Mixing
Random walk. Finding probabilities
$begingroup$
Suppose that $x_1, x_2,...$ are independent copies of random variable $xi$ having distribution $P(x=1)=P(x=-1)=frac{1}{2}.$ Let $S_0=0$, $S_k=x_1+x_2+...+x_k$, $k geq1,$ and $a,b$ be non negative integer numbers such that $a-b>1$. Find probabilities $$P(S_2=2, S_{a+b}=a-b), text{ } P(S_2=-2, S_{a+b}=a-b).$$
I know that $P(S_2=2)=P(x_1=1, x_2=1)=frac{1}{2}*frac{1}{2}$ and $P(S_2=-2)=P(x_1=-1, x_2=-1)=frac{1}{2}*frac{1}{2}.$ But what about integers $a$ and $b$. What should I do with them?
probability probability-theory probability-distributions random-variables random-walk
asked Jan 9 at 10:11
AtstovasAtstovas
1089
$endgroup$
add a comment |
$begingroup$
Suppose that $x_1, x_2,...$ are independent copies of random variable $xi$ having distribution $P(x=1)=P(x=-1)=frac{1}{2}.$ Let $S_0=0$, $S_k=x_1+x_2+...+x_k$, $k geq1,$ and $a,b$ be non negative integer numbers such that $a-b>1$. Find probabilities $$P(S_2=2, S_{a+b}=a-b), text{ } P(S_2=-2, S_{a+b}=a-b).$$
I know that $P(S_2=2)=P(x_1=1, x_2=1)=frac{1}{2}*frac{1}{2}$ and $P(S_2=-2)=P(x_1=-1, x_2=-1)=frac{1}{2}*frac{1}{2}.$ But what about integers $a$ and $b$. What should I do with them?
probability probability-theory probability-distributions random-variables random-walk
asked Jan 9 at 10:11
AtstovasAtstovas
1089
$endgroup$
add a comment |
$begingroup$
Suppose that $x_1, x_2,...$ are independent copies of random variable $xi$ having distribution $P(x=1)=P(x=-1)=frac{1}{2}.$ Let $S_0=0$, $S_k=x_1+x_2+...+x_k$, $k geq1,$ and $a,b$ be non negative integer numbers such that $a-b>1$. Find probabilities $$P(S_2=2, S_{a+b}=a-b), text{ } P(S_2=-2, S_{a+b}=a-b).$$
I know that $P(S_2=2)=P(x_1=1, x_2=1)=frac{1}{2}*frac{1}{2}$ and $P(S_2=-2)=P(x_1=-1, x_2=-1)=frac{1}{2}*frac{1}{2}.$ But what about integers $a$ and $b$. What should I do with them?
probability probability-theory probability-distributions random-variables random-walk
asked Jan 9 at 10:11
AtstovasAtstovas
1089
$endgroup$
Suppose that $x_1, x_2,...$ are independent copies of random variable $xi$ having distribution $P(x=1)=P(x=-1)=frac{1}{2}.$ Let $S_0=0$, $S_k=x_1+x_2+...+x_k$, $k geq1,$ and $a,b$ be non negative integer numbers such that $a-b>1$. Find probabilities $$P(S_2=2, S_{a+b}=a-b), text{ } P(S_2=-2, S_{a+b}=a-b).$$
I know that $P(S_2=2)=P(x_1=1, x_2=1)=frac{1}{2}*frac{1}{2}$ and $P(S_2=-2)=P(x_1=-1, x_2=-1)=frac{1}{2}*frac{1}{2}.$ But what about integers $a$ and $b$. What should I do with them?
probability probability-theory probability-distributions random-variables random-walk
probability probability-theory probability-distributions random-variables random-walk
asked Jan 9 at 10:11
AtstovasAtstovas
1089
asked Jan 9 at 10:11
AtstovasAtstovas
1089
asked Jan 9 at 10:11
AtstovasAtstovas
1089
asked Jan 9 at 10:11
AtstovasAtstovas
1089
asked Jan 9 at 10:11
AtstovasAtstovas
1089
1089
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can start with: $$P(S_2=2,S_{a+b}=a-b)=P(S=2)P(S_{a+b}=a-bmid S_2=2)=frac14P(S_{a+b-2}=a-b-2)$$
Now the question rises: how many ways are there from $0$ to $a-b-2$ taking $a+b-2$ steps?
