Combining Gaussian Distributions
$begingroup$
My question illustrated in a problem:
After giving and grading a statistics exam, the professor found the mean and standard deviation of points students got for each problem. One of the professors students had to miss the exam, and scheduled to take it later. With the calculated distributions only, what is the probability that the late student gets at least X% on the test?
The number of problems and percent cutoff are not predefined, and the points, mean, and standard deviation of each problem are independent to other problems.
You can assume each distribution is purely gaussian, if that helps.
For whatever reason, say that the professor doesn't have access to the individual students' data anymore, so he/she couldn't just count the number of students with a certain overall score, make a new distribution of overall scores, or any other workaround. The professor only has a set of independent gaussian distributions.
In short: summing together one random value per independent gaussian distribution, what is the probability of getting over a certain total value?
probability statistics
asked Dec 23 '18 at 1:42
AntonioAntonio
1
$endgroup$
add a comment |
$begingroup$
My question illustrated in a problem:
After giving and grading a statistics exam, the professor found the mean and standard deviation of points students got for each problem. One of the professors students had to miss the exam, and scheduled to take it later. With the calculated distributions only, what is the probability that the late student gets at least X% on the test?
The number of problems and percent cutoff are not predefined, and the points, mean, and standard deviation of each problem are independent to other problems.
You can assume each distribution is purely gaussian, if that helps.
For whatever reason, say that the professor doesn't have access to the individual students' data anymore, so he/she couldn't just count the number of students with a certain overall score, make a new distribution of overall scores, or any other workaround. The professor only has a set of independent gaussian distributions.
In short: summing together one random value per independent gaussian distribution, what is the probability of getting over a certain total value?
probability statistics
asked Dec 23 '18 at 1:42
AntonioAntonio
1
$endgroup$
$begingroup$
You seem to be suggests individual students' scores on each question are independent of their scores on other questions; that would be a curious exam which suggests that individuals' overall skills and efforts do not affect results. But if so, you should be aware that (a) the sum of independent Gaussian random variables have a Gaussian distribution, and (b) the sum of random variables has an expected value equal to the sum of the individual expected values, and (c) the sum of independent random variables has a variance equal to the sum of the individual variances
$endgroup$
– Henry
Dec 23 '18 at 1:50
$begingroup$
Haha yeah. This was just an example scenario to put my question into context; don't take the independent scores too seriously ;) (a) Is that so? Since each problem has a different number of points, the expected value isn't the same. So summing together the distributions could potentially be quite lumpy. (b) I am aware of that, but I'm not totally sure how that would help, unless all expected values are the same. (c) I did not know this. If the points for each problem WERE the same, could you then just combine them somehow? Multiply their standard deviations together or something?
$endgroup$
– Antonio
Dec 23 '18 at 2:04
$begingroup$
That's my bad; I mixed my vocab way up. I think I get what you're hinting at, and I think I got a solution. I did my best to validate it by generating random gaussians and adding them together, and it seems to be good. Please do let me know if I messed anything up.
$endgroup$
– Antonio
Dec 23 '18 at 5:28
add a comment |
$begingroup$
My question illustrated in a problem:
After giving and grading a statistics exam, the professor found the mean and standard deviation of points students got for each problem. One of the professors students had to miss the exam, and scheduled to take it later. With the calculated distributions only, what is the probability that the late student gets at least X% on the test?
The number of problems and percent cutoff are not predefined, and the points, mean, and standard deviation of each problem are independent to other problems.
You can assume each distribution is purely gaussian, if that helps.
For whatever reason, say that the professor doesn't have access to the individual students' data anymore, so he/she couldn't just count the number of students with a certain overall score, make a new distribution of overall scores, or any other workaround. The professor only has a set of independent gaussian distributions.
In short: summing together one random value per independent gaussian distribution, what is the probability of getting over a certain total value?
probability statistics
asked Dec 23 '18 at 1:42
AntonioAntonio
1
$endgroup$
My question illustrated in a problem:
After giving and grading a statistics exam, the professor found the mean and standard deviation of points students got for each problem. One of the professors students had to miss the exam, and scheduled to take it later. With the calculated distributions only, what is the probability that the late student gets at least X% on the test?
The number of problems and percent cutoff are not predefined, and the points, mean, and standard deviation of each problem are independent to other problems.
You can assume each distribution is purely gaussian, if that helps.
