Mixing
Conditional probability given only an average
$begingroup$
I’ve been working on this question, which I found on physics.SE. Unfortunately it was closed because it’s a homework question, but I’d like to get more of a hint than the original poster got.
I know that I’m supposed to use Bayes’ Theorem, but I don’t see how I’m supposed to use the fact that $bar{N}$ is known. Every effort so far has yielded an unwieldy fraction that can’t be simplified.
Edit: Apparently I'm supposed to copy/paste the question. Here it is:
This is one of the exercises of Barnett's book on quantum information.
A particle counter records counts with an efficiency $eta$. This means that each particle is detected with probability $eta$ and missed with probability $1-eta$. Let $N$ be the number of particles present and $n$ be the number of detected.
Then:
begin{equation}
P(n|N)=frac{N!}{n!(N-n)!}eta^n (1-eta)^{N-n}
end{equation}
I know the mean number of particle present is:
begin{equation}
bar{N}=sum N P(N)
end{equation}
I want to calculate $P(N|n)$. I'm stuck here by a while, so I do not know how to proceed.
Edit 2: I'm adding a screenshot of the question in question:
probability conditional-probability bayes-theorem quantum-information
asked Feb 3 at 0:25
AlexAlex
716418
$endgroup$
add a comment |
$begingroup$
I’ve been working on this question, which I found on physics.SE. Unfortunately it was closed because it’s a homework question, but I’d like to get more of a hint than the original poster got.
I know that I’m supposed to use Bayes’ Theorem, but I don’t see how I’m supposed to use the fact that $bar{N}$ is known. Every effort so far has yielded an unwieldy fraction that can’t be simplified.
Edit: Apparently I'm supposed to copy/paste the question. Here it is:
This is one of the exercises of Barnett's book on quantum information.
A particle counter records counts with an efficiency $eta$. This means that each particle is detected with probability $eta$ and missed with probability $1-eta$. Let $N$ be the number of particles present and $n$ be the number of detected.
Then:
begin{equation}
P(n|N)=frac{N!}{n!(N-n)!}eta^n (1-eta)^{N-n}
end{equation}
I know the mean number of particle present is:
begin{equation}
bar{N}=sum N P(N)
end{equation}
I want to calculate $P(N|n)$. I'm stuck here by a while, so I do not know how to proceed.
Edit 2: I'm adding a screenshot of the question in question:
probability conditional-probability bayes-theorem quantum-information
asked Feb 3 at 0:25
AlexAlex
716418
$endgroup$
$begingroup$
Please don't link a closed question. Copy the relevant (only math) parts here.
$endgroup$
– leonbloy
Feb 3 at 2:28
$begingroup$
@leonbloy I didn't know that was a rule, sorry. I've copied it now.
$endgroup$
– Alex
Feb 3 at 4:43
$begingroup$
"I want to calculate 𝑃(𝑁|𝑛)" To do so, you are missing some information, for example the (unconditional) distribution of N would do.
$endgroup$
– Did
Feb 3 at 9:19
add a comment |
$begingroup$
I’ve been working on this question, which I found on physics.SE. Unfortunately it was closed because it’s a homework question, but I’d like to get more of a hint than the original poster got.
I know that I’m supposed to use Bayes’ Theorem, but I don’t see how I’m supposed to use the fact that $bar{N}$ is known. Every effort so far has yielded an unwieldy fraction that can’t be simplified.
Edit: Apparently I'm supposed to copy/paste the question. Here it is:
This is one of the exercises of Barnett's book on quantum information.
A particle counter records counts with an efficiency $eta$. This means that each particle is detected with probability $eta$ and missed with probability $1-eta$. Let $N$ be the number of particles present and $n$ be the number of detected.
Then:
begin{equation}
P(n|N)=frac{N!}{n!(N-n)!}eta^n (1-eta)^{N-n}
end{equation}
I know the mean number of particle present is:
begin{equation}
bar{N}=sum N P(N)
end{equation}
I want to calculate $P(N|n)$. I'm stuck here by a while, so I do not know how to proceed.
Edit 2: I'm adding a screenshot of the question in question:
probability conditional-probability bayes-theorem quantum-information
asked Feb 3 at 0:25
AlexAlex
716418
$endgroup$
I’ve been working on this question, which I found on physics.SE. Unfortunately it was closed because it’s a homework question, but I’d like to get more of a hint than the original poster got.
I know that I’m supposed to use Bayes’ Theorem, but I don’t see how I’m supposed to use the fact that $bar{N}$ is known. Every effort so far has yielded an unwieldy fraction that can’t be simplified.
Edit: Apparently I'm supposed to copy/paste the question. Here it is:
This is one of the exercises of Barnett's book on quantum information.
A particle counter records counts with an efficiency $eta$. This means that each particle is detected with probability $eta$ and missed with probability $1-eta$. Let $N$ be the number of particles present and $n$ be the number of detected.
