Random walk. Finding probabilities $P_{2k,4}$ and $P_{2k,8}$
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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 $S_0=0$, $S_k=x_1+x_2+...+x_k$, $k geq1.$ Let $P_{2k,2n}$ be the probability that during the time interval $[0,2n]$ the particle spends $2k$ units of time on the positive side.
1) Find all possible probabilities $P_{2k,4}$ using only definition.
2) Find all possible probabilities $P_{2k,8}$ using formula $$ P_{2k,2n}=u_{2k}*u_{2n-2k}.$$
So, problem is that I don't know what I have to do in 1). Maybe only then I can do also 2). So how to start and what to use in 1)?
probability probability-theory probability-distributions random-walk
asked Jan 10 at 9:16
AtstovasAtstovas
1088
$endgroup$
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 $S_0=0$, $S_k=x_1+x_2+...+x_k$, $k geq1.$ Let $P_{2k,2n}$ be the probability that during the time interval $[0,2n]$ the particle spends $2k$ units of time on the positive side.
1) Find all possible probabilities $P_{2k,4}$ using only definition.
2) Find all possible probabilities $P_{2k,8}$ using formula $$ P_{2k,2n}=u_{2k}*u_{2n-2k}.$$
So, problem is that I don't know what I have to do in 1). Maybe only then I can do also 2). So how to start and what to use in 1)?
probability probability-theory probability-distributions random-walk
asked Jan 10 at 9:16
AtstovasAtstovas
1088
$endgroup$
$begingroup$
What is denoted by $u_{2k}$?
$endgroup$
– drhab
Jan 10 at 10:03
$begingroup$
It is formula from Shiryaev's Probability book page 100 (second edition)
$endgroup$
– Atstovas
Jan 10 at 10:13
$begingroup$
That does not answer my question. I have no disposal over the book.
$endgroup$
– drhab
Jan 10 at 10:21
$begingroup$
$u_{2k}=binom{2k}{k}*2^{-2k}$
$endgroup$
– Atstovas
Jan 10 at 10:26
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 $S_0=0$, $S_k=x_1+x_2+...+x_k$, $k geq1.$ Let $P_{2k,2n}$ be the probability that during the time interval $[0,2n]$ the particle spends $2k$ units of time on the positive side.
1) Find all possible probabilities $P_{2k,4}$ using only definition.
2) Find all possible probabilities $P_{2k,8}$ using formula $$ P_{2k,2n}=u_{2k}*u_{2n-2k}.$$
So, problem is that I don't know what I have to do in 1). Maybe only then I can do also 2). So how to start and what to use in 1)?
probability probability-theory probability-distributions random-walk
asked Jan 10 at 9:16
AtstovasAtstovas
1088
$endgroup$
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 $S_0=0$, $S_k=x_1+x_2+...+x_k$, $k geq1.$ Let $P_{2k,2n}$ be the probability that during the time interval $[0,2n]$ the particle spends $2k$ units of time on the positive side.
1) Find all possible probabilities $P_{2k,4}$ using only definition.
2) Find all possible probabilities $P_{2k,8}$ using formula $$ P_{2k,2n}=u_{2k}*u_{2n-2k}.$$
So, problem is that I don't know what I have to do in 1). Maybe only then I can do also 2). So how to start and what to use in 1)?
probability probability-theory probability-distributions random-walk
probability probability-theory probability-distributions random-walk
asked Jan 10 at 9:16
AtstovasAtstovas
1088
asked Jan 10 at 9:16
AtstovasAtstovas
1088
asked Jan 10 at 9:16
AtstovasAtstovas
1088
asked Jan 10 at 9:16
AtstovasAtstovas
1088
asked Jan 10 at 9:16
AtstovasAtstovas
1088
1088
$begingroup$
What is denoted by $u_{2k}$?
$endgroup$
– drhab
Jan 10 at 10:03
$begingroup$
It is formula from Shiryaev's Probability book page 100 (second edition)
$endgroup$
– Atstovas
Jan 10 at 10:13
$begingroup$
That does not answer my question. I have no disposal over the book.
$endgroup$
– drhab
Jan 10 at 10:21
$begingroup$
$u_{2k}=binom{2k}{k}*2^{-2k}$
$endgroup$
– Atstovas
Jan 10 at 10:26
add a comment |
$begingroup$
What is denoted by $u_{2k}$?
$endgroup$
– drhab
Jan 10 at 10:03
$begingroup$
It is formula from Shiryaev's Probability book page 100 (second edition)
$endgroup$
– Atstovas
Jan 10 at 10:13
$begingroup$
That does not answer my question. I have no disposal over the book.
