Average distance from center of circle

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Using calculus, we can show that the average distance of a point in a circle to the center is $2R/3$, where $R$ is the radius. However, I have a separate way of approaching this question through intuition that gives me a different answer, and I’d like to know why my intuition fails.

For each $thetain [0,2pi)$, we can consider the line segment of that angle from the center of the circle to the boundary. On this line segment, the average distance from the center should be $R/2$. Then the average distance from the center over all points in the circle should just be $R/2$ as well, since we can cover the circle with these line segments.

Why does this intuitive approach give the wrong answer? My best guess is that these line segments all share the origin, so this method counts the origin’s distance from itself multiple times, thereby throwing off the average by decreasing it, which agrees with the fact that we know the actual answer is greater.

However, couldn’t I just look at the average distance from the center for the open line segments that exclude the center? The average distance for these open line segments should still be $R/2$, and then I could apply the same argument for covering the circle with the open line segments. This time, I’d be missing the center, but missing a single point shouldn’t throw off the answer. Why does this argument not work?

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  • 5

    Related to Bertrand paradox: en.wikipedia.org/wiki/Bertrand_paradox_(probability)
    – z100
    Nov 29 at 20:57

  • Try your same argument, but with the diameter of the circle instead.
    – Bonnaduck
    Dec 3 at 15:51

up vote
18
down vote

favorite

3

Using calculus, we can show that the average distance of a point in a circle to the center is $2R/3$, where $R$ is the radius. However, I have a separate way of approaching this question through intuition that gives me a different answer, and I’d like to know why my intuition fails.

For each $thetain [0,2pi)$, we can consider the line segment of that angle from the center of the circle to the boundary. On this line segment, the average distance from the center should be $R/2$. Then the average distance from the center over all points in the circle should just be $R/2$ as well, since we can cover the circle with these line segments.

Why does this intuitive approach give the wrong answer? My best guess is that these line segments all share the origin, so this method counts the origin’s distance from itself multiple times, thereby throwing off the average by decreasing it, which agrees with the fact that we know the actual answer is greater.

However, couldn’t I just look at the average distance from the center for the open line segments that exclude the center? The average distance for these open line segments should still be $R/2$, and then I could apply the same argument for covering the circle with the open line segments. This time, I’d be missing the center, but missing a single point shouldn’t throw off the answer. Why does this argument not work?

share|cite|improve this question

  • 5

    Related to Bertrand paradox: en.wikipedia.org/wiki/Bertrand_paradox_(probability)
    – z100
    Nov 29 at 20:57

  • Try your same argument, but with the diameter of the circle instead.
    – Bonnaduck
    Dec 3 at 15:51

up vote
18
down vote

favorite

3

up vote
18
down vote

favorite

3
3

Using calculus, we can show that the average distance of a point in a circle to the center is $2R/3$, where $R$ is the radius. However, I have a separate way of approaching this question through intuition that gives me a different answer, and I’d like to know why my intuition fails.

For each $thetain [0,2pi)$, we can consider the line segment of that angle from the center of the circle to the boundary. On this line segment, the average distance from the center should be $R/2$. Then the average distance from the center over all points in the circle should just be $R/2$ as well, since we can cover the circle with these line segments.

Why does this intuitive approach give the wrong answer? My best guess is that these line segments all share the origin, so this method counts the origin’s distance from itself multiple times, thereby throwing off the average by decreasing it, which agrees with the fact that we know the actual answer is greater.

However, couldn’t I just look at the average distance from the center for the open line segments that exclude the center? The average distance for these open line segments should still be $R/2$, and then I could apply the same argument for covering the circle with the open line segments. This time, I’d be missing the center, but missing a single point shouldn’t throw off the answer. Why does this argument not work?

share|cite|improve this question

Using calculus, we can show that the average distance of a point in a circle to the center is $2R/3$, where $R$ is the radius. However, I have a separate way of approaching this question through intuition that gives me a different answer, and I’d like to know why my intuition fails.

For each $thetain [0,2pi)$, we can consider the line segment of that angle from the center of the circle to the boundary. On this line segment, the average distance from the center should be $R/2$. Then the average distance from the center over all points in the circle should just be $R/2$ as well, since we can cover the circle with these line segments.

