Mixing
Conditional probability and recursion [closed]
$begingroup$
There are n unstable molecules in a row, $m_1,m_2,...,m_n$. One of the $n− 1$ pairs of neighbours, chosen at random, combines to form a stable dimer; this process continues until there remain $U_n$ isolated molecules.
a) Show that the probability that $m_1$ remains isolated is
$sum_{k=0}^{n-1} frac{(−1)^{k}}{k!}$
conditional-probability
asked Jan 3 at 12:35
phw.phw.
472
$endgroup$
closed as off-topic by Did, amWhy, Davide Giraudo, KReiser, Shailesh Jan 4 at 0:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, amWhy, Davide Giraudo, KReiser, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
There are n unstable molecules in a row, $m_1,m_2,...,m_n$. One of the $n− 1$ pairs of neighbours, chosen at random, combines to form a stable dimer; this process continues until there remain $U_n$ isolated molecules.
a) Show that the probability that $m_1$ remains isolated is
$sum_{k=0}^{n-1} frac{(−1)^{k}}{k!}$
conditional-probability
asked Jan 3 at 12:35
phw.phw.
472
$endgroup$
closed as off-topic by Did, amWhy, Davide Giraudo, KReiser, Shailesh Jan 4 at 0:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, amWhy, Davide Giraudo, KReiser, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
There are n unstable molecules in a row, $m_1,m_2,...,m_n$. One of the $n− 1$ pairs of neighbours, chosen at random, combines to form a stable dimer; this process continues until there remain $U_n$ isolated molecules.
a) Show that the probability that $m_1$ remains isolated is
$sum_{k=0}^{n-1} frac{(−1)^{k}}{k!}$
conditional-probability
asked Jan 3 at 12:35
phw.phw.
472
$endgroup$
There are n unstable molecules in a row, $m_1,m_2,...,m_n$. One of the $n− 1$ pairs of neighbours, chosen at random, combines to form a stable dimer; this process continues until there remain $U_n$ isolated molecules.
a) Show that the probability that $m_1$ remains isolated is
$sum_{k=0}^{n-1} frac{(−1)^{k}}{k!}$
conditional-probability
conditional-probability
asked Jan 3 at 12:35
phw.phw.
472
asked Jan 3 at 12:35
phw.phw.
472
asked Jan 3 at 12:35
phw.phw.
472
asked Jan 3 at 12:35
phw.phw.
472
asked Jan 3 at 12:35
phw.phw.
472
472
closed as off-topic by Did, amWhy, Davide Giraudo, KReiser, Shailesh Jan 4 at 0:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, amWhy, Davide Giraudo, KReiser, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, amWhy, Davide Giraudo, KReiser, Shailesh Jan 4 at 0:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, amWhy, Davide Giraudo, KReiser, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is probably a more elegant way to do this, but here is one method that works:
Let $p_n$ be the desired probability.
At each choice we imagine that the left most member of the pair is being specified.
On the first move you choose any of the first $n-1$ points as the left most element of your first pair. Equal probability, of course. Then we can ignore what happens to the right and we have a shorter block in the front. It follows that the probability satisfies the recursion $$p_n=frac 1{n-1}sum_{i=1}^{n-1}p_{i-1}=frac 1{n-1}sum_{i=0}^{n-2}p_i$$
where we use the convention $p_0=0$.
Easy to see that your expression matches this for small $n$. To see that it matches for all $n$ note that our expression quickly implies the recursion $$(n-1)p_n-(n-2)p_{n-1}=p_{n-2}$$ Letting $a_n=sum_{k=0}^{n-1}frac {(-1)^k}{k!}$ we now seek to prove that the $a_n$ satisfy the same recursion. We check that
$$(n-1)a_n-(n-2)a_{n-2}=(n-1)times sum_{k=0}^{n-1}frac {(-1)^k}{k!}-(n-2)times sum_{k=0}^{n-2}frac {(-1)^k}{k!}$$$$=(n-1)times frac {(-1)^n}{(n-1)!}+sum_{k=0}^{n-2}frac {(-1)^k}{k!}
=frac {(-1)^{n}}{(n-2)!}+a_{n-3}$$ And, since $(-1)^{n-2}=(-1)^n$ we are done.
answered Jan 3 at 16:08
lulululu
39.6k24677
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is probably a more elegant way to do this, but here is one method that works:
Let $p_n$ be the desired probability.
At each choice we imagine that the left most member of the pair is being specified.
On the first move you choose any of the first $n-1$ points as the left most element of your first pair. Equal probability, of course. Then we can ignore what happens to the right and we have a shorter block in the front. It follows that the probability satisfies the recursion $$p_n=frac 1{n-1}sum_{i=1}^{n-1}p_{i-1}=frac 1{n-1}sum_{i=0}^{n-2}p_i$$
where we use the convention $p_0=0$.