If there are $r$ steps to right and $l$ to left then we have the equalities:
- $r+l=a+b-2$
- $r-l=a-b-2$
leading to $r=a-2$ and $l=b$ and showing that there are $binom{a+b-2}{b}$ routes.
I leave the rest to you.
Edit:
If e.g. $a=3$ and $b=2$ then:
$begin{aligned}Pleft(S_{a+b}=a-bmid S_{2}=2right) & =Pleft(S_{5}=1mid S_{2}=2right)\
& =Pleft(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=1mid x_{1}+x_{2}=2right)\
& =Pleft(2+x_{3}+x_{4}+x_{5}=1mid x_{1}+x_{2}=2right)\
& =Pleft(x_{3}+x_{4}+x_{5}=-1mid x_{1}+x_{2}=2right)\
& =Pleft(x_{3}+x_{4}+x_{5}=-1right)\
& =Pleft(x_{1}+x_{2}+x_{3}=-1right)\
& =Pleft(S_{3}=-1right)\
& =Pleft(S_{a+b-2}=a-b-2right)
end{aligned}
$
The $5$-th equality rests on independence.
The $6$-th equality on the fact that $x_1+x_2+x_3$ and $x_3+x_4+x_5$ have the same distribution.
answered Jan 9 at 10:26
drhabdrhab
100k544130
$endgroup$
$begingroup$
Can you explain me this $P(S_{a+b}=a-b|S_2=2)=P(S_{a+b-2}=a-b-2)$?
$endgroup$
– Atstovas
Jan 10 at 7:52
$begingroup$
I have edit an example ($a=3$ and $b=2$) to make things more clear.
$endgroup$
– drhab
Jan 10 at 9:17
$begingroup$
Now I see! and I have another question. From where I should now that there are $binom{a+b-2}{b}$ routes?
$endgroup$
– Atstovas
Jan 10 at 9:26
$begingroup$
Calculating $P(S_{a+b-2}=a-b-2)$ we must consider the question how many routes exist from $0$ to $a-b-2$ when $a+b-2$ steps are taken. In my answer I showed that there must be $b$ steps to the left. So we must select exactly $b$ of the $a+b-2$ steps to be the steps to the left. There are $binom{a+b-2}{b}$ such selections.
$endgroup$
– drhab
Jan 10 at 9:36
add a comment |
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1 Answer
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$begingroup$
You can start with: $$P(S_2=2,S_{a+b}=a-b)=P(S=2)P(S_{a+b}=a-bmid S_2=2)=frac14P(S_{a+b-2}=a-b-2)$$
Now the question rises: how many ways are there from $0$ to $a-b-2$ taking $a+b-2$ steps?
If there are $r$ steps to right and $l$ to left then we have the equalities:
- $r+l=a+b-2$
- $r-l=a-b-2$
leading to $r=a-2$ and $l=b$ and showing that there are $binom{a+b-2}{b}$ routes.
I leave the rest to you.
Edit:
If e.g. $a=3$ and $b=2$ then:
$begin{aligned}Pleft(S_{a+b}=a-bmid S_{2}=2right) & =Pleft(S_{5}=1mid S_{2}=2right)\
& =Pleft(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=1mid x_{1}+x_{2}=2right)\
& =Pleft(2+x_{3}+x_{4}+x_{5}=1mid x_{1}+x_{2}=2right)\
& =Pleft(x_{3}+x_{4}+x_{5}=-1mid x_{1}+x_{2}=2right)\
& =Pleft(x_{3}+x_{4}+x_{5}=-1right)\
& =Pleft(x_{1}+x_{2}+x_{3}=-1right)\
& =Pleft(S_{3}=-1right)\
& =Pleft(S_{a+b-2}=a-b-2right)
end{aligned}
$
The $5$-th equality rests on independence.