Why does this intuitive approach give the wrong answer? My best guess is that these line segments all share the origin, so this method counts the origin’s distance from itself multiple times, thereby throwing off the average by decreasing it, which agrees with the fact that we know the actual answer is greater.

However, couldn’t I just look at the average distance from the center for the open line segments that exclude the center? The average distance for these open line segments should still be $R/2$, and then I could apply the same argument for covering the circle with the open line segments. This time, I’d be missing the center, but missing a single point shouldn’t throw off the answer. Why does this argument not work?

calculus probability geometry intuition

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edited Nov 29 at 21:21

asked Nov 29 at 20:30

Kevin Long

3,36121330

3,36121330

  • 5

    Related to Bertrand paradox: en.wikipedia.org/wiki/Bertrand_paradox_(probability)
    – z100
    Nov 29 at 20:57

  • Try your same argument, but with the diameter of the circle instead.
    – Bonnaduck
    Dec 3 at 15:51

  • 5

    Related to Bertrand paradox: en.wikipedia.org/wiki/Bertrand_paradox_(probability)
    – z100
    Nov 29 at 20:57

  • Try your same argument, but with the diameter of the circle instead.
    – Bonnaduck
    Dec 3 at 15:51

5

5

Related to Bertrand paradox: en.wikipedia.org/wiki/Bertrand_paradox_(probability)
– z100
Nov 29 at 20:57

Related to Bertrand paradox: en.wikipedia.org/wiki/Bertrand_paradox_(probability)
– z100
Nov 29 at 20:57

Try your same argument, but with the diameter of the circle instead.
– Bonnaduck
Dec 3 at 15:51

Try your same argument, but with the diameter of the circle instead.
– Bonnaduck
Dec 3 at 15:51

4 Answers
4

active

oldest

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up vote
14
down vote

accepted

Try actually drawing a circle and then drawing about twenty or thirty of your radial segments with their outer endpoints evenly spaced around the circumference of the circle. But don’t draw so many segments that there is no space between the segments.

Does your diagram look darker near the center of the circle than near the circumference?
It should look that way if you drew dark-colored lines on a light-colored surface.

This darker appearance reflects the notion that your distribution of “random” points is denser near the center of the circle.

If you draw two congruent circles within the larger circle, each much smaller than the larger circle, putting one of the small circles near the larger circle’s circumference and one near the center,
more of your radial lines will pass through the circle near the center than through the circle near the circumference.
As a result, your probability distribution is more likely to produce a point inside the small circle near the center than inside the small circle near the circumference.

When people speak of a uniform distribution over the area of a circle, they generally mean that any two congruent regions within the circle are equally likely to be “hit.”
That is, in a uniform distribution over the area of a circle, you would be equally likely to choose a point in either of the smaller circles in the previous paragraph.

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  • 3

    To be more mathematically precise: OP used polar coordinates and calculated $int_0^1 r dr = 1/2$ the average distance from one end on a strip. Higher “Density” is related to a smaller Jacobian – which would be $r$. If we take into account this correction factor and weight all points by this factor we find the correct answer: $int_0^1 int_0^{2pi} r r drdphi / pi = 2/3$
    – WorldSEnder
    Nov 29 at 23:26

up vote
9
down vote

You are looking at two different probability spaces. In the first case, you pick a point at random in the disk, sort of like throwing a dart at a dartboard, if we assume that the probability of landing in any small region is proportional to the area of the region.

In the second we first pick a radius, say by spinning the disk and choosing the vertical radius in the upper half-disk, as in wheel of fortune, and then we pick a point uniformly at random on that radius.

It’s somewhat surprising that we get different answers, I agree, but there’s no a priori reason why we should get the same answer, is there?

