What is the probability distribution of random samples of apparent diameter of a oblate spheroid?
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What is the probability distribution of random samples of apparent diameter of a oblate spheroid?
If I am asked to measure the “diameter” with an oblate spheroid, (a surface of revolution of an ellipse around its minor axis), I will get a value between the minor diameter and the major diameter of the generating ellipse. If I measure a lot, randomizing the samples, I will get a distribution of values which ranges between the minor and major diameters. What will the distribution look like?
I imagine that since there is a line around the equator where the maximum measurement can be made and a single point at the poles where the minimum measurement can be made, the distribution will be higher at the maximum end than at the minimum end. Meanwhile, if we measure from, say 20 degrees north latitude across to 20 degrees south latitude on the other side, we will get a slightly lower than maximum measurement and since there are two, their probability will be slightly higher than that of the maximum. That is about where I stall out.
I considered trying a surface integral over the area of the spheroid, integrating the diameter of the spheroid at each point. If I were to plot how much area corresponds to each diameter, normalized so the area under the curve is one, I think I would have what I want, I think.
I was hoping somebody anybody could throw some light, either telling me where I am going wrong or give some pointers to send me in the right direction.
The back story:
When I was in engineering school they had us repeatedly measure a glass marble with a micrometer to get a collection of values. We then made histograms of the data and tried to make it fit a normal distribution. Since a normal distribution goes from negative infinity to positive infinity (with an area under the curve of one) and the source population of our samples was constrained by the size of the marble, the data was anything but normally distributed.
I saw a calculation of the probability distribution of samples from a sine wave and wondered about the distribution for samples from a spheroid.
As we all know, continuous probability distributions can be used to make guesses about ranges of values. The area under the Probability Distribution Function (PDF) curve is one because if you integrate from negative to positive infinity, all values will be represented; the probability of a sample being between negative and positive infinity is one. The probability of ranges of values can be found by integrating the PDF between the range lower and maximum values. The probability of getting exactly X is zero. The probability of getting between X-5% and X+5% is finite and depends upon the distribution.
probability probability-distributions
asked Dec 15 '18 at 22:52
user1683793user1683793
13614
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add a comment |
$begingroup$
What is the probability distribution of random samples of apparent diameter of a oblate spheroid?
If I am asked to measure the “diameter” with an oblate spheroid, (a surface of revolution of an ellipse around its minor axis), I will get a value between the minor diameter and the major diameter of the generating ellipse. If I measure a lot, randomizing the samples, I will get a distribution of values which ranges between the minor and major diameters. What will the distribution look like?
I imagine that since there is a line around the equator where the maximum measurement can be made and a single point at the poles where the minimum measurement can be made, the distribution will be higher at the maximum end than at the minimum end. Meanwhile, if we measure from, say 20 degrees north latitude across to 20 degrees south latitude on the other side, we will get a slightly lower than maximum measurement and since there are two, their probability will be slightly higher than that of the maximum. That is about where I stall out.
I considered trying a surface integral over the area of the spheroid, integrating the diameter of the spheroid at each point. If I were to plot how much area corresponds to each diameter, normalized so the area under the curve is one, I think I would have what I want, I think.
I was hoping somebody anybody could throw some light, either telling me where I am going wrong or give some pointers to send me in the right direction.
The back story:
When I was in engineering school they had us repeatedly measure a glass marble with a micrometer to get a collection of values. We then made histograms of the data and tried to make it fit a normal distribution. Since a normal distribution goes from negative infinity to positive infinity (with an area under the curve of one) and the source population of our samples was constrained by the size of the marble, the data was anything but normally distributed.
I saw a calculation of the probability distribution of samples from a sine wave and wondered about the distribution for samples from a spheroid.
