Mixing
Drawing random realisation from quantity with poisson error
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How would I draw a random realisation of a variable with an upper and lower error determined from Poisson statistics using the Gehrels 1986 formula?
See: http://adsabs.harvard.edu/abs/1986ApJ...303..336G
For example, I have a count of:
$x = 3^{+5.9}_{-1.3}$
and I want to draw many random realisations of this to use in a Monte Carlo simulation. How do I do this, please?
If it were a Gaussian variable I'd use a random Gaussian (or split-normal) distribution centred on $mu=3$ with a standard deviation $sigma^+=5.9$ and $sigma^-=1.3$.
Thanks for any help here.
random poisson-distribution standard-deviation monte-carlo error-propagation
asked Jan 31 at 13:25
alex_lewisalex_lewis
206
$endgroup$
add a comment |
$begingroup$
How would I draw a random realisation of a variable with an upper and lower error determined from Poisson statistics using the Gehrels 1986 formula?
See: http://adsabs.harvard.edu/abs/1986ApJ...303..336G
For example, I have a count of:
$x = 3^{+5.9}_{-1.3}$
and I want to draw many random realisations of this to use in a Monte Carlo simulation. How do I do this, please?
If it were a Gaussian variable I'd use a random Gaussian (or split-normal) distribution centred on $mu=3$ with a standard deviation $sigma^+=5.9$ and $sigma^-=1.3$.
Thanks for any help here.
random poisson-distribution standard-deviation monte-carlo error-propagation
asked Jan 31 at 13:25
alex_lewisalex_lewis
206
$endgroup$
$begingroup$
If you want to simulate that random variable, I don't see what is the connection to the confidence limits. Why wouldn't you just use the unbiased estimator for $lambda$ which is the number of events divided by the length of the interval measured?
$endgroup$
– Ron Kaminsky
Feb 10 at 22:05
$begingroup$
Possibly what is confusing you is that the normal distribution has two parameters and therefore the variance can be "reverse engineered" using the confidence limits. Poisson only has one parameter, so there is no second parameter to "reverse engineer", nor any necessity to do so.
$endgroup$
– Ron Kaminsky
Feb 10 at 22:14
add a comment |
$begingroup$
How would I draw a random realisation of a variable with an upper and lower error determined from Poisson statistics using the Gehrels 1986 formula?
See: http://adsabs.harvard.edu/abs/1986ApJ...303..336G
For example, I have a count of:
$x = 3^{+5.9}_{-1.3}$
and I want to draw many random realisations of this to use in a Monte Carlo simulation. How do I do this, please?
If it were a Gaussian variable I'd use a random Gaussian (or split-normal) distribution centred on $mu=3$ with a standard deviation $sigma^+=5.9$ and $sigma^-=1.3$.
Thanks for any help here.
random poisson-distribution standard-deviation monte-carlo error-propagation
asked Jan 31 at 13:25
alex_lewisalex_lewis
206
$endgroup$
How would I draw a random realisation of a variable with an upper and lower error determined from Poisson statistics using the Gehrels 1986 formula?
See: http://adsabs.harvard.edu/abs/1986ApJ...303..336G
For example, I have a count of:
$x = 3^{+5.9}_{-1.3}$
and I want to draw many random realisations of this to use in a Monte Carlo simulation. How do I do this, please?
If it were a Gaussian variable I'd use a random Gaussian (or split-normal) distribution centred on $mu=3$ with a standard deviation $sigma^+=5.9$ and $sigma^-=1.3$.
Thanks for any help here.
random poisson-distribution standard-deviation monte-carlo error-propagation
random poisson-distribution standard-deviation monte-carlo error-propagation
asked Jan 31 at 13:25
alex_lewisalex_lewis
206
asked Jan 31 at 13:25
alex_lewisalex_lewis
206
edited Feb 5 at 23:37
alex_lewis
asked Jan 31 at 13:25
alex_lewisalex_lewis
206
asked Jan 31 at 13:25
alex_lewisalex_lewis
206
asked Jan 31 at 13:25
alex_lewisalex_lewis
206
206
$begingroup$
If you want to simulate that random variable, I don't see what is the connection to the confidence limits. Why wouldn't you just use the unbiased estimator for $lambda$ which is the number of events divided by the length of the interval measured?