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  • I’m not too knowledgeable on probability, but isn’t the first case a uniform distribution of all points in the circle? If that’s the case, isn’t that equivalent to picking a radius uniformly, then a point on the radius uniformly?
    – Kevin Long
    Nov 29 at 20:44

  • 1

    @Kevin Long, no, you may simply use a simulation picking random points in both cases, in first case the density would look uniform on the disc, in 2nd one density would decrease from center to the border. However you cannot say the 1st is better.
    – z100
    Nov 29 at 20:52

  • 1

    @KevinLong No. Try it for a discrete distribution: draw a circle, then some number of evenly spaced (in angle) radii, and on each radius, some number of evenly spaced points. Look at the distribution of the points, and you will see that points near the center are clustered while those near the edge are spread out.
    – AlexanderJ93
    Nov 29 at 20:57

  • 1

    @z100 Thanks, I see now. I was working on a misunderstanding of uniform distribution.
    – Kevin Long
    Nov 29 at 21:14

  • 2

    @KevinLong I made a visual representation of the distribution here jsfiddle.net/13r60zdn
    – Kruga
    Nov 30 at 10:03

up vote
3
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The short answer is that you can’t cut a 2-dimensional object into 1-dimensional objects that have zero width. You could try cutting your circle into thick line segments, but that wouldn’t work because they would all crowd together near the center (and overlap, so you would be double counting).

Imagine you have a fat marker. You can easily draw a line from the center to the edge). But now, the second line that you will try to draw from the center has to start a little farther away from the true center, otherwise you’ll overlap. And after you’ve drawn about 4 lines (in the shape of a cross), the 5th line is going to be even harder to draw close to the center. My sketch shows the fat marker lines, and the green stars are the centers of the fat lines, and you can see they are not all the same distance.

enter image description here

If you want to cut your circle into smaller objects, then you could consider cutting it into triangles, and then calculating the average distance from the apex of the triangle. The more triangles you have, the less error you have near the circumference where the triangle’s straight edge underestimates the curved edge.

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  • 1

    Actually, your point about how thick line segments overlap is an intuitive explanation for why the expected value is greater than half. It’s exactly that the segments toward the centre have their probability “discounted” to adjust for it, while those toward the perimeter do not, and maintain “full value” as it were. So the points in each segment toward the perimeter have greater likelihood than those toward the centre.
    – Nij
    Nov 30 at 3:01

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0
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If you draw radii with uniform direction, then draw points uniformly along them, you don’t obtain a uniform distribution in the disk, and this biases the estimate in favor of shorter radii.

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    4 Answers
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    4 Answers
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    up vote
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    accepted

    Try actually drawing a circle and then drawing about twenty or thirty of your radial segments with their outer endpoints evenly spaced around the circumference of the circle. But don’t draw so many segments that there is no space between the segments.

    Does your diagram look darker near the center of the circle than near the circumference?
    It should look that way if you drew dark-colored lines on a light-colored surface.

    This darker appearance reflects the notion that your distribution of “random” points is denser near the center of the circle.

    If you draw two congruent circles within the larger circle, each much smaller than the larger circle, putting one of the small circles near the larger circle’s circumference and one near the center,
    more of your radial lines will pass through the circle near the center than through the circle near the circumference.
    As a result, your probability distribution is more likely to produce a point inside the small circle near the center than inside the small circle near the circumference.

    When people speak of a uniform distribution over the area of a circle, they generally mean that any two congruent regions within the circle are equally likely to be “hit.”
    That is, in a uniform distribution over the area of a circle, you would be equally likely to choose a point in either of the smaller circles in the previous paragraph.

    share|cite|improve this answer

    • 3

      To be more mathematically precise: OP used polar coordinates and calculated $int_0^1 r dr = 1/2$ the average distance from one end on a strip. Higher “Density” is related to a smaller Jacobian – which would be $r$. If we take into account this correction factor and weight all points by this factor we find the correct answer: $int_0^1 int_0^{2pi} r r drdphi / pi = 2/3$
      – WorldSEnder
      Nov 29 at 23:26

    up vote
    14
    down vote

    accepted

    Try actually drawing a circle and then drawing about twenty or thirty of your radial segments with their outer endpoints evenly spaced around the circumference of the circle. But don’t draw so many segments that there is no space between the segments.

    Does your diagram look darker near the center of the circle than near the circumference?
    It should look that way if you drew dark-colored lines on a light-colored surface.

    This darker appearance reflects the notion that your distribution of “random” points is denser near the center of the circle.