For whatever reason, say that the professor doesn't have access to the individual students' data anymore, so he/she couldn't just count the number of students with a certain overall score, make a new distribution of overall scores, or any other workaround. The professor only has a set of independent gaussian distributions.
In short: summing together one random value per independent gaussian distribution, what is the probability of getting over a certain total value?
probability statistics
probability statistics
asked Dec 23 '18 at 1:42
AntonioAntonio
1
asked Dec 23 '18 at 1:42
AntonioAntonio
1
asked Dec 23 '18 at 1:42
AntonioAntonio
1
asked Dec 23 '18 at 1:42
AntonioAntonio
1
asked Dec 23 '18 at 1:42
AntonioAntonio
1
1
$begingroup$
You seem to be suggests individual students' scores on each question are independent of their scores on other questions; that would be a curious exam which suggests that individuals' overall skills and efforts do not affect results. But if so, you should be aware that (a) the sum of independent Gaussian random variables have a Gaussian distribution, and (b) the sum of random variables has an expected value equal to the sum of the individual expected values, and (c) the sum of independent random variables has a variance equal to the sum of the individual variances
$endgroup$
– Henry
Dec 23 '18 at 1:50
$begingroup$
Haha yeah. This was just an example scenario to put my question into context; don't take the independent scores too seriously ;) (a) Is that so? Since each problem has a different number of points, the expected value isn't the same. So summing together the distributions could potentially be quite lumpy. (b) I am aware of that, but I'm not totally sure how that would help, unless all expected values are the same. (c) I did not know this. If the points for each problem WERE the same, could you then just combine them somehow? Multiply their standard deviations together or something?
$endgroup$
– Antonio
Dec 23 '18 at 2:04
$begingroup$
That's my bad; I mixed my vocab way up. I think I get what you're hinting at, and I think I got a solution. I did my best to validate it by generating random gaussians and adding them together, and it seems to be good. Please do let me know if I messed anything up.
$endgroup$
– Antonio
Dec 23 '18 at 5:28
add a comment |
$begingroup$
You seem to be suggests individual students' scores on each question are independent of their scores on other questions; that would be a curious exam which suggests that individuals' overall skills and efforts do not affect results. But if so, you should be aware that (a) the sum of independent Gaussian random variables have a Gaussian distribution, and (b) the sum of random variables has an expected value equal to the sum of the individual expected values, and (c) the sum of independent random variables has a variance equal to the sum of the individual variances
$endgroup$
– Henry
Dec 23 '18 at 1:50
$begingroup$
Haha yeah. This was just an example scenario to put my question into context; don't take the independent scores too seriously ;) (a) Is that so? Since each problem has a different number of points, the expected value isn't the same. So summing together the distributions could potentially be quite lumpy. (b) I am aware of that, but I'm not totally sure how that would help, unless all expected values are the same. (c) I did not know this. If the points for each problem WERE the same, could you then just combine them somehow? Multiply their standard deviations together or something?
$endgroup$
– Antonio
Dec 23 '18 at 2:04
$begingroup$
That's my bad; I mixed my vocab way up. I think I get what you're hinting at, and I think I got a solution. I did my best to validate it by generating random gaussians and adding them together, and it seems to be good. Please do let me know if I messed anything up.
$endgroup$
– Antonio
Dec 23 '18 at 5:28
$begingroup$
You seem to be suggests individual students' scores on each question are independent of their scores on other questions; that would be a curious exam which suggests that individuals' overall skills and efforts do not affect results. But if so, you should be aware that (a) the sum of independent Gaussian random variables have a Gaussian distribution, and (b) the sum of random variables has an expected value equal to the sum of the individual expected values, and (c) the sum of independent random variables has a variance equal to the sum of the individual variances
$endgroup$
– Henry
Dec 23 '18 at 1:50
$begingroup$
You seem to be suggests individual students' scores on each question are independent of their scores on other questions; that would be a curious exam which suggests that individuals' overall skills and efforts do not affect results. But if so, you should be aware that (a) the sum of independent Gaussian random variables have a Gaussian distribution, and (b) the sum of random variables has an expected value equal to the sum of the individual expected values, and (c) the sum of independent random variables has a variance equal to the sum of the individual variances
$endgroup$
– Henry
Dec 23 '18 at 1:50
$begingroup$
Haha yeah. This was just an example scenario to put my question into context; don't take the independent scores too seriously ;) (a) Is that so? Since each problem has a different number of points, the expected value isn't the same. So summing together the distributions could potentially be quite lumpy. (b) I am aware of that, but I'm not totally sure how that would help, unless all expected values are the same. (c) I did not know this. If the points for each problem WERE the same, could you then just combine them somehow? Multiply their standard deviations together or something?
