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How to solve the probability of the binomial distribution sequence below?
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Let $x_0$, $x_1$...$x_n$ be a sequence of independent random variables. $x_i = 1$ has probability $p$ and $x_i = 0$ has probability $1-p$. Let $k$ be the smallest integer such that $x_k = x_k+1$. Find probability that $x_k = 1$.
Example: $1,0,0,1,1,cdotsquad k = 1$ and $x_k = 0$.
My idea is that the probability of $x_k = 1$ is equal to the probability that no consecutive zero occurs before, which is $1-x_k = 0$ and then this question becomes the same question.
probability random-variables binomial-distribution
edited Aug 6 at 8:08
TheSimpliFire
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asked Aug 6 at 7:59
neverflow
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Let $x_0$, $x_1$...$x_n$ be a sequence of independent random variables. $x_i = 1$ has probability $p$ and $x_i = 0$ has probability $1-p$. Let $k$ be the smallest integer such that $x_k = x_k+1$. Find probability that $x_k = 1$.
Example: $1,0,0,1,1,cdotsquad k = 1$ and $x_k = 0$.
My idea is that the probability of $x_k = 1$ is equal to the probability that no consecutive zero occurs before, which is $1-x_k = 0$ and then this question becomes the same question.
probability random-variables binomial-distribution
edited Aug 6 at 8:08
TheSimpliFire
9,69261952
asked Aug 6 at 7:59
neverflow
12
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
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Aug 6 at 8:01
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up vote
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up vote
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down vote
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Let $x_0$, $x_1$...$x_n$ be a sequence of independent random variables. $x_i = 1$ has probability $p$ and $x_i = 0$ has probability $1-p$. Let $k$ be the smallest integer such that $x_k = x_k+1$. Find probability that $x_k = 1$.
Example: $1,0,0,1,1,cdotsquad k = 1$ and $x_k = 0$.
My idea is that the probability of $x_k = 1$ is equal to the probability that no consecutive zero occurs before, which is $1-x_k = 0$ and then this question becomes the same question.
probability random-variables binomial-distribution
edited Aug 6 at 8:08
TheSimpliFire
9,69261952
asked Aug 6 at 7:59
neverflow
12
Let $x_0$, $x_1$...$x_n$ be a sequence of independent random variables. $x_i = 1$ has probability $p$ and $x_i = 0$ has probability $1-p$. Let $k$ be the smallest integer such that $x_k = x_k+1$. Find probability that $x_k = 1$.
Example: $1,0,0,1,1,cdotsquad k = 1$ and $x_k = 0$.
My idea is that the probability of $x_k = 1$ is equal to the probability that no consecutive zero occurs before, which is $1-x_k = 0$ and then this question becomes the same question.
probability random-variables binomial-distribution
edited Aug 6 at 8:08
TheSimpliFire
9,69261952
asked Aug 6 at 7:59
neverflow
12
edited Aug 6 at 8:08
TheSimpliFire
9,69261952
edited Aug 6 at 8:08
TheSimpliFire
9,69261952
edited Aug 6 at 8:08
TheSimpliFire
9,69261952
9,69261952
asked Aug 6 at 7:59
neverflow
12
asked Aug 6 at 7:59
neverflow
12
asked Aug 6 at 7:59
neverflow
12
12
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 8:01
add a comment |Â
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 8:01
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 8:01
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 8:01
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2 Answers
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Denote $r:=P(x_k=1)$ and $r_0:=P(x_k=1mid x_0=0)$ and $r_1:=P(x_k=1mid x_0=1)$.
Then we have the equalities:
- $r=pr_1+(1-p)r_0$
- $r_0=pr_1$
- $r_1=p+(1-p)r_0$
Do you see why, and can you take it from here?
edit
$beginarrayccccc
& & 0\
& 1-pnearrow & & nwarrow\
textstart & & pdownarrow & & uparrow1-p\
& psearrow & & nearrow\
& & 1\
& & & psearrow\
& & & & x_k=1
endarray$
answered Aug 6 at 8:14
drhab
86.8k541118
Thank you! I have questions on the equalities, I think r = r1+r0 and I couldn't see why r0=pr1, could you explain it more? Thank you in advance.
– neverflow
Aug 6 at 8:33
I added a (clumsy) picture of the situation in the hope that things become more clear. If $x_0$ has taken its value then we land in "status" $0$ or "status"1. If we are in status $0$ then we can only end up with $x_k=1$ if the next $x_i$ gives a $1$ and secondly from status $1$ we go on to status $x_k=1$. That gives $r_0=pr_1$.
– drhab
Aug 6 at 8:57
$rneq r_1+r_0$ (as you think) but $r=r_1p_1+r_0(1-p)$. In general we do not have $P(A)=P(Amid B)+P(Amid B^complement)$ but $P(A)=P(Amid B)P(B)+P(Amid B^complement)P(B^complement)$
– drhab
Aug 6 at 9:13
Thanks for the picture and explanation! I think I get it!
