Mixing
Expected number of transitions in a binary sequence
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Consider a randomly generated (iid) binary sequence of length $N$. What is the expected number of transitions in that sequence?
A transition occurs when a bit changes from $0$ to $1$ or from $1$ to $0$.
For example, consider the following binary sequence of length $5$.
$$10010$$
There are 3 transitions in the above sequence.
probability combinatorics markov-chains
asked Apr 22 '17 at 1:45
JustinJustin
1747
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add a comment |
$begingroup$
Consider a randomly generated (iid) binary sequence of length $N$. What is the expected number of transitions in that sequence?
A transition occurs when a bit changes from $0$ to $1$ or from $1$ to $0$.
For example, consider the following binary sequence of length $5$.
$$10010$$
There are 3 transitions in the above sequence.
probability combinatorics markov-chains
asked Apr 22 '17 at 1:45
JustinJustin
1747
$endgroup$
$begingroup$
If you're assuming that each bit is $0$ or $1$ with probability $1/2$ each, this needs to be stated explicitly.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:48
$begingroup$
If that's the case, then you can proceed by generating a $N-1$ bit sequence, in which the $k$th bit is $0$ if the $k$th and $k+1$th bits of the original sequence are the same, and $1$ if they're different. Show that this derived sequence also consists of bits that are independently $0$ or $1$ with probability $1/2$ each, and figure out the expected number of $1$'s in the derived sequence.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:50
add a comment |
$begingroup$
Consider a randomly generated (iid) binary sequence of length $N$. What is the expected number of transitions in that sequence?
A transition occurs when a bit changes from $0$ to $1$ or from $1$ to $0$.
For example, consider the following binary sequence of length $5$.
$$10010$$
There are 3 transitions in the above sequence.
probability combinatorics markov-chains
asked Apr 22 '17 at 1:45
JustinJustin
1747
$endgroup$
Consider a randomly generated (iid) binary sequence of length $N$. What is the expected number of transitions in that sequence?
A transition occurs when a bit changes from $0$ to $1$ or from $1$ to $0$.
For example, consider the following binary sequence of length $5$.
$$10010$$
There are 3 transitions in the above sequence.
probability combinatorics markov-chains
probability combinatorics markov-chains
asked Apr 22 '17 at 1:45
JustinJustin
1747
asked Apr 22 '17 at 1:45
JustinJustin
1747
asked Apr 22 '17 at 1:45
JustinJustin
1747
asked Apr 22 '17 at 1:45
JustinJustin
1747
asked Apr 22 '17 at 1:45
JustinJustin
1747
1747
$begingroup$
If you're assuming that each bit is $0$ or $1$ with probability $1/2$ each, this needs to be stated explicitly.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:48
$begingroup$
If that's the case, then you can proceed by generating a $N-1$ bit sequence, in which the $k$th bit is $0$ if the $k$th and $k+1$th bits of the original sequence are the same, and $1$ if they're different. Show that this derived sequence also consists of bits that are independently $0$ or $1$ with probability $1/2$ each, and figure out the expected number of $1$'s in the derived sequence.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:50
add a comment |
$begingroup$
If you're assuming that each bit is $0$ or $1$ with probability $1/2$ each, this needs to be stated explicitly.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:48
$begingroup$
If that's the case, then you can proceed by generating a $N-1$ bit sequence, in which the $k$th bit is $0$ if the $k$th and $k+1$th bits of the original sequence are the same, and $1$ if they're different. Show that this derived sequence also consists of bits that are independently $0$ or $1$ with probability $1/2$ each, and figure out the expected number of $1$'s in the derived sequence.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:50
$begingroup$
If you're assuming that each bit is $0$ or $1$ with probability $1/2$ each, this needs to be stated explicitly.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:48
$begingroup$
If you're assuming that each bit is $0$ or $1$ with probability $1/2$ each, this needs to be stated explicitly.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:48
$begingroup$
If that's the case, then you can proceed by generating a $N-1$ bit sequence, in which the $k$th bit is $0$ if the $k$th and $k+1$th bits of the original sequence are the same, and $1$ if they're different. Show that this derived sequence also consists of bits that are independently $0$ or $1$ with probability $1/2$ each, and figure out the expected number of $1$'s in the derived sequence.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:50
$begingroup$
If that's the case, then you can proceed by generating a $N-1$ bit sequence, in which the $k$th bit is $0$ if the $k$th and $k+1$th bits of the original sequence are the same, and $1$ if they're different. Show that this derived sequence also consists of bits that are independently $0$ or $1$ with probability $1/2$ each, and figure out the expected number of $1$'s in the derived sequence.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:50
add a comment |
2 Answers
2
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There are $N-1$ adjacent pairs of bits, and each pair is a transition with probability $frac12$ (or, in general, with probability $2p(1-p)$, where $p$ is the probability of a $1$), so the expected number of transitions is $frac{N-1}{2}$.
In the case where $p = frac12$, the distribution is binomial; in general, it's not, because adjacent transitions are not independent. (Imagine $p=0.01$; then any transition from $0$ to $1$ is overwhelmingly likely to be followed by a transition back to $0$.)
answered Apr 22 '17 at 1:49
Misha LavrovMisha Lavrov
47.9k657107
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add a comment |
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Assuming the probability of 0 and 1 is equal, and the sequence is memory-less, then the average number of transitions will be (n-1)/2. For each transition for one bit to the next the probability of the next bit being different from the current bit is 1/2, and therefore it counts as 1/2 the transition. For n bits you have (n-1) possible transitions, with a probability of 1/2 for each.
answered Jan 27 at 8:03
Yishai KaganYishai Kagan
111
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
There are $N-1$ adjacent pairs of bits, and each pair is a transition with probability $frac12$ (or, in general, with probability $2p(1-p)$, where $p$ is the probability of a $1$), so the expected number of transitions is $frac{N-1}{2}$.