$endgroup$
– Antonio
Dec 23 '18 at 2:04
$begingroup$
Haha yeah. This was just an example scenario to put my question into context; don't take the independent scores too seriously ;) (a) Is that so? Since each problem has a different number of points, the expected value isn't the same. So summing together the distributions could potentially be quite lumpy. (b) I am aware of that, but I'm not totally sure how that would help, unless all expected values are the same. (c) I did not know this. If the points for each problem WERE the same, could you then just combine them somehow? Multiply their standard deviations together or something?
$endgroup$
– Antonio
Dec 23 '18 at 2:04
$begingroup$
That's my bad; I mixed my vocab way up. I think I get what you're hinting at, and I think I got a solution. I did my best to validate it by generating random gaussians and adding them together, and it seems to be good. Please do let me know if I messed anything up.
$endgroup$
– Antonio
Dec 23 '18 at 5:28
$begingroup$
That's my bad; I mixed my vocab way up. I think I get what you're hinting at, and I think I got a solution. I did my best to validate it by generating random gaussians and adding them together, and it seems to be good. Please do let me know if I messed anything up.
$endgroup$
– Antonio
Dec 23 '18 at 5:28
add a comment |
1 Answer
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oldest
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$begingroup$
Alright, I think I figured it out.
I'm not marking this as accepted, though (or at least not for a while), just in case I did something wrong and someone can correct me, as my statistical skills are borderline nonexistent.
You can combine all gaussian distributions into a single gaussian where $mu = sum_{i=1}^nmu_i$ and $sigma = sqrt{sum_{i=0}^nsigma_i^2}$, where $n$ denotes the number of gaussian distributions to sum over, and $mu_i$ and $sigma_i$ denote the $i^{th}$ gaussians mean and standard deviation, respectively.
This is all I really care about, but just for completeness, to find the probability of the student getting above $X$%, you need to do a z table lookup. To do that, you must first get your z score. This is done via $Z = frac{X - mu}{sigma}$. After finding your $P$ with the corresponding $Z$ in the z table, because we want everything ABOVE $X$%, not below, we get our final probability as $1 - P$.
In one equation, the solution for $n$ gaussian distributions with a total score of at least $X$% is:$$1-Phi(frac{X - sum_{i=1}^nmu_i}{sqrt{sum_{i=0}^nsigma_i^2}})$$
answered Dec 23 '18 at 5:26
AntonioAntonio
1
$endgroup$
add a comment |
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$begingroup$
Alright, I think I figured it out.
I'm not marking this as accepted, though (or at least not for a while), just in case I did something wrong and someone can correct me, as my statistical skills are borderline nonexistent.
You can combine all gaussian distributions into a single gaussian where $mu = sum_{i=1}^nmu_i$ and $sigma = sqrt{sum_{i=0}^nsigma_i^2}$, where $n$ denotes the number of gaussian distributions to sum over, and $mu_i$ and $sigma_i$ denote the $i^{th}$ gaussians mean and standard deviation, respectively.
This is all I really care about, but just for completeness, to find the probability of the student getting above $X$%, you need to do a z table lookup. To do that, you must first get your z score. This is done via $Z = frac{X - mu}{sigma}$. After finding your $P$ with the corresponding $Z$ in the z table, because we want everything ABOVE $X$%, not below, we get our final probability as $1 - P$.
In one equation, the solution for $n$ gaussian distributions with a total score of at least $X$% is:$$1-Phi(frac{X - sum_{i=1}^nmu_i}{sqrt{sum_{i=0}^nsigma_i^2}})$$
answered Dec 23 '18 at 5:26
AntonioAntonio
1
$endgroup$
add a comment |
$begingroup$
Alright, I think I figured it out.
I'm not marking this as accepted, though (or at least not for a while), just in case I did something wrong and someone can correct me, as my statistical skills are borderline nonexistent.
You can combine all gaussian distributions into a single gaussian where $mu = sum_{i=1}^nmu_i$ and $sigma = sqrt{sum_{i=0}^nsigma_i^2}$, where $n$ denotes the number of gaussian distributions to sum over, and $mu_i$ and $sigma_i$ denote the $i^{th}$ gaussians mean and standard deviation, respectively.