– neverflow
Aug 6 at 12:28
add a comment |Â
up vote
1
down vote
For $x_k=1$, the first $k-1$ samples must either be: an alternating sequence of $1,0$ (and $k$ is odd) or $0$ followed by an alternating sequence of $1,0$ (and $k$ is even); and of course samples $x_k, x_k+1$ must then be both $1$.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Denote $r:=P(x_k=1)$ and $r_0:=P(x_k=1mid x_0=0)$ and $r_1:=P(x_k=1mid x_0=1)$.
Then we have the equalities:
- $r=pr_1+(1-p)r_0$
- $r_0=pr_1$
- $r_1=p+(1-p)r_0$
Do you see why, and can you take it from here?
edit
$beginarrayccccc
& & 0\
& 1-pnearrow & & nwarrow\
textstart & & pdownarrow & & uparrow1-p\
& psearrow & & nearrow\
& & 1\
& & & psearrow\
& & & & x_k=1
endarray$
answered Aug 6 at 8:14
drhab
86.8k541118
Thank you! I have questions on the equalities, I think r = r1+r0 and I couldn't see why r0=pr1, could you explain it more? Thank you in advance.
– neverflow
Aug 6 at 8:33
I added a (clumsy) picture of the situation in the hope that things become more clear. If $x_0$ has taken its value then we land in "status" $0$ or "status"1. If we are in status $0$ then we can only end up with $x_k=1$ if the next $x_i$ gives a $1$ and secondly from status $1$ we go on to status $x_k=1$. That gives $r_0=pr_1$.
– drhab
Aug 6 at 8:57
$rneq r_1+r_0$ (as you think) but $r=r_1p_1+r_0(1-p)$. In general we do not have $P(A)=P(Amid B)+P(Amid B^complement)$ but $P(A)=P(Amid B)P(B)+P(Amid B^complement)P(B^complement)$
– drhab
Aug 6 at 9:13
Thanks for the picture and explanation! I think I get it!
– neverflow
Aug 6 at 12:28
add a comment |Â
up vote
2
down vote
Denote $r:=P(x_k=1)$ and $r_0:=P(x_k=1mid x_0=0)$ and $r_1:=P(x_k=1mid x_0=1)$.
Then we have the equalities:
- $r=pr_1+(1-p)r_0$
- $r_0=pr_1$
- $r_1=p+(1-p)r_0$
Do you see why, and can you take it from here?
edit
$beginarrayccccc
& & 0\
& 1-pnearrow & & nwarrow\
textstart & & pdownarrow & & uparrow1-p\
& psearrow & & nearrow\
& & 1\
& & & psearrow\
& & & & x_k=1
endarray$
answered Aug 6 at 8:14
drhab
86.8k541118
Thank you! I have questions on the equalities, I think r = r1+r0 and I couldn't see why r0=pr1, could you explain it more? Thank you in advance.
– neverflow
Aug 6 at 8:33
I added a (clumsy) picture of the situation in the hope that things become more clear. If $x_0$ has taken its value then we land in "status" $0$ or "status"1. If we are in status $0$ then we can only end up with $x_k=1$ if the next $x_i$ gives a $1$ and secondly from status $1$ we go on to status $x_k=1$. That gives $r_0=pr_1$.
– drhab
Aug 6 at 8:57
$rneq r_1+r_0$ (as you think) but $r=r_1p_1+r_0(1-p)$. In general we do not have $P(A)=P(Amid B)+P(Amid B^complement)$ but $P(A)=P(Amid B)P(B)+P(Amid B^complement)P(B^complement)$
– drhab
Aug 6 at 9:13
Thanks for the picture and explanation! I think I get it!
– neverflow
Aug 6 at 12:28
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Denote $r:=P(x_k=1)$ and $r_0:=P(x_k=1mid x_0=0)$ and $r_1:=P(x_k=1mid x_0=1)$.
Then we have the equalities:
- $r=pr_1+(1-p)r_0$
- $r_0=pr_1$
- $r_1=p+(1-p)r_0$
Do you see why, and can you take it from here?
edit
$beginarrayccccc
& & 0\
& 1-pnearrow & & nwarrow\
textstart & & pdownarrow & & uparrow1-p\
& psearrow & & nearrow\
& & 1\
& & & psearrow\
& & & & x_k=1
endarray$
answered Aug 6 at 8:14
drhab
86.8k541118
Denote $r:=P(x_k=1)$ and $r_0:=P(x_k=1mid x_0=0)$ and $r_1:=P(x_k=1mid x_0=1)$.
Then we have the equalities:
- $r=pr_1+(1-p)r_0$
- $r_0=pr_1$
- $r_1=p+(1-p)r_0$
Do you see why, and can you take it from here?
edit
$beginarrayccccc
& & 0\
& 1-pnearrow & & nwarrow\
textstart & & pdownarrow & & uparrow1-p\
& psearrow & & nearrow\
& & 1\
& & & psearrow\
& & & & x_k=1
endarray$
answered Aug 6 at 8:14
drhab
86.8k541118
edited Aug 6 at 8:58
answered Aug 6 at 8:14
drhab
86.8k541118
answered Aug 6 at 8:14
drhab
86.8k541118
answered Aug 6 at 8:14
drhab
86.8k541118
86.8k541118
Thank you! I have questions on the equalities, I think r = r1+r0 and I couldn't see why r0=pr1, could you explain it more? Thank you in advance.