As we all know, continuous probability distributions can be used to make guesses about ranges of values. The area under the Probability Distribution Function (PDF) curve is one because if you integrate from negative to positive infinity, all values will be represented; the probability of a sample being between negative and positive infinity is one. The probability of ranges of values can be found by integrating the PDF between the range lower and maximum values. The probability of getting exactly X is zero. The probability of getting between X-5% and X+5% is finite and depends upon the distribution.
probability probability-distributions
asked Dec 15 '18 at 22:52
user1683793user1683793
13614
$endgroup$
1
$begingroup$
You could approach it geometrically: Calculate the diameters parametrized by the spherical coordinates $phi$ and $theta$, being "angle with the x-y-plane" and "angle in x-y-plane with positive x-axis", respectively. Then you could use calculus to calculate a c.d.f. of the "diameter" random variable $D$: $P_D(x)=Pr(D leq x)$ as a geometric probability based on the areas of regions in $phi-theta$ space.
$endgroup$
– Matthias
Dec 15 '18 at 23:04
1
$begingroup$
Very good. (VERY good.) That is kind of like my surface integral idea but you actually put it together. I believe that will work... Excuse me, I have some calculus to do.
$endgroup$
– user1683793
Dec 15 '18 at 23:28
$begingroup$
Your note speaks of "randomizing the samples". You need to define the method you are using to define randomizing.
$endgroup$
– herb steinberg
Dec 16 '18 at 0:51
$begingroup$
The idea is I want to measure many, random, places on the object. In school, they had us place the marble on the desk between each sample. I believe that introduced a bias since the marble rolled to lower the center of gravity. Maybe samples could be randomized by, say, throwing the sphere in air and catching before measuring. The objective is to find a distribution that might result from truly random measurements.
$endgroup$
– user1683793
Dec 16 '18 at 1:06
add a comment |
$begingroup$
What is the probability distribution of random samples of apparent diameter of a oblate spheroid?
If I am asked to measure the “diameter” with an oblate spheroid, (a surface of revolution of an ellipse around its minor axis), I will get a value between the minor diameter and the major diameter of the generating ellipse. If I measure a lot, randomizing the samples, I will get a distribution of values which ranges between the minor and major diameters. What will the distribution look like?
I imagine that since there is a line around the equator where the maximum measurement can be made and a single point at the poles where the minimum measurement can be made, the distribution will be higher at the maximum end than at the minimum end. Meanwhile, if we measure from, say 20 degrees north latitude across to 20 degrees south latitude on the other side, we will get a slightly lower than maximum measurement and since there are two, their probability will be slightly higher than that of the maximum. That is about where I stall out.
I considered trying a surface integral over the area of the spheroid, integrating the diameter of the spheroid at each point. If I were to plot how much area corresponds to each diameter, normalized so the area under the curve is one, I think I would have what I want, I think.
I was hoping somebody anybody could throw some light, either telling me where I am going wrong or give some pointers to send me in the right direction.
The back story:
When I was in engineering school they had us repeatedly measure a glass marble with a micrometer to get a collection of values. We then made histograms of the data and tried to make it fit a normal distribution. Since a normal distribution goes from negative infinity to positive infinity (with an area under the curve of one) and the source population of our samples was constrained by the size of the marble, the data was anything but normally distributed.
I saw a calculation of the probability distribution of samples from a sine wave and wondered about the distribution for samples from a spheroid.
As we all know, continuous probability distributions can be used to make guesses about ranges of values. The area under the Probability Distribution Function (PDF) curve is one because if you integrate from negative to positive infinity, all values will be represented; the probability of a sample being between negative and positive infinity is one. The probability of ranges of values can be found by integrating the PDF between the range lower and maximum values. The probability of getting exactly X is zero. The probability of getting between X-5% and X+5% is finite and depends upon the distribution.
probability probability-distributions
asked Dec 15 '18 at 22:52
user1683793user1683793
13614
$endgroup$
What is the probability distribution of random samples of apparent diameter of a oblate spheroid?
If I am asked to measure the “diameter” with an oblate spheroid, (a surface of revolution of an ellipse around its minor axis), I will get a value between the minor diameter and the major diameter of the generating ellipse. If I measure a lot, randomizing the samples, I will get a distribution of values which ranges between the minor and major diameters. What will the distribution look like?
I imagine that since there is a line around the equator where the maximum measurement can be made and a single point at the poles where the minimum measurement can be made, the distribution will be higher at the maximum end than at the minimum end. Meanwhile, if we measure from, say 20 degrees north latitude across to 20 degrees south latitude on the other side, we will get a slightly lower than maximum measurement and since there are two, their probability will be slightly higher than that of the maximum. That is about where I stall out.