The $6$-th equality on the fact that $x_1+x_2+x_3$ and $x_3+x_4+x_5$ have the same distribution.
answered Jan 9 at 10:26
drhabdrhab
100k544130
$endgroup$
$begingroup$
Can you explain me this $P(S_{a+b}=a-b|S_2=2)=P(S_{a+b-2}=a-b-2)$?
$endgroup$
– Atstovas
Jan 10 at 7:52
$begingroup$
I have edit an example ($a=3$ and $b=2$) to make things more clear.
$endgroup$
– drhab
Jan 10 at 9:17
$begingroup$
Now I see! and I have another question. From where I should now that there are $binom{a+b-2}{b}$ routes?
$endgroup$
– Atstovas
Jan 10 at 9:26
$begingroup$
Calculating $P(S_{a+b-2}=a-b-2)$ we must consider the question how many routes exist from $0$ to $a-b-2$ when $a+b-2$ steps are taken. In my answer I showed that there must be $b$ steps to the left. So we must select exactly $b$ of the $a+b-2$ steps to be the steps to the left. There are $binom{a+b-2}{b}$ such selections.
$endgroup$
– drhab
Jan 10 at 9:36
add a comment |
$begingroup$
You can start with: $$P(S_2=2,S_{a+b}=a-b)=P(S=2)P(S_{a+b}=a-bmid S_2=2)=frac14P(S_{a+b-2}=a-b-2)$$
Now the question rises: how many ways are there from $0$ to $a-b-2$ taking $a+b-2$ steps?
If there are $r$ steps to right and $l$ to left then we have the equalities:
- $r+l=a+b-2$
- $r-l=a-b-2$
leading to $r=a-2$ and $l=b$ and showing that there are $binom{a+b-2}{b}$ routes.
I leave the rest to you.
Edit:
If e.g. $a=3$ and $b=2$ then:
$begin{aligned}Pleft(S_{a+b}=a-bmid S_{2}=2right) & =Pleft(S_{5}=1mid S_{2}=2right)\
& =Pleft(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=1mid x_{1}+x_{2}=2right)\
& =Pleft(2+x_{3}+x_{4}+x_{5}=1mid x_{1}+x_{2}=2right)\
& =Pleft(x_{3}+x_{4}+x_{5}=-1mid x_{1}+x_{2}=2right)\
& =Pleft(x_{3}+x_{4}+x_{5}=-1right)\
& =Pleft(x_{1}+x_{2}+x_{3}=-1right)\
& =Pleft(S_{3}=-1right)\
& =Pleft(S_{a+b-2}=a-b-2right)
end{aligned}
$
The $5$-th equality rests on independence.
The $6$-th equality on the fact that $x_1+x_2+x_3$ and $x_3+x_4+x_5$ have the same distribution.
answered Jan 9 at 10:26
drhabdrhab
100k544130
$endgroup$
$begingroup$
Can you explain me this $P(S_{a+b}=a-b|S_2=2)=P(S_{a+b-2}=a-b-2)$?
$endgroup$
– Atstovas
Jan 10 at 7:52
$begingroup$
I have edit an example ($a=3$ and $b=2$) to make things more clear.
$endgroup$
– drhab
Jan 10 at 9:17
$begingroup$
Now I see! and I have another question. From where I should now that there are $binom{a+b-2}{b}$ routes?