Then:
begin{equation}
P(n|N)=frac{N!}{n!(N-n)!}eta^n (1-eta)^{N-n}
end{equation}
I know the mean number of particle present is:
begin{equation}
bar{N}=sum N P(N)
end{equation}
I want to calculate $P(N|n)$. I'm stuck here by a while, so I do not know how to proceed.
Edit 2: I'm adding a screenshot of the question in question:
probability conditional-probability bayes-theorem quantum-information
probability conditional-probability bayes-theorem quantum-information
asked Feb 3 at 0:25
AlexAlex
716418
asked Feb 3 at 0:25
AlexAlex
716418
edited Feb 3 at 15:29
Alex
asked Feb 3 at 0:25
AlexAlex
716418
asked Feb 3 at 0:25
AlexAlex
716418
asked Feb 3 at 0:25
AlexAlex
716418
716418
$begingroup$
Please don't link a closed question. Copy the relevant (only math) parts here.
$endgroup$
– leonbloy
Feb 3 at 2:28
$begingroup$
@leonbloy I didn't know that was a rule, sorry. I've copied it now.
$endgroup$
– Alex
Feb 3 at 4:43
$begingroup$
"I want to calculate 𝑃(𝑁|𝑛)" To do so, you are missing some information, for example the (unconditional) distribution of N would do.
$endgroup$
– Did
Feb 3 at 9:19
add a comment |
$begingroup$
Please don't link a closed question. Copy the relevant (only math) parts here.
$endgroup$
– leonbloy
Feb 3 at 2:28
$begingroup$
@leonbloy I didn't know that was a rule, sorry. I've copied it now.
$endgroup$
– Alex
Feb 3 at 4:43
$begingroup$
"I want to calculate 𝑃(𝑁|𝑛)" To do so, you are missing some information, for example the (unconditional) distribution of N would do.
$endgroup$
– Did
Feb 3 at 9:19
$begingroup$
Please don't link a closed question. Copy the relevant (only math) parts here.
$endgroup$
– leonbloy
Feb 3 at 2:28
$begingroup$
Please don't link a closed question. Copy the relevant (only math) parts here.
$endgroup$
– leonbloy
Feb 3 at 2:28
$begingroup$
@leonbloy I didn't know that was a rule, sorry. I've copied it now.
$endgroup$
– Alex
Feb 3 at 4:43
$begingroup$
@leonbloy I didn't know that was a rule, sorry. I've copied it now.
$endgroup$
– Alex
Feb 3 at 4:43
$begingroup$
"I want to calculate 𝑃(𝑁|𝑛)" To do so, you are missing some information, for example the (unconditional) distribution of N would do.
$endgroup$
– Did
Feb 3 at 9:19
$begingroup$
"I want to calculate 𝑃(𝑁|𝑛)" To do so, you are missing some information, for example the (unconditional) distribution of N would do.
$endgroup$
– Did
Feb 3 at 9:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You surely know that
$$ P(N|n) = frac{P(n|N) P(N)}{P(n)}=frac{P(n|N) P(N)}{sum_N P(n|N) P(N)}$$
This tells you that you $P(n|N)$ is not enough, you need $P(N)$. To understand this should be your starting point.
Now, here, you only give us the mean of $N$ (actually you wrote the general formula of the expected value, but I guess you meant that you know $bar{N}$). Still not enough.
Perhaps there is some implicit statistical model for $N$ that you've missed? Perhaps (just guessing) it follows a Poisson distribution? If so, you are done, because the Poisson distribution depends on a single parameter (which is the mean), hence you indeed can write $P(N)$.
answered Feb 3 at 13:49
leonbloyleonbloy
42.4k647108
$endgroup$
$begingroup$
I've attached a screenshot of the exact question from the textbook. Unfortunately, doing this same calculation for a Poisson distribution was part (a) of the question. Do you have any other thoughts, given this?
$endgroup$
– Alex
Feb 3 at 15:31
$begingroup$
No, I don't see how one could do with the mean alone.
$endgroup$
– leonbloy
Feb 4 at 15:10
add a comment |
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1 Answer
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$begingroup$
You surely know that
$$ P(N|n) = frac{P(n|N) P(N)}{P(n)}=frac{P(n|N) P(N)}{sum_N P(n|N) P(N)}$$
This tells you that you $P(n|N)$ is not enough, you need $P(N)$. To understand this should be your starting point.
Now, here, you only give us the mean of $N$ (actually you wrote the general formula of the expected value, but I guess you meant that you know $bar{N}$). Still not enough.
Perhaps there is some implicit statistical model for $N$ that you've missed? Perhaps (just guessing) it follows a Poisson distribution? If so, you are done, because the Poisson distribution depends on a single parameter (which is the mean), hence you indeed can write $P(N)$.
answered Feb 3 at 13:49
leonbloyleonbloy
42.4k647108
$endgroup$
$begingroup$
I've attached a screenshot of the exact question from the textbook. Unfortunately, doing this same calculation for a Poisson distribution was part (a) of the question. Do you have any other thoughts, given this?
$endgroup$
– Alex
Feb 3 at 15:31
$begingroup$
No, I don't see how one could do with the mean alone.