    If you draw two congruent circles within the larger circle, each much smaller than the larger circle, putting one of the small circles near the larger circle’s circumference and one near the center,
    more of your radial lines will pass through the circle near the center than through the circle near the circumference.
    As a result, your probability distribution is more likely to produce a point inside the small circle near the center than inside the small circle near the circumference.

    When people speak of a uniform distribution over the area of a circle, they generally mean that any two congruent regions within the circle are equally likely to be “hit.”
    That is, in a uniform distribution over the area of a circle, you would be equally likely to choose a point in either of the smaller circles in the previous paragraph.

    share|cite|improve this answer

    • 3

      To be more mathematically precise: OP used polar coordinates and calculated $int_0^1 r dr = 1/2$ the average distance from one end on a strip. Higher “Density” is related to a smaller Jacobian – which would be $r$. If we take into account this correction factor and weight all points by this factor we find the correct answer: $int_0^1 int_0^{2pi} r r drdphi / pi = 2/3$
      – WorldSEnder
      Nov 29 at 23:26

    up vote
    14
    down vote

    accepted

    up vote
    14
    down vote

    accepted

    Try actually drawing a circle and then drawing about twenty or thirty of your radial segments with their outer endpoints evenly spaced around the circumference of the circle. But don’t draw so many segments that there is no space between the segments.

    Does your diagram look darker near the center of the circle than near the circumference?
    It should look that way if you drew dark-colored lines on a light-colored surface.

    This darker appearance reflects the notion that your distribution of “random” points is denser near the center of the circle.

    If you draw two congruent circles within the larger circle, each much smaller than the larger circle, putting one of the small circles near the larger circle’s circumference and one near the center,
    more of your radial lines will pass through the circle near the center than through the circle near the circumference.
    As a result, your probability distribution is more likely to produce a point inside the small circle near the center than inside the small circle near the circumference.

    When people speak of a uniform distribution over the area of a circle, they generally mean that any two congruent regions within the circle are equally likely to be “hit.”
    That is, in a uniform distribution over the area of a circle, you would be equally likely to choose a point in either of the smaller circles in the previous paragraph.

    share|cite|improve this answer

    Try actually drawing a circle and then drawing about twenty or thirty of your radial segments with their outer endpoints evenly spaced around the circumference of the circle. But don’t draw so many segments that there is no space between the segments.

    Does your diagram look darker near the center of the circle than near the circumference?
    It should look that way if you drew dark-colored lines on a light-colored surface.

    This darker appearance reflects the notion that your distribution of “random” points is denser near the center of the circle.

    If you draw two congruent circles within the larger circle, each much smaller than the larger circle, putting one of the small circles near the larger circle’s circumference and one near the center,
    more of your radial lines will pass through the circle near the center than through the circle near the circumference.
    As a result, your probability distribution is more likely to produce a point inside the small circle near the center than inside the small circle near the circumference.

    When people speak of a uniform distribution over the area of a circle, they generally mean that any two congruent regions within the circle are equally likely to be “hit.”
    That is, in a uniform distribution over the area of a circle, you would be equally likely to choose a point in either of the smaller circles in the previous paragraph.

    share|cite|improve this answer

    share|cite|improve this answer

    share|cite|improve this answer

    answered Nov 29 at 21:03

    David K

    51.8k340114

    51.8k340114

    • 3

      To be more mathematically precise: OP used polar coordinates and calculated $int_0^1 r dr = 1/2$ the average distance from one end on a strip. Higher “Density” is related to a smaller Jacobian – which would be $r$. If we take into account this correction factor and weight all points by this factor we find the correct answer: $int_0^1 int_0^{2pi} r r drdphi / pi = 2/3$
      – WorldSEnder
      Nov 29 at 23:26

    • 3

      To be more mathematically precise: OP used polar coordinates and calculated $int_0^1 r dr = 1/2$ the average distance from one end on a strip. Higher “Density” is related to a smaller Jacobian – which would be $r$. If we take into account this correction factor and weight all points by this factor we find the correct answer: $int_0^1 int_0^{2pi} r r drdphi / pi = 2/3$
      – WorldSEnder
      Nov 29 at 23:26