In the case where $p = frac12$, the distribution is binomial; in general, it's not, because adjacent transitions are not independent. (Imagine $p=0.01$; then any transition from $0$ to $1$ is overwhelmingly likely to be followed by a transition back to $0$.)
answered Apr 22 '17 at 1:49
Misha LavrovMisha Lavrov
47.9k657107
$endgroup$
add a comment |
$begingroup$
There are $N-1$ adjacent pairs of bits, and each pair is a transition with probability $frac12$ (or, in general, with probability $2p(1-p)$, where $p$ is the probability of a $1$), so the expected number of transitions is $frac{N-1}{2}$.
In the case where $p = frac12$, the distribution is binomial; in general, it's not, because adjacent transitions are not independent. (Imagine $p=0.01$; then any transition from $0$ to $1$ is overwhelmingly likely to be followed by a transition back to $0$.)
answered Apr 22 '17 at 1:49
Misha LavrovMisha Lavrov
47.9k657107
$endgroup$
add a comment |
$begingroup$
There are $N-1$ adjacent pairs of bits, and each pair is a transition with probability $frac12$ (or, in general, with probability $2p(1-p)$, where $p$ is the probability of a $1$), so the expected number of transitions is $frac{N-1}{2}$.
In the case where $p = frac12$, the distribution is binomial; in general, it's not, because adjacent transitions are not independent. (Imagine $p=0.01$; then any transition from $0$ to $1$ is overwhelmingly likely to be followed by a transition back to $0$.)
answered Apr 22 '17 at 1:49
Misha LavrovMisha Lavrov
47.9k657107
$endgroup$
There are $N-1$ adjacent pairs of bits, and each pair is a transition with probability $frac12$ (or, in general, with probability $2p(1-p)$, where $p$ is the probability of a $1$), so the expected number of transitions is $frac{N-1}{2}$.
In the case where $p = frac12$, the distribution is binomial; in general, it's not, because adjacent transitions are not independent. (Imagine $p=0.01$; then any transition from $0$ to $1$ is overwhelmingly likely to be followed by a transition back to $0$.)
answered Apr 22 '17 at 1:49
Misha LavrovMisha Lavrov
47.9k657107
edited Apr 22 '17 at 2:04
answered Apr 22 '17 at 1:49
Misha LavrovMisha Lavrov
47.9k657107
answered Apr 22 '17 at 1:49
Misha LavrovMisha Lavrov
47.9k657107
answered Apr 22 '17 at 1:49
Misha LavrovMisha Lavrov
47.9k657107
47.9k657107
add a comment |
add a comment |
$begingroup$
Assuming the probability of 0 and 1 is equal, and the sequence is memory-less, then the average number of transitions will be (n-1)/2. For each transition for one bit to the next the probability of the next bit being different from the current bit is 1/2, and therefore it counts as 1/2 the transition. For n bits you have (n-1) possible transitions, with a probability of 1/2 for each.
answered Jan 27 at 8:03
Yishai KaganYishai Kagan
111
$endgroup$
add a comment |
$begingroup$
Assuming the probability of 0 and 1 is equal, and the sequence is memory-less, then the average number of transitions will be (n-1)/2. For each transition for one bit to the next the probability of the next bit being different from the current bit is 1/2, and therefore it counts as 1/2 the transition. For n bits you have (n-1) possible transitions, with a probability of 1/2 for each.
answered Jan 27 at 8:03
Yishai KaganYishai Kagan
111
$endgroup$
add a comment |
$begingroup$
Assuming the probability of 0 and 1 is equal, and the sequence is memory-less, then the average number of transitions will be (n-1)/2. For each transition for one bit to the next the probability of the next bit being different from the current bit is 1/2, and therefore it counts as 1/2 the transition. For n bits you have (n-1) possible transitions, with a probability of 1/2 for each.
answered Jan 27 at 8:03
Yishai KaganYishai Kagan
111
$endgroup$
Assuming the probability of 0 and 1 is equal, and the sequence is memory-less, then the average number of transitions will be (n-1)/2. For each transition for one bit to the next the probability of the next bit being different from the current bit is 1/2, and therefore it counts as 1/2 the transition. For n bits you have (n-1) possible transitions, with a probability of 1/2 for each.
answered Jan 27 at 8:03
Yishai KaganYishai Kagan
111
answered Jan 27 at 8:03
Yishai KaganYishai Kagan
111
answered Jan 27 at 8:03
Yishai KaganYishai Kagan
111
answered Jan 27 at 8:03
Yishai KaganYishai Kagan
111
111
add a comment |
add a comment |
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$begingroup$
If you're assuming that each bit is $0$ or $1$ with probability $1/2$ each, this needs to be stated explicitly.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:48
$begingroup$
If that's the case, then you can proceed by generating a $N-1$ bit sequence, in which the $k$th bit is $0$ if the $k$th and $k+1$th bits of the original sequence are the same, and $1$ if they're different. Show that this derived sequence also consists of bits that are independently $0$ or $1$ with probability $1/2$ each, and figure out the expected number of $1$'s in the derived sequence.
$endgroup$
– Brian Tung
Apr 22 '17 at 1:50