$endgroup$
– leonbloy
Feb 4 at 15:10
add a comment |
$begingroup$
You surely know that
$$ P(N|n) = frac{P(n|N) P(N)}{P(n)}=frac{P(n|N) P(N)}{sum_N P(n|N) P(N)}$$
This tells you that you $P(n|N)$ is not enough, you need $P(N)$. To understand this should be your starting point.
Now, here, you only give us the mean of $N$ (actually you wrote the general formula of the expected value, but I guess you meant that you know $bar{N}$). Still not enough.
Perhaps there is some implicit statistical model for $N$ that you've missed? Perhaps (just guessing) it follows a Poisson distribution? If so, you are done, because the Poisson distribution depends on a single parameter (which is the mean), hence you indeed can write $P(N)$.
answered Feb 3 at 13:49
leonbloyleonbloy
42.4k647108
$endgroup$
$begingroup$
I've attached a screenshot of the exact question from the textbook. Unfortunately, doing this same calculation for a Poisson distribution was part (a) of the question. Do you have any other thoughts, given this?
$endgroup$
– Alex
Feb 3 at 15:31
$begingroup$
No, I don't see how one could do with the mean alone.
$endgroup$
– leonbloy
Feb 4 at 15:10
add a comment |
$begingroup$
You surely know that
$$ P(N|n) = frac{P(n|N) P(N)}{P(n)}=frac{P(n|N) P(N)}{sum_N P(n|N) P(N)}$$
This tells you that you $P(n|N)$ is not enough, you need $P(N)$. To understand this should be your starting point.
Now, here, you only give us the mean of $N$ (actually you wrote the general formula of the expected value, but I guess you meant that you know $bar{N}$). Still not enough.
Perhaps there is some implicit statistical model for $N$ that you've missed? Perhaps (just guessing) it follows a Poisson distribution? If so, you are done, because the Poisson distribution depends on a single parameter (which is the mean), hence you indeed can write $P(N)$.
answered Feb 3 at 13:49
leonbloyleonbloy
42.4k647108
$endgroup$
You surely know that
$$ P(N|n) = frac{P(n|N) P(N)}{P(n)}=frac{P(n|N) P(N)}{sum_N P(n|N) P(N)}$$
This tells you that you $P(n|N)$ is not enough, you need $P(N)$. To understand this should be your starting point.
Now, here, you only give us the mean of $N$ (actually you wrote the general formula of the expected value, but I guess you meant that you know $bar{N}$). Still not enough.
Perhaps there is some implicit statistical model for $N$ that you've missed? Perhaps (just guessing) it follows a Poisson distribution? If so, you are done, because the Poisson distribution depends on a single parameter (which is the mean), hence you indeed can write $P(N)$.
answered Feb 3 at 13:49
leonbloyleonbloy
42.4k647108
answered Feb 3 at 13:49
leonbloyleonbloy
42.4k647108
answered Feb 3 at 13:49
leonbloyleonbloy
42.4k647108
answered Feb 3 at 13:49
leonbloyleonbloy
42.4k647108
42.4k647108
$begingroup$
I've attached a screenshot of the exact question from the textbook. Unfortunately, doing this same calculation for a Poisson distribution was part (a) of the question. Do you have any other thoughts, given this?
$endgroup$
– Alex
Feb 3 at 15:31
$begingroup$
No, I don't see how one could do with the mean alone.
$endgroup$
– leonbloy
Feb 4 at 15:10
add a comment |
$begingroup$
I've attached a screenshot of the exact question from the textbook. Unfortunately, doing this same calculation for a Poisson distribution was part (a) of the question. Do you have any other thoughts, given this?
$endgroup$
– Alex
Feb 3 at 15:31
$begingroup$
No, I don't see how one could do with the mean alone.
$endgroup$
– leonbloy
Feb 4 at 15:10
$begingroup$
I've attached a screenshot of the exact question from the textbook. Unfortunately, doing this same calculation for a Poisson distribution was part (a) of the question. Do you have any other thoughts, given this?
$endgroup$
– Alex
Feb 3 at 15:31
$begingroup$
I've attached a screenshot of the exact question from the textbook. Unfortunately, doing this same calculation for a Poisson distribution was part (a) of the question. Do you have any other thoughts, given this?
$endgroup$
– Alex
Feb 3 at 15:31
$begingroup$
No, I don't see how one could do with the mean alone.
$endgroup$
– leonbloy
Feb 4 at 15:10
$begingroup$
No, I don't see how one could do with the mean alone.
$endgroup$
– leonbloy
Feb 4 at 15:10
add a comment |
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$begingroup$
Please don't link a closed question. Copy the relevant (only math) parts here.
$endgroup$
– leonbloy
Feb 3 at 2:28
$begingroup$
@leonbloy I didn't know that was a rule, sorry. I've copied it now.
$endgroup$
– Alex
Feb 3 at 4:43
$begingroup$
"I want to calculate 𝑃(𝑁|𝑛)" To do so, you are missing some information, for example the (unconditional) distribution of N would do.
$endgroup$
– Did
Feb 3 at 9:19