    3

    3

    To be more mathematically precise: OP used polar coordinates and calculated $int_0^1 r dr = 1/2$ the average distance from one end on a strip. Higher “Density” is related to a smaller Jacobian – which would be $r$. If we take into account this correction factor and weight all points by this factor we find the correct answer: $int_0^1 int_0^{2pi} r r drdphi / pi = 2/3$
    – WorldSEnder
    Nov 29 at 23:26

    To be more mathematically precise: OP used polar coordinates and calculated $int_0^1 r dr = 1/2$ the average distance from one end on a strip. Higher “Density” is related to a smaller Jacobian – which would be $r$. If we take into account this correction factor and weight all points by this factor we find the correct answer: $int_0^1 int_0^{2pi} r r drdphi / pi = 2/3$
    – WorldSEnder
    Nov 29 at 23:26

    up vote
    9
    down vote

    You are looking at two different probability spaces. In the first case, you pick a point at random in the disk, sort of like throwing a dart at a dartboard, if we assume that the probability of landing in any small region is proportional to the area of the region.

    In the second we first pick a radius, say by spinning the disk and choosing the vertical radius in the upper half-disk, as in wheel of fortune, and then we pick a point uniformly at random on that radius.

    It’s somewhat surprising that we get different answers, I agree, but there’s no a priori reason why we should get the same answer, is there?

    share|cite|improve this answer

    • I’m not too knowledgeable on probability, but isn’t the first case a uniform distribution of all points in the circle? If that’s the case, isn’t that equivalent to picking a radius uniformly, then a point on the radius uniformly?
      – Kevin Long
      Nov 29 at 20:44

    • 1

      @Kevin Long, no, you may simply use a simulation picking random points in both cases, in first case the density would look uniform on the disc, in 2nd one density would decrease from center to the border. However you cannot say the 1st is better.
      – z100
      Nov 29 at 20:52

    • 1

      @KevinLong No. Try it for a discrete distribution: draw a circle, then some number of evenly spaced (in angle) radii, and on each radius, some number of evenly spaced points. Look at the distribution of the points, and you will see that points near the center are clustered while those near the edge are spread out.
      – AlexanderJ93
      Nov 29 at 20:57

    • 1

      @z100 Thanks, I see now. I was working on a misunderstanding of uniform distribution.
      – Kevin Long
      Nov 29 at 21:14

    • 2

      @KevinLong I made a visual representation of the distribution here jsfiddle.net/13r60zdn
      – Kruga
      Nov 30 at 10:03

    up vote
    9
    down vote

    You are looking at two different probability spaces. In the first case, you pick a point at random in the disk, sort of like throwing a dart at a dartboard, if we assume that the probability of landing in any small region is proportional to the area of the region.

    In the second we first pick a radius, say by spinning the disk and choosing the vertical radius in the upper half-disk, as in wheel of fortune, and then we pick a point uniformly at random on that radius.

    It’s somewhat surprising that we get different answers, I agree, but there’s no a priori reason why we should get the same answer, is there?

    share|cite|improve this answer

    • I’m not too knowledgeable on probability, but isn’t the first case a uniform distribution of all points in the circle? If that’s the case, isn’t that equivalent to picking a radius uniformly, then a point on the radius uniformly?
      – Kevin Long
      Nov 29 at 20:44

    • 1

      @Kevin Long, no, you may simply use a simulation picking random points in both cases, in first case the density would look uniform on the disc, in 2nd one density would decrease from center to the border. However you cannot say the 1st is better.
      – z100
      Nov 29 at 20:52

    • 1

      @KevinLong No. Try it for a discrete distribution: draw a circle, then some number of evenly spaced (in angle) radii, and on each radius, some number of evenly spaced points. Look at the distribution of the points, and you will see that points near the center are clustered while those near the edge are spread out.
      – AlexanderJ93
      Nov 29 at 20:57

    • 1

      @z100 Thanks, I see now. I was working on a misunderstanding of uniform distribution.
      – Kevin Long
      Nov 29 at 21:14

    • 2

      @KevinLong I made a visual representation of the distribution here jsfiddle.net/13r60zdn
      – Kruga
      Nov 30 at 10:03

    up vote
    9
    down vote

    up vote
    9
    down vote

    You are looking at two different probability spaces. In the first case, you pick a point at random in the disk, sort of like throwing a dart at a dartboard, if we assume that the probability of landing in any small region is proportional to the area of the region.