I considered trying a surface integral over the area of the spheroid, integrating the diameter of the spheroid at each point. If I were to plot how much area corresponds to each diameter, normalized so the area under the curve is one, I think I would have what I want, I think.
I was hoping somebody anybody could throw some light, either telling me where I am going wrong or give some pointers to send me in the right direction.
The back story:
When I was in engineering school they had us repeatedly measure a glass marble with a micrometer to get a collection of values. We then made histograms of the data and tried to make it fit a normal distribution. Since a normal distribution goes from negative infinity to positive infinity (with an area under the curve of one) and the source population of our samples was constrained by the size of the marble, the data was anything but normally distributed.
I saw a calculation of the probability distribution of samples from a sine wave and wondered about the distribution for samples from a spheroid.
As we all know, continuous probability distributions can be used to make guesses about ranges of values. The area under the Probability Distribution Function (PDF) curve is one because if you integrate from negative to positive infinity, all values will be represented; the probability of a sample being between negative and positive infinity is one. The probability of ranges of values can be found by integrating the PDF between the range lower and maximum values. The probability of getting exactly X is zero. The probability of getting between X-5% and X+5% is finite and depends upon the distribution.
probability probability-distributions
probability probability-distributions
asked Dec 15 '18 at 22:52
user1683793user1683793
13614
asked Dec 15 '18 at 22:52
user1683793user1683793
13614
edited Dec 15 '18 at 23:32
user1683793
asked Dec 15 '18 at 22:52
user1683793user1683793
13614
asked Dec 15 '18 at 22:52
user1683793user1683793
13614
asked Dec 15 '18 at 22:52
user1683793user1683793
13614
13614
1
$begingroup$
You could approach it geometrically: Calculate the diameters parametrized by the spherical coordinates $phi$ and $theta$, being "angle with the x-y-plane" and "angle in x-y-plane with positive x-axis", respectively. Then you could use calculus to calculate a c.d.f. of the "diameter" random variable $D$: $P_D(x)=Pr(D leq x)$ as a geometric probability based on the areas of regions in $phi-theta$ space.
$endgroup$
– Matthias
Dec 15 '18 at 23:04
1
$begingroup$
Very good. (VERY good.) That is kind of like my surface integral idea but you actually put it together. I believe that will work... Excuse me, I have some calculus to do.
$endgroup$
– user1683793
Dec 15 '18 at 23:28
$begingroup$
Your note speaks of "randomizing the samples". You need to define the method you are using to define randomizing.
$endgroup$
– herb steinberg
Dec 16 '18 at 0:51
$begingroup$
The idea is I want to measure many, random, places on the object. In school, they had us place the marble on the desk between each sample. I believe that introduced a bias since the marble rolled to lower the center of gravity. Maybe samples could be randomized by, say, throwing the sphere in air and catching before measuring. The objective is to find a distribution that might result from truly random measurements.
$endgroup$
– user1683793
Dec 16 '18 at 1:06
add a comment |
1
$begingroup$
You could approach it geometrically: Calculate the diameters parametrized by the spherical coordinates $phi$ and $theta$, being "angle with the x-y-plane" and "angle in x-y-plane with positive x-axis", respectively. Then you could use calculus to calculate a c.d.f. of the "diameter" random variable $D$: $P_D(x)=Pr(D leq x)$ as a geometric probability based on the areas of regions in $phi-theta$ space.
$endgroup$
– Matthias
Dec 15 '18 at 23:04
1
$begingroup$
Very good. (VERY good.) That is kind of like my surface integral idea but you actually put it together. I believe that will work... Excuse me, I have some calculus to do.
$endgroup$
– user1683793
Dec 15 '18 at 23:28
$begingroup$
Your note speaks of "randomizing the samples". You need to define the method you are using to define randomizing.
