Number of k consecutive 1's subsequences in a binary string.
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Say a sequence ${X_1, X_2,ldots ,X_n}$ is given, where $X_p$ is either one or zero ($0 < p < n$) with
probability $frac{1}{2}$ each. Let $N_k$ be the number of consecutive 1's subsequences of length k in ${X_1, X_2,ldots ,X_n}$,what is $Bbb E[N_k]$?
Example: in the sequence $1110011$ we have $N_2=3.$
probability expected-value
asked Dec 19 '18 at 12:10
AdddisonAdddison
846
$endgroup$
add a comment |
$begingroup$
Say a sequence ${X_1, X_2,ldots ,X_n}$ is given, where $X_p$ is either one or zero ($0 < p < n$) with
probability $frac{1}{2}$ each. Let $N_k$ be the number of consecutive 1's subsequences of length k in ${X_1, X_2,ldots ,X_n}$,what is $Bbb E[N_k]$?
Example: in the sequence $1110011$ we have $N_2=3.$
probability expected-value
asked Dec 19 '18 at 12:10
AdddisonAdddison
846
$endgroup$
$begingroup$
Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
$endgroup$
– lulu
Dec 19 '18 at 12:13
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If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
$endgroup$
– player100
Dec 19 '18 at 12:24
$begingroup$
It does not have to be exactly k,I added an example for clearance.@player100
$endgroup$
– Adddison
Dec 19 '18 at 12:29
add a comment |
$begingroup$
Say a sequence ${X_1, X_2,ldots ,X_n}$ is given, where $X_p$ is either one or zero ($0 < p < n$) with
probability $frac{1}{2}$ each. Let $N_k$ be the number of consecutive 1's subsequences of length k in ${X_1, X_2,ldots ,X_n}$,what is $Bbb E[N_k]$?
Example: in the sequence $1110011$ we have $N_2=3.$
probability expected-value
asked Dec 19 '18 at 12:10
AdddisonAdddison
846
$endgroup$
Say a sequence ${X_1, X_2,ldots ,X_n}$ is given, where $X_p$ is either one or zero ($0 < p < n$) with
probability $frac{1}{2}$ each. Let $N_k$ be the number of consecutive 1's subsequences of length k in ${X_1, X_2,ldots ,X_n}$,what is $Bbb E[N_k]$?
Example: in the sequence $1110011$ we have $N_2=3.$
probability expected-value
probability expected-value
asked Dec 19 '18 at 12:10
AdddisonAdddison
846
asked Dec 19 '18 at 12:10
AdddisonAdddison
846
edited Dec 19 '18 at 12:28
Adddison
asked Dec 19 '18 at 12:10
AdddisonAdddison
846
asked Dec 19 '18 at 12:10
AdddisonAdddison
846
asked Dec 19 '18 at 12:10
AdddisonAdddison
846
846
$begingroup$
Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
$endgroup$
– lulu
Dec 19 '18 at 12:13
$begingroup$
If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
$endgroup$
– player100
Dec 19 '18 at 12:24
$begingroup$
It does not have to be exactly k,I added an example for clearance.@player100
$endgroup$
– Adddison
Dec 19 '18 at 12:29
add a comment |
$begingroup$
Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
$endgroup$
– lulu
Dec 19 '18 at 12:13
$begingroup$
If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
$endgroup$
– player100
Dec 19 '18 at 12:24
$begingroup$
It does not have to be exactly k,I added an example for clearance.@player100
$endgroup$
– Adddison
Dec 19 '18 at 12:29
$begingroup$
Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
$endgroup$
– lulu
Dec 19 '18 at 12:13
$begingroup$
Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
$endgroup$
– lulu
Dec 19 '18 at 12:13
$begingroup$
If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
$endgroup$
– player100
Dec 19 '18 at 12:24
$begingroup$
If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
$endgroup$
– player100
Dec 19 '18 at 12:24
$begingroup$
It does not have to be exactly k,I added an example for clearance.@player100
$endgroup$
– Adddison
Dec 19 '18 at 12:29
$begingroup$
It does not have to be exactly k,I added an example for clearance.@player100
$endgroup$
– Adddison
Dec 19 '18 at 12:29
add a comment |
1 Answer
1
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$N_k=sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1}.$ Hence $E(N_k)=E(sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1})=(n-k+1)E(X_1,X_2ldots X_k)=frac{n-k+1}{2^k}$ if we have independence among $X_i$'s.
answered Dec 19 '18 at 13:24
John_WickJohn_Wick
1,596111
$endgroup$
$begingroup$
add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
$endgroup$
– Adddison
Dec 19 '18 at 16:21
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$N_k=sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1}.$ Hence $E(N_k)=E(sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1})=(n-k+1)E(X_1,X_2ldots X_k)=frac{n-k+1}{2^k}$ if we have independence among $X_i$'s.
answered Dec 19 '18 at 13:24
John_WickJohn_Wick
1,596111
$endgroup$
$begingroup$
add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
$endgroup$
– Adddison
Dec 19 '18 at 16:21
add a comment |
$begingroup$
$N_k=sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1}.$ Hence $E(N_k)=E(sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1})=(n-k+1)E(X_1,X_2ldots X_k)=frac{n-k+1}{2^k}$ if we have independence among $X_i$'s.
answered Dec 19 '18 at 13:24
John_WickJohn_Wick
1,596111
$endgroup$
$begingroup$
add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
$endgroup$
– Adddison
Dec 19 '18 at 16:21
add a comment |
$begingroup$
$N_k=sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1}.$ Hence $E(N_k)=E(sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1})=(n-k+1)E(X_1,X_2ldots X_k)=frac{n-k+1}{2^k}$ if we have independence among $X_i$'s.
answered Dec 19 '18 at 13:24
John_WickJohn_Wick
1,596111
$endgroup$
$N_k=sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1}.$ Hence $E(N_k)=E(sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1})=(n-k+1)E(X_1,X_2ldots X_k)=frac{n-k+1}{2^k}$ if we have independence among $X_i$'s.
answered Dec 19 '18 at 13:24
John_WickJohn_Wick
1,596111
answered Dec 19 '18 at 13:24
John_WickJohn_Wick
1,596111
answered Dec 19 '18 at 13:24
John_WickJohn_Wick
1,596111
answered Dec 19 '18 at 13:24
John_WickJohn_Wick
1,596111
1,596111
$begingroup$
add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
$endgroup$
– Adddison
Dec 19 '18 at 16:21
add a comment |
$begingroup$
add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
$endgroup$
– Adddison
Dec 19 '18 at 16:21
$begingroup$
add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
$endgroup$
– Adddison
Dec 19 '18 at 16:21
$begingroup$
add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
$endgroup$
– Adddison
Dec 19 '18 at 16:21
add a comment |
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$begingroup$
Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
$endgroup$
– lulu
Dec 19 '18 at 12:13
$begingroup$
If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
$endgroup$
– player100
Dec 19 '18 at 12:24
$begingroup$
It does not have to be exactly k,I added an example for clearance.@player100
$endgroup$
– Adddison
Dec 19 '18 at 12:29