    In the second we first pick a radius, say by spinning the disk and choosing the vertical radius in the upper half-disk, as in wheel of fortune, and then we pick a point uniformly at random on that radius.

    It’s somewhat surprising that we get different answers, I agree, but there’s no a priori reason why we should get the same answer, is there?

    share|cite|improve this answer

    You are looking at two different probability spaces. In the first case, you pick a point at random in the disk, sort of like throwing a dart at a dartboard, if we assume that the probability of landing in any small region is proportional to the area of the region.

    In the second we first pick a radius, say by spinning the disk and choosing the vertical radius in the upper half-disk, as in wheel of fortune, and then we pick a point uniformly at random on that radius.

    It’s somewhat surprising that we get different answers, I agree, but there’s no a priori reason why we should get the same answer, is there?

    share|cite|improve this answer

    share|cite|improve this answer

    share|cite|improve this answer

    answered Nov 29 at 20:39

    saulspatz

    13.5k21327

    13.5k21327

    • I’m not too knowledgeable on probability, but isn’t the first case a uniform distribution of all points in the circle? If that’s the case, isn’t that equivalent to picking a radius uniformly, then a point on the radius uniformly?
      – Kevin Long
      Nov 29 at 20:44

    • 1

      @Kevin Long, no, you may simply use a simulation picking random points in both cases, in first case the density would look uniform on the disc, in 2nd one density would decrease from center to the border. However you cannot say the 1st is better.
      – z100
      Nov 29 at 20:52

    • 1

      @KevinLong No. Try it for a discrete distribution: draw a circle, then some number of evenly spaced (in angle) radii, and on each radius, some number of evenly spaced points. Look at the distribution of the points, and you will see that points near the center are clustered while those near the edge are spread out.
      – AlexanderJ93
      Nov 29 at 20:57

    • 1

      @z100 Thanks, I see now. I was working on a misunderstanding of uniform distribution.
      – Kevin Long
      Nov 29 at 21:14

    • 2

      @KevinLong I made a visual representation of the distribution here jsfiddle.net/13r60zdn
      – Kruga
      Nov 30 at 10:03

    • I’m not too knowledgeable on probability, but isn’t the first case a uniform distribution of all points in the circle? If that’s the case, isn’t that equivalent to picking a radius uniformly, then a point on the radius uniformly?
      – Kevin Long
      Nov 29 at 20:44

    • 1

      @Kevin Long, no, you may simply use a simulation picking random points in both cases, in first case the density would look uniform on the disc, in 2nd one density would decrease from center to the border. However you cannot say the 1st is better.
      – z100
      Nov 29 at 20:52

    • 1

      @KevinLong No. Try it for a discrete distribution: draw a circle, then some number of evenly spaced (in angle) radii, and on each radius, some number of evenly spaced points. Look at the distribution of the points, and you will see that points near the center are clustered while those near the edge are spread out.
      – AlexanderJ93
      Nov 29 at 20:57

    • 1

      @z100 Thanks, I see now. I was working on a misunderstanding of uniform distribution.
      – Kevin Long
      Nov 29 at 21:14

    • 2

      @KevinLong I made a visual representation of the distribution here jsfiddle.net/13r60zdn
      – Kruga
      Nov 30 at 10:03

    I’m not too knowledgeable on probability, but isn’t the first case a uniform distribution of all points in the circle? If that’s the case, isn’t that equivalent to picking a radius uniformly, then a point on the radius uniformly?
    – Kevin Long
    Nov 29 at 20:44

    I’m not too knowledgeable on probability, but isn’t the first case a uniform distribution of all points in the circle? If that’s the case, isn’t that equivalent to picking a radius uniformly, then a point on the radius uniformly?
    – Kevin Long
    Nov 29 at 20:44