$endgroup$
– drhab
Jan 10 at 10:21
$begingroup$
$u_{2k}=binom{2k}{k}*2^{-2k}$
$endgroup$
– Atstovas
Jan 10 at 10:26
$begingroup$
What is denoted by $u_{2k}$?
$endgroup$
– drhab
Jan 10 at 10:03
$begingroup$
What is denoted by $u_{2k}$?
$endgroup$
– drhab
Jan 10 at 10:03
$begingroup$
It is formula from Shiryaev's Probability book page 100 (second edition)
$endgroup$
– Atstovas
Jan 10 at 10:13
$begingroup$
It is formula from Shiryaev's Probability book page 100 (second edition)
$endgroup$
– Atstovas
Jan 10 at 10:13
$begingroup$
That does not answer my question. I have no disposal over the book.
$endgroup$
– drhab
Jan 10 at 10:21
$begingroup$
That does not answer my question. I have no disposal over the book.
$endgroup$
– drhab
Jan 10 at 10:21
$begingroup$
$u_{2k}=binom{2k}{k}*2^{-2k}$
$endgroup$
– Atstovas
Jan 10 at 10:26
$begingroup$
$u_{2k}=binom{2k}{k}*2^{-2k}$
$endgroup$
– Atstovas
Jan 10 at 10:26
add a comment |
1 Answer
1
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$begingroup$
1)
In time interval $[0,4]$ there are $2^4=16$ routes that can be taken. The routes are equiprobable.
Now you must pose yourself the question: "how many of these routes are such that the particle spends $2k$ units of time on the positive axis?" This for $k=0$, $k=1$ and $k=2$. It is just a matter of counting.
If there are $m_k$ such routes then: $$P_{2k,4}=frac{m_k}{2^4}=frac{m_k}{16}$$
answered Jan 10 at 9:57
drhabdrhab
99.4k544130
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add a comment |
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1 Answer
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$begingroup$
1)
In time interval $[0,4]$ there are $2^4=16$ routes that can be taken. The routes are equiprobable.
Now you must pose yourself the question: "how many of these routes are such that the particle spends $2k$ units of time on the positive axis?" This for $k=0$, $k=1$ and $k=2$. It is just a matter of counting.
If there are $m_k$ such routes then: $$P_{2k,4}=frac{m_k}{2^4}=frac{m_k}{16}$$
answered Jan 10 at 9:57
drhabdrhab
99.4k544130
$endgroup$
add a comment |
$begingroup$
1)
In time interval $[0,4]$ there are $2^4=16$ routes that can be taken. The routes are equiprobable.
Now you must pose yourself the question: "how many of these routes are such that the particle spends $2k$ units of time on the positive axis?" This for $k=0$, $k=1$ and $k=2$. It is just a matter of counting.
If there are $m_k$ such routes then: $$P_{2k,4}=frac{m_k}{2^4}=frac{m_k}{16}$$
answered Jan 10 at 9:57
drhabdrhab
99.4k544130
$endgroup$
add a comment |
$begingroup$
1)
In time interval $[0,4]$ there are $2^4=16$ routes that can be taken. The routes are equiprobable.
Now you must pose yourself the question: "how many of these routes are such that the particle spends $2k$ units of time on the positive axis?" This for $k=0$, $k=1$ and $k=2$. It is just a matter of counting.
If there are $m_k$ such routes then: $$P_{2k,4}=frac{m_k}{2^4}=frac{m_k}{16}$$
answered Jan 10 at 9:57
drhabdrhab
99.4k544130
$endgroup$
1)
In time interval $[0,4]$ there are $2^4=16$ routes that can be taken. The routes are equiprobable.
Now you must pose yourself the question: "how many of these routes are such that the particle spends $2k$ units of time on the positive axis?" This for $k=0$, $k=1$ and $k=2$. It is just a matter of counting.
If there are $m_k$ such routes then: $$P_{2k,4}=frac{m_k}{2^4}=frac{m_k}{16}$$
answered Jan 10 at 9:57
drhabdrhab
99.4k544130
edited Jan 10 at 10:22
answered Jan 10 at 9:57
drhabdrhab
99.4k544130
answered Jan 10 at 9:57
drhabdrhab
99.4k544130
answered Jan 10 at 9:57
drhabdrhab
99.4k544130
99.4k544130
add a comment |
add a comment |
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$begingroup$
What is denoted by $u_{2k}$?
$endgroup$
– drhab
Jan 10 at 10:03
$begingroup$
It is formula from Shiryaev's Probability book page 100 (second edition)
$endgroup$
– Atstovas
Jan 10 at 10:13
$begingroup$
That does not answer my question. I have no disposal over the book.
$endgroup$
– drhab
Jan 10 at 10:21
$begingroup$
$u_{2k}=binom{2k}{k}*2^{-2k}$
$endgroup$
– Atstovas
Jan 10 at 10:26