$endgroup$
– herb steinberg
Dec 16 '18 at 0:51
$begingroup$
The idea is I want to measure many, random, places on the object. In school, they had us place the marble on the desk between each sample. I believe that introduced a bias since the marble rolled to lower the center of gravity. Maybe samples could be randomized by, say, throwing the sphere in air and catching before measuring. The objective is to find a distribution that might result from truly random measurements.
$endgroup$
– user1683793
Dec 16 '18 at 1:06
1
1
$begingroup$
You could approach it geometrically: Calculate the diameters parametrized by the spherical coordinates $phi$ and $theta$, being "angle with the x-y-plane" and "angle in x-y-plane with positive x-axis", respectively. Then you could use calculus to calculate a c.d.f. of the "diameter" random variable $D$: $P_D(x)=Pr(D leq x)$ as a geometric probability based on the areas of regions in $phi-theta$ space.
$endgroup$
– Matthias
Dec 15 '18 at 23:04
$begingroup$
You could approach it geometrically: Calculate the diameters parametrized by the spherical coordinates $phi$ and $theta$, being "angle with the x-y-plane" and "angle in x-y-plane with positive x-axis", respectively. Then you could use calculus to calculate a c.d.f. of the "diameter" random variable $D$: $P_D(x)=Pr(D leq x)$ as a geometric probability based on the areas of regions in $phi-theta$ space.
$endgroup$
– Matthias
Dec 15 '18 at 23:04
1
1
$begingroup$
Very good. (VERY good.) That is kind of like my surface integral idea but you actually put it together. I believe that will work... Excuse me, I have some calculus to do.
$endgroup$
– user1683793
Dec 15 '18 at 23:28
$begingroup$
Very good. (VERY good.) That is kind of like my surface integral idea but you actually put it together. I believe that will work... Excuse me, I have some calculus to do.
$endgroup$
– user1683793
Dec 15 '18 at 23:28
$begingroup$
Your note speaks of "randomizing the samples". You need to define the method you are using to define randomizing.
$endgroup$
– herb steinberg
Dec 16 '18 at 0:51
$begingroup$
Your note speaks of "randomizing the samples". You need to define the method you are using to define randomizing.
$endgroup$
– herb steinberg
Dec 16 '18 at 0:51
$begingroup$
The idea is I want to measure many, random, places on the object. In school, they had us place the marble on the desk between each sample. I believe that introduced a bias since the marble rolled to lower the center of gravity. Maybe samples could be randomized by, say, throwing the sphere in air and catching before measuring. The objective is to find a distribution that might result from truly random measurements.
$endgroup$
– user1683793
Dec 16 '18 at 1:06
$begingroup$
The idea is I want to measure many, random, places on the object. In school, they had us place the marble on the desk between each sample. I believe that introduced a bias since the marble rolled to lower the center of gravity. Maybe samples could be randomized by, say, throwing the sphere in air and catching before measuring. The objective is to find a distribution that might result from truly random measurements.
$endgroup$
– user1683793
Dec 16 '18 at 1:06
add a comment |
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You could approach it geometrically: Calculate the diameters parametrized by the spherical coordinates $phi$ and $theta$, being "angle with the x-y-plane" and "angle in x-y-plane with positive x-axis", respectively. Then you could use calculus to calculate a c.d.f. of the "diameter" random variable $D$: $P_D(x)=Pr(D leq x)$ as a geometric probability based on the areas of regions in $phi-theta$ space.
$endgroup$
– Matthias
Dec 15 '18 at 23:04
1
$begingroup$
Very good. (VERY good.) That is kind of like my surface integral idea but you actually put it together. I believe that will work... Excuse me, I have some calculus to do.
$endgroup$
– user1683793
Dec 15 '18 at 23:28
$begingroup$
Your note speaks of "randomizing the samples". You need to define the method you are using to define randomizing.
$endgroup$
– herb steinberg
Dec 16 '18 at 0:51
$begingroup$
The idea is I want to measure many, random, places on the object. In school, they had us place the marble on the desk between each sample. I believe that introduced a bias since the marble rolled to lower the center of gravity. Maybe samples could be randomized by, say, throwing the sphere in air and catching before measuring. The objective is to find a distribution that might result from truly random measurements.
$endgroup$
– user1683793
Dec 16 '18 at 1:06