This is all I really care about, but just for completeness, to find the probability of the student getting above $X$%, you need to do a z table lookup. To do that, you must first get your z score. This is done via $Z = frac{X - mu}{sigma}$. After finding your $P$ with the corresponding $Z$ in the z table, because we want everything ABOVE $X$%, not below, we get our final probability as $1 - P$.
In one equation, the solution for $n$ gaussian distributions with a total score of at least $X$% is:$$1-Phi(frac{X - sum_{i=1}^nmu_i}{sqrt{sum_{i=0}^nsigma_i^2}})$$
answered Dec 23 '18 at 5:26
AntonioAntonio
1
$endgroup$
add a comment |
$begingroup$
Alright, I think I figured it out.
I'm not marking this as accepted, though (or at least not for a while), just in case I did something wrong and someone can correct me, as my statistical skills are borderline nonexistent.
You can combine all gaussian distributions into a single gaussian where $mu = sum_{i=1}^nmu_i$ and $sigma = sqrt{sum_{i=0}^nsigma_i^2}$, where $n$ denotes the number of gaussian distributions to sum over, and $mu_i$ and $sigma_i$ denote the $i^{th}$ gaussians mean and standard deviation, respectively.
This is all I really care about, but just for completeness, to find the probability of the student getting above $X$%, you need to do a z table lookup. To do that, you must first get your z score. This is done via $Z = frac{X - mu}{sigma}$. After finding your $P$ with the corresponding $Z$ in the z table, because we want everything ABOVE $X$%, not below, we get our final probability as $1 - P$.
In one equation, the solution for $n$ gaussian distributions with a total score of at least $X$% is:$$1-Phi(frac{X - sum_{i=1}^nmu_i}{sqrt{sum_{i=0}^nsigma_i^2}})$$
answered Dec 23 '18 at 5:26
AntonioAntonio
1
$endgroup$
Alright, I think I figured it out.
I'm not marking this as accepted, though (or at least not for a while), just in case I did something wrong and someone can correct me, as my statistical skills are borderline nonexistent.
You can combine all gaussian distributions into a single gaussian where $mu = sum_{i=1}^nmu_i$ and $sigma = sqrt{sum_{i=0}^nsigma_i^2}$, where $n$ denotes the number of gaussian distributions to sum over, and $mu_i$ and $sigma_i$ denote the $i^{th}$ gaussians mean and standard deviation, respectively.
This is all I really care about, but just for completeness, to find the probability of the student getting above $X$%, you need to do a z table lookup. To do that, you must first get your z score. This is done via $Z = frac{X - mu}{sigma}$. After finding your $P$ with the corresponding $Z$ in the z table, because we want everything ABOVE $X$%, not below, we get our final probability as $1 - P$.
In one equation, the solution for $n$ gaussian distributions with a total score of at least $X$% is:$$1-Phi(frac{X - sum_{i=1}^nmu_i}{sqrt{sum_{i=0}^nsigma_i^2}})$$
answered Dec 23 '18 at 5:26
AntonioAntonio
1
answered Dec 23 '18 at 5:26
AntonioAntonio
1
answered Dec 23 '18 at 5:26
AntonioAntonio
1
answered Dec 23 '18 at 5:26
AntonioAntonio
1
1
add a comment |
add a comment |
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$begingroup$
You seem to be suggests individual students' scores on each question are independent of their scores on other questions; that would be a curious exam which suggests that individuals' overall skills and efforts do not affect results. But if so, you should be aware that (a) the sum of independent Gaussian random variables have a Gaussian distribution, and (b) the sum of random variables has an expected value equal to the sum of the individual expected values, and (c) the sum of independent random variables has a variance equal to the sum of the individual variances
$endgroup$
– Henry
Dec 23 '18 at 1:50
$begingroup$
Haha yeah. This was just an example scenario to put my question into context; don't take the independent scores too seriously ;) (a) Is that so? Since each problem has a different number of points, the expected value isn't the same. So summing together the distributions could potentially be quite lumpy. (b) I am aware of that, but I'm not totally sure how that would help, unless all expected values are the same. (c) I did not know this. If the points for each problem WERE the same, could you then just combine them somehow? Multiply their standard deviations together or something?
$endgroup$
– Antonio
Dec 23 '18 at 2:04
$begingroup$
That's my bad; I mixed my vocab way up. I think I get what you're hinting at, and I think I got a solution. I did my best to validate it by generating random gaussians and adding them together, and it seems to be good. Please do let me know if I messed anything up.
$endgroup$
– Antonio
Dec 23 '18 at 5:28