$endgroup$
– Ron Kaminsky
Feb 10 at 22:05
$begingroup$
Possibly what is confusing you is that the normal distribution has two parameters and therefore the variance can be "reverse engineered" using the confidence limits. Poisson only has one parameter, so there is no second parameter to "reverse engineer", nor any necessity to do so.
$endgroup$
– Ron Kaminsky
Feb 10 at 22:14
add a comment |
$begingroup$
If you want to simulate that random variable, I don't see what is the connection to the confidence limits. Why wouldn't you just use the unbiased estimator for $lambda$ which is the number of events divided by the length of the interval measured?
$endgroup$
– Ron Kaminsky
Feb 10 at 22:05
$begingroup$
Possibly what is confusing you is that the normal distribution has two parameters and therefore the variance can be "reverse engineered" using the confidence limits. Poisson only has one parameter, so there is no second parameter to "reverse engineer", nor any necessity to do so.
$endgroup$
– Ron Kaminsky
Feb 10 at 22:14
$begingroup$
If you want to simulate that random variable, I don't see what is the connection to the confidence limits. Why wouldn't you just use the unbiased estimator for $lambda$ which is the number of events divided by the length of the interval measured?
$endgroup$
– Ron Kaminsky
Feb 10 at 22:05
$begingroup$
If you want to simulate that random variable, I don't see what is the connection to the confidence limits. Why wouldn't you just use the unbiased estimator for $lambda$ which is the number of events divided by the length of the interval measured?
$endgroup$
– Ron Kaminsky
Feb 10 at 22:05
$begingroup$
Possibly what is confusing you is that the normal distribution has two parameters and therefore the variance can be "reverse engineered" using the confidence limits. Poisson only has one parameter, so there is no second parameter to "reverse engineer", nor any necessity to do so.
$endgroup$
– Ron Kaminsky
Feb 10 at 22:14
$begingroup$
Possibly what is confusing you is that the normal distribution has two parameters and therefore the variance can be "reverse engineered" using the confidence limits. Poisson only has one parameter, so there is no second parameter to "reverse engineer", nor any necessity to do so.
$endgroup$
– Ron Kaminsky
Feb 10 at 22:14
add a comment |
1 Answer
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$begingroup$
Since you give no information how the confidence interval should affect your simulation (said confidence interval being something you have arbitrarily chosen via choosing its probability cutoffs), I see only two possible answers to your question.
The first is that you should just use the rate which was observed as the parameter for a random variable with a Poisson distribution (since the observed rate is an unbiased estimator for $lambda$).
The second and more interesting possibility is Bayesian in nature. Given an observation of $k$ events, the probability distribution for $lambda$ is a gamma distribution with parameters $alpha = k+1$ and $beta = 1$. Therefore, you can use a two-step method: generate a value for $lambda$ from that distribution, and then generate a random value for $k$ from the Poisson distribution with that choice for $lambda$. (I suppose if you really want your choice of confidence interval to have some effect, you could limit the possible values for $lambda$ generated by that gamma distribution to your confidence interval, by rejecting any values outside of the interval.)
answered Feb 11 at 21:03
Ron KaminskyRon Kaminsky
1678
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1 Answer
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$begingroup$
Since you give no information how the confidence interval should affect your simulation (said confidence interval being something you have arbitrarily chosen via choosing its probability cutoffs), I see only two possible answers to your question.
The first is that you should just use the rate which was observed as the parameter for a random variable with a Poisson distribution (since the observed rate is an unbiased estimator for $lambda$).