Easy to see that your expression matches this for small $n$. To see that it matches for all $n$ note that our expression quickly implies the recursion $$(n-1)p_n-(n-2)p_{n-1}=p_{n-2}$$ Letting $a_n=sum_{k=0}^{n-1}frac {(-1)^k}{k!}$ we now seek to prove that the $a_n$ satisfy the same recursion. We check that
$$(n-1)a_n-(n-2)a_{n-2}=(n-1)times sum_{k=0}^{n-1}frac {(-1)^k}{k!}-(n-2)times sum_{k=0}^{n-2}frac {(-1)^k}{k!}$$$$=(n-1)times frac {(-1)^n}{(n-1)!}+sum_{k=0}^{n-2}frac {(-1)^k}{k!}
=frac {(-1)^{n}}{(n-2)!}+a_{n-3}$$ And, since $(-1)^{n-2}=(-1)^n$ we are done.
answered Jan 3 at 16:08
lulululu
39.6k24677
$endgroup$
add a comment |
$begingroup$
There is probably a more elegant way to do this, but here is one method that works:
Let $p_n$ be the desired probability.
At each choice we imagine that the left most member of the pair is being specified.
On the first move you choose any of the first $n-1$ points as the left most element of your first pair. Equal probability, of course. Then we can ignore what happens to the right and we have a shorter block in the front. It follows that the probability satisfies the recursion $$p_n=frac 1{n-1}sum_{i=1}^{n-1}p_{i-1}=frac 1{n-1}sum_{i=0}^{n-2}p_i$$
where we use the convention $p_0=0$.
Easy to see that your expression matches this for small $n$. To see that it matches for all $n$ note that our expression quickly implies the recursion $$(n-1)p_n-(n-2)p_{n-1}=p_{n-2}$$ Letting $a_n=sum_{k=0}^{n-1}frac {(-1)^k}{k!}$ we now seek to prove that the $a_n$ satisfy the same recursion. We check that
$$(n-1)a_n-(n-2)a_{n-2}=(n-1)times sum_{k=0}^{n-1}frac {(-1)^k}{k!}-(n-2)times sum_{k=0}^{n-2}frac {(-1)^k}{k!}$$$$=(n-1)times frac {(-1)^n}{(n-1)!}+sum_{k=0}^{n-2}frac {(-1)^k}{k!}
=frac {(-1)^{n}}{(n-2)!}+a_{n-3}$$ And, since $(-1)^{n-2}=(-1)^n$ we are done.
answered Jan 3 at 16:08
lulululu
39.6k24677
$endgroup$
add a comment |
$begingroup$
There is probably a more elegant way to do this, but here is one method that works:
Let $p_n$ be the desired probability.
At each choice we imagine that the left most member of the pair is being specified.
On the first move you choose any of the first $n-1$ points as the left most element of your first pair. Equal probability, of course. Then we can ignore what happens to the right and we have a shorter block in the front. It follows that the probability satisfies the recursion $$p_n=frac 1{n-1}sum_{i=1}^{n-1}p_{i-1}=frac 1{n-1}sum_{i=0}^{n-2}p_i$$
where we use the convention $p_0=0$.
Easy to see that your expression matches this for small $n$. To see that it matches for all $n$ note that our expression quickly implies the recursion $$(n-1)p_n-(n-2)p_{n-1}=p_{n-2}$$ Letting $a_n=sum_{k=0}^{n-1}frac {(-1)^k}{k!}$ we now seek to prove that the $a_n$ satisfy the same recursion. We check that
$$(n-1)a_n-(n-2)a_{n-2}=(n-1)times sum_{k=0}^{n-1}frac {(-1)^k}{k!}-(n-2)times sum_{k=0}^{n-2}frac {(-1)^k}{k!}$$$$=(n-1)times frac {(-1)^n}{(n-1)!}+sum_{k=0}^{n-2}frac {(-1)^k}{k!}
=frac {(-1)^{n}}{(n-2)!}+a_{n-3}$$ And, since $(-1)^{n-2}=(-1)^n$ we are done.
answered Jan 3 at 16:08
lulululu
39.6k24677
$endgroup$
There is probably a more elegant way to do this, but here is one method that works:
Let $p_n$ be the desired probability.
At each choice we imagine that the left most member of the pair is being specified.
On the first move you choose any of the first $n-1$ points as the left most element of your first pair. Equal probability, of course. Then we can ignore what happens to the right and we have a shorter block in the front. It follows that the probability satisfies the recursion $$p_n=frac 1{n-1}sum_{i=1}^{n-1}p_{i-1}=frac 1{n-1}sum_{i=0}^{n-2}p_i$$
where we use the convention $p_0=0$.
Easy to see that your expression matches this for small $n$. To see that it matches for all $n$ note that our expression quickly implies the recursion $$(n-1)p_n-(n-2)p_{n-1}=p_{n-2}$$ Letting $a_n=sum_{k=0}^{n-1}frac {(-1)^k}{k!}$ we now seek to prove that the $a_n$ satisfy the same recursion. We check that
$$(n-1)a_n-(n-2)a_{n-2}=(n-1)times sum_{k=0}^{n-1}frac {(-1)^k}{k!}-(n-2)times sum_{k=0}^{n-2}frac {(-1)^k}{k!}$$$$=(n-1)times frac {(-1)^n}{(n-1)!}+sum_{k=0}^{n-2}frac {(-1)^k}{k!}
=frac {(-1)^{n}}{(n-2)!}+a_{n-3}$$ And, since $(-1)^{n-2}=(-1)^n$ we are done.
answered Jan 3 at 16:08
lulululu
39.6k24677
edited Jan 10 at 23:52
answered Jan 3 at 16:08
lulululu
39.6k24677
answered Jan 3 at 16:08
lulululu
39.6k24677
answered Jan 3 at 16:08
lulululu
39.6k24677
39.6k24677
add a comment |
add a comment |
