=k+1}^np^{-i}$ for $k=0,1,dots,n$ and substitution in $(1)$ confirms that conjecture.
The the outcome is:$$mathbb EX=sum_{i=1}^np^{-i}$$
addendum:
Another way to look at it is this:
Let $nu_k$ denote the expectation of the number of throws needed to achieve exactly $k$ tails in a row.
Then $nu_0=0$ and to be found is $nu_n$.
Now let it be that after $X$ steps we have a consecutive row of exactly $k-1$ tails. Then by throwing tails we need $X+1$ throws to get a consecutive row of exactly $k$ tails. By throwing heads we come back in start position and from there $X+1+Y$ steps will be needed to get a consecutive row of exactly $k$ tails. This with $mathbb EX=nu_{k-1}$ and $mathbb EY=nu_{k}$ hence giving the equality:$$nu_k=p(1+nu_{k-1})+q(1+nu_{k-1}+nu_k)$$or shorter:$$pnu_k=1+nu_{k-1}$$
This leads easily to:$$nu_k=sum_{i=1}^kp^{-i}text{ for }k=1,2,dots$$so that: $$nu_n=sum_{i=1}^np^{-i}$$
Credit for the second solution (definitely the most elegant one) goes to @lulu.
I was on a path that was nice but not completely okay.
Someone attended me on that (thank you @saulspatz).
Then someone said to me: take the original path, but with an adaption (thank you @lulu).
answered 22 hours ago
drhab
94.2k543125
edited 19 hours ago
answered 22 hours ago
drhab
94.2k543125
answered 22 hours ago
drhab
94.2k543125
answered 22 hours ago
drhab
94.2k543125
94.2k543125
1
I did this too, but on reflection, I don't think it's right. If we toss a tail, we can't say we just need $n-1$ tails in a row. If the next toss is heads, we're back to needing $n$ tails in a row.
– saulspatz
22 hours ago
@saulspatz You are correct. I will delete this within some minutes. Thank you.
– drhab
22 hours ago
add a comment |
1
I did this too, but on reflection, I don't think it's right. If we toss a tail, we can't say we just need $n-1$ tails in a row. If the next toss is heads, we're back to needing $n$ tails in a row.
– saulspatz
22 hours ago
@saulspatz You are correct. I will delete this within some minutes. Thank you.
– drhab
22 hours ago
1
1
I did this too, but on reflection, I don't think it's right. If we toss a tail, we can't say we just need $n-1$ tails in a row. If the next toss is heads, we're back to needing $n$ tails in a row.
– saulspatz
22 hours ago
I did this too, but on reflection, I don't think it's right. If we toss a tail, we can't say we just need $n-1$ tails in a row. If the next toss is heads, we're back to needing $n$ tails in a row.
– saulspatz
22 hours ago
@saulspatz You are correct. I will delete this within some minutes. Thank you.
– drhab
22 hours ago
@saulspatz You are correct. I will delete this within some minutes. Thank you.
– drhab
22 hours ago
add a comment |
up vote
0
down vote
This is a finite-state absorbing Markov chain. The state is the number of consecutive tails we have tossed so far. We start in state $0$ and the absorbing state is $n$. If we are in state $k<n$ then we transition to state $k+1$ with probability $p$ and we transition to state $0$ with probability $q,$ so that the first column on the matrix $Q$ described in the wiki has $q$ in every position, and the entries on the superdiagonal are all $p.$
You need to figure out $(I-Q)^{-1}$
answered 21 hours ago
saulspatz
12.8k21327
add a comment |
up vote
0
down vote
This is a finite-state absorbing Markov chain. The state is the number of consecutive tails we have tossed so far. We start in state $0$ and the absorbing state is $n$. If we are in state $k<n$ then we transition to state $k+1$ with probability $p$ and we transition to state $0$ with probability $q,$ so that the first column on the matrix $Q$ described in the wiki has $q$ in every position, and the entries on the superdiagonal are all $p.$
You need to figure out $(I-Q)^{-1}$
answered 21 hours ago
saulspatz
12.8k21327
add a comment |
up vote
0
down vote
up vote
0
down vote
This is a finite-state absorbing Markov chain. The state is the number of consecutive tails we have tossed so far. We start in state $0$ and the absorbing state is $n$. If we are in state $k<n$ then we transition to state $k+1$ with probability $p$ and we transition to state $0$ with probability $q,$ so that the first column on the matrix $Q$ described in the wiki has $q$ in every position, and the entries on the superdiagonal are all $p.$
You need to figure out $(I-Q)^{-1}$
answered 21 hours ago
saulspatz
12.8k21327
This is a finite-state absorbing Markov chain. The state is the number of consecutive tails we have tossed so far. We start in state $0$ and the absorbing state is $n$. If we are in state $k<n$ then we transition to state $k+1$ with probability $p$ and we transition to state $0$ with probability $q,$ so that the first column on the matrix $Q$ described in the wiki has $q$ in every position, and the entries on the superdiagonal are all $p.$
You need to figure out $(I-Q)^{-1}$
answered 21 hours ago
saulspatz
12.8k21327
answered 21 hours ago
saulspatz
12.8k21327
answered 21 hours ago
saulspatz
12.8k21327
answered 21 hours ago
saulspatz
12.8k21327
12.8k21327
add a comment |
add a comment |
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@drhab I think your proposed solution was nearly correct. To get $k$ tails in a row I must first get $k-1$ tails in a row. So, we go out $mu_{k-1}$ turns and toss. either I win or I restart. If we restart I then expect $1+mu_{k-1}+mu_k$ tosses (including the $1+mu_{k-1}$ tosses I have already done). Thus $mu_k=p(1+mu_{k-1})+q(1+mu_{k-1}+mu_k)$. No?
– lulu
22 hours ago
@lulu That gives $mu_k=p^{-1}+mu_{k-1}$ and seems not to agree with what I found now after some struggling. On the other hand I have not the guts to say it is wrong, since it looks good. Good chance that my revised is answer is wrong also.
– drhab
20 hours ago
@drhab Does it? I get $mu_k=mu_{k-1}+1+qmu_kimplies pmu_k=mu_{k-1}+1$ Thus $mu_k=frac 1ptimes (1+mu_{k-1})$ Which seems to harmonize with your new answer.
– lulu
20 hours ago
@lulu Ah, yes. I misunderstood at first hand because in my new answer $mu_k$ is defined on a different way than in my first answer, $mu_n$ and $mu_0$ somehow "switched" of meaning.
– drhab
20 hours ago
@drhab I think your first answer (correctly modified) is the way to go. Comparatively clear.
– lulu
20 hours ago