$endgroup$
– Atstovas
Jan 10 at 9:26
$begingroup$
Calculating $P(S_{a+b-2}=a-b-2)$ we must consider the question how many routes exist from $0$ to $a-b-2$ when $a+b-2$ steps are taken. In my answer I showed that there must be $b$ steps to the left. So we must select exactly $b$ of the $a+b-2$ steps to be the steps to the left. There are $binom{a+b-2}{b}$ such selections.
$endgroup$
– drhab
Jan 10 at 9:36
add a comment |
$begingroup$
You can start with: $$P(S_2=2,S_{a+b}=a-b)=P(S=2)P(S_{a+b}=a-bmid S_2=2)=frac14P(S_{a+b-2}=a-b-2)$$
Now the question rises: how many ways are there from $0$ to $a-b-2$ taking $a+b-2$ steps?
If there are $r$ steps to right and $l$ to left then we have the equalities:
- $r+l=a+b-2$
- $r-l=a-b-2$
leading to $r=a-2$ and $l=b$ and showing that there are $binom{a+b-2}{b}$ routes.
I leave the rest to you.
Edit:
If e.g. $a=3$ and $b=2$ then:
$begin{aligned}Pleft(S_{a+b}=a-bmid S_{2}=2right) & =Pleft(S_{5}=1mid S_{2}=2right)\
& =Pleft(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=1mid x_{1}+x_{2}=2right)\
& =Pleft(2+x_{3}+x_{4}+x_{5}=1mid x_{1}+x_{2}=2right)\
& =Pleft(x_{3}+x_{4}+x_{5}=-1mid x_{1}+x_{2}=2right)\
& =Pleft(x_{3}+x_{4}+x_{5}=-1right)\
& =Pleft(x_{1}+x_{2}+x_{3}=-1right)\
& =Pleft(S_{3}=-1right)\
& =Pleft(S_{a+b-2}=a-b-2right)
end{aligned}
$
The $5$-th equality rests on independence.
The $6$-th equality on the fact that $x_1+x_2+x_3$ and $x_3+x_4+x_5$ have the same distribution.
answered Jan 9 at 10:26
drhabdrhab
100k544130
$endgroup$
You can start with: $$P(S_2=2,S_{a+b}=a-b)=P(S=2)P(S_{a+b}=a-bmid S_2=2)=frac14P(S_{a+b-2}=a-b-2)$$
Now the question rises: how many ways are there from $0$ to $a-b-2$ taking $a+b-2$ steps?
If there are $r$ steps to right and $l$ to left then we have the equalities:
- $r+l=a+b-2$
- $r-l=a-b-2$
leading to $r=a-2$ and $l=b$ and showing that there are $binom{a+b-2}{b}$ routes.
I leave the rest to you.
Edit:
If e.g. $a=3$ and $b=2$ then:
$begin{aligned}Pleft(S_{a+b}=a-bmid S_{2}=2right) & =Pleft(S_{5}=1mid S_{2}=2right)\
& =Pleft(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=1mid x_{1}+x_{2}=2right)\
& =Pleft(2+x_{3}+x_{4}+x_{5}=1mid x_{1}+x_{2}=2right)\
& =Pleft(x_{3}+x_{4}+x_{5}=-1mid x_{1}+x_{2}=2right)\
& =Pleft(x_{3}+x_{4}+x_{5}=-1right)\
& =Pleft(x_{1}+x_{2}+x_{3}=-1right)\
& =Pleft(S_{3}=-1right)\
& =Pleft(S_{a+b-2}=a-b-2right)
end{aligned}
$
The $5$-th equality rests on independence.
The $6$-th equality on the fact that $x_1+x_2+x_3$ and $x_3+x_4+x_5$ have the same distribution.
answered Jan 9 at 10:26
drhabdrhab
100k544130
edited Jan 10 at 9:16
answered Jan 9 at 10:26
drhabdrhab
100k544130
answered Jan 9 at 10:26
drhabdrhab
100k544130
answered Jan 9 at 10:26
drhabdrhab
100k544130
100k544130
$begingroup$
Can you explain me this $P(S_{a+b}=a-b|S_2=2)=P(S_{a+b-2}=a-b-2)$?