The second and more interesting possibility is Bayesian in nature. Given an observation of $k$ events, the probability distribution for $lambda$ is a gamma distribution with parameters $alpha = k+1$ and $beta = 1$. Therefore, you can use a two-step method: generate a value for $lambda$ from that distribution, and then generate a random value for $k$ from the Poisson distribution with that choice for $lambda$. (I suppose if you really want your choice of confidence interval to have some effect, you could limit the possible values for $lambda$ generated by that gamma distribution to your confidence interval, by rejecting any values outside of the interval.)
answered Feb 11 at 21:03
Ron KaminskyRon Kaminsky
1678
$endgroup$
add a comment |
$begingroup$
Since you give no information how the confidence interval should affect your simulation (said confidence interval being something you have arbitrarily chosen via choosing its probability cutoffs), I see only two possible answers to your question.
The first is that you should just use the rate which was observed as the parameter for a random variable with a Poisson distribution (since the observed rate is an unbiased estimator for $lambda$).
The second and more interesting possibility is Bayesian in nature. Given an observation of $k$ events, the probability distribution for $lambda$ is a gamma distribution with parameters $alpha = k+1$ and $beta = 1$. Therefore, you can use a two-step method: generate a value for $lambda$ from that distribution, and then generate a random value for $k$ from the Poisson distribution with that choice for $lambda$. (I suppose if you really want your choice of confidence interval to have some effect, you could limit the possible values for $lambda$ generated by that gamma distribution to your confidence interval, by rejecting any values outside of the interval.)
answered Feb 11 at 21:03
Ron KaminskyRon Kaminsky
1678
$endgroup$
add a comment |
$begingroup$
Since you give no information how the confidence interval should affect your simulation (said confidence interval being something you have arbitrarily chosen via choosing its probability cutoffs), I see only two possible answers to your question.
The first is that you should just use the rate which was observed as the parameter for a random variable with a Poisson distribution (since the observed rate is an unbiased estimator for $lambda$).
The second and more interesting possibility is Bayesian in nature. Given an observation of $k$ events, the probability distribution for $lambda$ is a gamma distribution with parameters $alpha = k+1$ and $beta = 1$. Therefore, you can use a two-step method: generate a value for $lambda$ from that distribution, and then generate a random value for $k$ from the Poisson distribution with that choice for $lambda$. (I suppose if you really want your choice of confidence interval to have some effect, you could limit the possible values for $lambda$ generated by that gamma distribution to your confidence interval, by rejecting any values outside of the interval.)
answered Feb 11 at 21:03
Ron KaminskyRon Kaminsky
1678
$endgroup$
Since you give no information how the confidence interval should affect your simulation (said confidence interval being something you have arbitrarily chosen via choosing its probability cutoffs), I see only two possible answers to your question.
The first is that you should just use the rate which was observed as the parameter for a random variable with a Poisson distribution (since the observed rate is an unbiased estimator for $lambda$).
The second and more interesting possibility is Bayesian in nature. Given an observation of $k$ events, the probability distribution for $lambda$ is a gamma distribution with parameters $alpha = k+1$ and $beta = 1$. Therefore, you can use a two-step method: generate a value for $lambda$ from that distribution, and then generate a random value for $k$ from the Poisson distribution with that choice for $lambda$. (I suppose if you really want your choice of confidence interval to have some effect, you could limit the possible values for $lambda$ generated by that gamma distribution to your confidence interval, by rejecting any values outside of the interval.)
answered Feb 11 at 21:03
Ron KaminskyRon Kaminsky
1678
edited Feb 12 at 19:14
answered Feb 11 at 21:03
Ron KaminskyRon Kaminsky
1678
answered Feb 11 at 21:03
Ron KaminskyRon Kaminsky
1678
answered Feb 11 at 21:03
Ron KaminskyRon Kaminsky
1678
1678
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$begingroup$
If you want to simulate that random variable, I don't see what is the connection to the confidence limits. Why wouldn't you just use the unbiased estimator for $lambda$ which is the number of events divided by the length of the interval measured?
$endgroup$
– Ron Kaminsky
Feb 10 at 22:05
$begingroup$
Possibly what is confusing you is that the normal distribution has two parameters and therefore the variance can be "reverse engineered" using the confidence limits. Poisson only has one parameter, so there is no second parameter to "reverse engineer", nor any necessity to do so.
$endgroup$
– Ron Kaminsky
Feb 10 at 22:14