    1

    1

    @Kevin Long, no, you may simply use a simulation picking random points in both cases, in first case the density would look uniform on the disc, in 2nd one density would decrease from center to the border. However you cannot say the 1st is better.
    – z100
    Nov 29 at 20:52

    @Kevin Long, no, you may simply use a simulation picking random points in both cases, in first case the density would look uniform on the disc, in 2nd one density would decrease from center to the border. However you cannot say the 1st is better.
    – z100
    Nov 29 at 20:52

    1

    1

    @KevinLong No. Try it for a discrete distribution: draw a circle, then some number of evenly spaced (in angle) radii, and on each radius, some number of evenly spaced points. Look at the distribution of the points, and you will see that points near the center are clustered while those near the edge are spread out.
    – AlexanderJ93
    Nov 29 at 20:57

    @KevinLong No. Try it for a discrete distribution: draw a circle, then some number of evenly spaced (in angle) radii, and on each radius, some number of evenly spaced points. Look at the distribution of the points, and you will see that points near the center are clustered while those near the edge are spread out.
    – AlexanderJ93
    Nov 29 at 20:57

    1

    1

    @z100 Thanks, I see now. I was working on a misunderstanding of uniform distribution.
    – Kevin Long
    Nov 29 at 21:14

    @z100 Thanks, I see now. I was working on a misunderstanding of uniform distribution.
    – Kevin Long
    Nov 29 at 21:14

    2

    2

    @KevinLong I made a visual representation of the distribution here jsfiddle.net/13r60zdn
    – Kruga
    Nov 30 at 10:03

    @KevinLong I made a visual representation of the distribution here jsfiddle.net/13r60zdn
    – Kruga
    Nov 30 at 10:03

    up vote
    3
    down vote

    The short answer is that you can’t cut a 2-dimensional object into 1-dimensional objects that have zero width. You could try cutting your circle into thick line segments, but that wouldn’t work because they would all crowd together near the center (and overlap, so you would be double counting).

    Imagine you have a fat marker. You can easily draw a line from the center to the edge). But now, the second line that you will try to draw from the center has to start a little farther away from the true center, otherwise you’ll overlap. And after you’ve drawn about 4 lines (in the shape of a cross), the 5th line is going to be even harder to draw close to the center. My sketch shows the fat marker lines, and the green stars are the centers of the fat lines, and you can see they are not all the same distance.

    enter image description here

    If you want to cut your circle into smaller objects, then you could consider cutting it into triangles, and then calculating the average distance from the apex of the triangle. The more triangles you have, the less error you have near the circumference where the triangle’s straight edge underestimates the curved edge.

    share|cite|improve this answer

    • 1

      Actually, your point about how thick line segments overlap is an intuitive explanation for why the expected value is greater than half. It’s exactly that the segments toward the centre have their probability “discounted” to adjust for it, while those toward the perimeter do not, and maintain “full value” as it were. So the points in each segment toward the perimeter have greater likelihood than those toward the centre.
      – Nij
      Nov 30 at 3:01

    up vote
    3
    down vote

    The short answer is that you can’t cut a 2-dimensional object into 1-dimensional objects that have zero width. You could try cutting your circle into thick line segments, but that wouldn’t work because they would all crowd together near the center (and overlap, so you would be double counting).

    Imagine you have a fat marker. You can easily draw a line from the center to the edge). But now, the second line that you will try to draw from the center has to start a little farther away from the true center, otherwise you’ll overlap. And after you’ve drawn about 4 lines (in the shape of a cross), the 5th line is going to be even harder to draw close to the center. My sketch shows the fat marker lines, and the green stars are the centers of the fat lines, and you can see they are not all the same distance.

    enter image description here

    If you want to cut your circle into smaller objects, then you could consider cutting it into triangles, and then calculating the average distance from the apex of the triangle. The more triangles you have, the less error you have near the circumference where the triangle’s straight edge underestimates the curved edge.

    share|cite|improve this answer

    • 1

      Actually, your point about how thick line segments overlap is an intuitive explanation for why the expected value is greater than half. It’s exactly that the segments toward the centre have their probability “discounted” to adjust for it, while those toward the perimeter do not, and maintain “full value” as it were. So the points in each segment toward the perimeter have greater likelihood than those toward the centre.
      – Nij
      Nov 30 at 3:01

    up vote
    3
    down vote

    up vote
    3
    down vote

    The short answer is that you can’t cut a 2-dimensional object into 1-dimensional objects that have zero width. You could try cutting your circle into thick line segments, but that wouldn’t work because they would all crowd together near the center (and overlap, so you would be double counting).