$endgroup$
– Atstovas
Jan 10 at 7:52
$begingroup$
I have edit an example ($a=3$ and $b=2$) to make things more clear.
$endgroup$
– drhab
Jan 10 at 9:17
$begingroup$
Now I see! and I have another question. From where I should now that there are $binom{a+b-2}{b}$ routes?
$endgroup$
– Atstovas
Jan 10 at 9:26
$begingroup$
Calculating $P(S_{a+b-2}=a-b-2)$ we must consider the question how many routes exist from $0$ to $a-b-2$ when $a+b-2$ steps are taken. In my answer I showed that there must be $b$ steps to the left. So we must select exactly $b$ of the $a+b-2$ steps to be the steps to the left. There are $binom{a+b-2}{b}$ such selections.
$endgroup$
– drhab
Jan 10 at 9:36
add a comment |
$begingroup$
Can you explain me this $P(S_{a+b}=a-b|S_2=2)=P(S_{a+b-2}=a-b-2)$?
$endgroup$
– Atstovas
Jan 10 at 7:52
$begingroup$
I have edit an example ($a=3$ and $b=2$) to make things more clear.
$endgroup$
– drhab
Jan 10 at 9:17
$begingroup$
Now I see! and I have another question. From where I should now that there are $binom{a+b-2}{b}$ routes?
$endgroup$
– Atstovas
Jan 10 at 9:26
$begingroup$
Calculating $P(S_{a+b-2}=a-b-2)$ we must consider the question how many routes exist from $0$ to $a-b-2$ when $a+b-2$ steps are taken. In my answer I showed that there must be $b$ steps to the left. So we must select exactly $b$ of the $a+b-2$ steps to be the steps to the left. There are $binom{a+b-2}{b}$ such selections.
$endgroup$
– drhab
Jan 10 at 9:36
$begingroup$
Can you explain me this $P(S_{a+b}=a-b|S_2=2)=P(S_{a+b-2}=a-b-2)$?
$endgroup$
– Atstovas
Jan 10 at 7:52
$begingroup$
Can you explain me this $P(S_{a+b}=a-b|S_2=2)=P(S_{a+b-2}=a-b-2)$?
$endgroup$
– Atstovas
Jan 10 at 7:52
$begingroup$
I have edit an example ($a=3$ and $b=2$) to make things more clear.
$endgroup$
– drhab
Jan 10 at 9:17
$begingroup$
I have edit an example ($a=3$ and $b=2$) to make things more clear.
$endgroup$
– drhab
Jan 10 at 9:17
$begingroup$
Now I see! and I have another question. From where I should now that there are $binom{a+b-2}{b}$ routes?
$endgroup$
– Atstovas
Jan 10 at 9:26
$begingroup$
Now I see! and I have another question. From where I should now that there are $binom{a+b-2}{b}$ routes?
$endgroup$
– Atstovas
Jan 10 at 9:26
$begingroup$
Calculating $P(S_{a+b-2}=a-b-2)$ we must consider the question how many routes exist from $0$ to $a-b-2$ when $a+b-2$ steps are taken. In my answer I showed that there must be $b$ steps to the left. So we must select exactly $b$ of the $a+b-2$ steps to be the steps to the left. There are $binom{a+b-2}{b}$ such selections.
$endgroup$
– drhab
Jan 10 at 9:36
$begingroup$
Calculating $P(S_{a+b-2}=a-b-2)$ we must consider the question how many routes exist from $0$ to $a-b-2$ when $a+b-2$ steps are taken. In my answer I showed that there must be $b$ steps to the left. So we must select exactly $b$ of the $a+b-2$ steps to be the steps to the left. There are $binom{a+b-2}{b}$ such selections.
$endgroup$
– drhab
Jan 10 at 9:36
add a comment |
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