– neverflow
Aug 6 at 8:33
I added a (clumsy) picture of the situation in the hope that things become more clear. If $x_0$ has taken its value then we land in "status" $0$ or "status"1. If we are in status $0$ then we can only end up with $x_k=1$ if the next $x_i$ gives a $1$ and secondly from status $1$ we go on to status $x_k=1$. That gives $r_0=pr_1$.
– drhab
Aug 6 at 8:57
$rneq r_1+r_0$ (as you think) but $r=r_1p_1+r_0(1-p)$. In general we do not have $P(A)=P(Amid B)+P(Amid B^complement)$ but $P(A)=P(Amid B)P(B)+P(Amid B^complement)P(B^complement)$
– drhab
Aug 6 at 9:13
Thanks for the picture and explanation! I think I get it!
– neverflow
Aug 6 at 12:28
add a comment |Â
Thank you! I have questions on the equalities, I think r = r1+r0 and I couldn't see why r0=pr1, could you explain it more? Thank you in advance.
– neverflow
Aug 6 at 8:33
I added a (clumsy) picture of the situation in the hope that things become more clear. If $x_0$ has taken its value then we land in "status" $0$ or "status"1. If we are in status $0$ then we can only end up with $x_k=1$ if the next $x_i$ gives a $1$ and secondly from status $1$ we go on to status $x_k=1$. That gives $r_0=pr_1$.
– drhab
Aug 6 at 8:57
$rneq r_1+r_0$ (as you think) but $r=r_1p_1+r_0(1-p)$. In general we do not have $P(A)=P(Amid B)+P(Amid B^complement)$ but $P(A)=P(Amid B)P(B)+P(Amid B^complement)P(B^complement)$
– drhab
Aug 6 at 9:13
Thanks for the picture and explanation! I think I get it!
– neverflow
Aug 6 at 12:28
Thank you! I have questions on the equalities, I think r = r1+r0 and I couldn't see why r0=pr1, could you explain it more? Thank you in advance.
– neverflow
Aug 6 at 8:33
Thank you! I have questions on the equalities, I think r = r1+r0 and I couldn't see why r0=pr1, could you explain it more? Thank you in advance.
– neverflow
Aug 6 at 8:33
I added a (clumsy) picture of the situation in the hope that things become more clear. If $x_0$ has taken its value then we land in "status" $0$ or "status"1. If we are in status $0$ then we can only end up with $x_k=1$ if the next $x_i$ gives a $1$ and secondly from status $1$ we go on to status $x_k=1$. That gives $r_0=pr_1$.
– drhab
Aug 6 at 8:57
I added a (clumsy) picture of the situation in the hope that things become more clear. If $x_0$ has taken its value then we land in "status" $0$ or "status"1. If we are in status $0$ then we can only end up with $x_k=1$ if the next $x_i$ gives a $1$ and secondly from status $1$ we go on to status $x_k=1$. That gives $r_0=pr_1$.
– drhab
Aug 6 at 8:57
$rneq r_1+r_0$ (as you think) but $r=r_1p_1+r_0(1-p)$. In general we do not have $P(A)=P(Amid B)+P(Amid B^complement)$ but $P(A)=P(Amid B)P(B)+P(Amid B^complement)P(B^complement)$
– drhab
Aug 6 at 9:13
$rneq r_1+r_0$ (as you think) but $r=r_1p_1+r_0(1-p)$. In general we do not have $P(A)=P(Amid B)+P(Amid B^complement)$ but $P(A)=P(Amid B)P(B)+P(Amid B^complement)P(B^complement)$
– drhab
Aug 6 at 9:13
Thanks for the picture and explanation! I think I get it!
– neverflow
Aug 6 at 12:28
Thanks for the picture and explanation! I think I get it!
– neverflow
Aug 6 at 12:28
add a comment |Â
up vote
1
down vote
For $x_k=1$, the first $k-1$ samples must either be: an alternating sequence of $1,0$ (and $k$ is odd) or $0$ followed by an alternating sequence of $1,0$ (and $k$ is even); and of course samples $x_k, x_k+1$ must then be both $1$.
add a comment |Â
up vote
1
down vote
For $x_k=1$, the first $k-1$ samples must either be: an alternating sequence of $1,0$ (and $k$ is odd) or $0$ followed by an alternating sequence of $1,0$ (and $k$ is even); and of course samples $x_k, x_k+1$ must then be both $1$.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For $x_k=1$, the first $k-1$ samples must either be: an alternating sequence of $1,0$ (and $k$ is odd) or $0$ followed by an alternating sequence of $1,0$ (and $k$ is even); and of course samples $x_k, x_k+1$ must then be both $1$.
For $x_k=1$, the first $k-1$ samples must either be: an alternating sequence of $1,0$ (and $k$ is odd) or $0$ followed by an alternating sequence of $1,0$ (and $k$ is even); and of course samples $x_k, x_k+1$ must then be both $1$.
answered Aug 6 at 8:17
community wiki
Graham Kemp
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 6 at 8:01