    Imagine you have a fat marker. You can easily draw a line from the center to the edge). But now, the second line that you will try to draw from the center has to start a little farther away from the true center, otherwise you’ll overlap. And after you’ve drawn about 4 lines (in the shape of a cross), the 5th line is going to be even harder to draw close to the center. My sketch shows the fat marker lines, and the green stars are the centers of the fat lines, and you can see they are not all the same distance.

    enter image description here

    If you want to cut your circle into smaller objects, then you could consider cutting it into triangles, and then calculating the average distance from the apex of the triangle. The more triangles you have, the less error you have near the circumference where the triangle’s straight edge underestimates the curved edge.

    share|cite|improve this answer

    The short answer is that you can’t cut a 2-dimensional object into 1-dimensional objects that have zero width. You could try cutting your circle into thick line segments, but that wouldn’t work because they would all crowd together near the center (and overlap, so you would be double counting).

    Imagine you have a fat marker. You can easily draw a line from the center to the edge). But now, the second line that you will try to draw from the center has to start a little farther away from the true center, otherwise you’ll overlap. And after you’ve drawn about 4 lines (in the shape of a cross), the 5th line is going to be even harder to draw close to the center. My sketch shows the fat marker lines, and the green stars are the centers of the fat lines, and you can see they are not all the same distance.

    enter image description here

    If you want to cut your circle into smaller objects, then you could consider cutting it into triangles, and then calculating the average distance from the apex of the triangle. The more triangles you have, the less error you have near the circumference where the triangle’s straight edge underestimates the curved edge.

    share|cite|improve this answer

    share|cite|improve this answer

    share|cite|improve this answer

    edited Nov 30 at 23:17

    answered Nov 29 at 22:20

    Mark Lakata

    98057

    98057

    • 1

      Actually, your point about how thick line segments overlap is an intuitive explanation for why the expected value is greater than half. It’s exactly that the segments toward the centre have their probability “discounted” to adjust for it, while those toward the perimeter do not, and maintain “full value” as it were. So the points in each segment toward the perimeter have greater likelihood than those toward the centre.
      – Nij
      Nov 30 at 3:01

    • 1

      Actually, your point about how thick line segments overlap is an intuitive explanation for why the expected value is greater than half. It’s exactly that the segments toward the centre have their probability “discounted” to adjust for it, while those toward the perimeter do not, and maintain “full value” as it were. So the points in each segment toward the perimeter have greater likelihood than those toward the centre.
      – Nij
      Nov 30 at 3:01

    1

    1

    Actually, your point about how thick line segments overlap is an intuitive explanation for why the expected value is greater than half. It’s exactly that the segments toward the centre have their probability “discounted” to adjust for it, while those toward the perimeter do not, and maintain “full value” as it were. So the points in each segment toward the perimeter have greater likelihood than those toward the centre.
    – Nij
    Nov 30 at 3:01

    Actually, your point about how thick line segments overlap is an intuitive explanation for why the expected value is greater than half. It’s exactly that the segments toward the centre have their probability “discounted” to adjust for it, while those toward the perimeter do not, and maintain “full value” as it were. So the points in each segment toward the perimeter have greater likelihood than those toward the centre.
    – Nij
    Nov 30 at 3:01

    up vote
    0
    down vote

    If you draw radii with uniform direction, then draw points uniformly along them, you don’t obtain a uniform distribution in the disk, and this biases the estimate in favor of shorter radii.

    share|cite|improve this answer

      up vote
      0
      down vote

      If you draw radii with uniform direction, then draw points uniformly along them, you don’t obtain a uniform distribution in the disk, and this biases the estimate in favor of shorter radii.

      share|cite|improve this answer

        up vote
        0
        down vote

        up vote
        0
        down vote

        If you draw radii with uniform direction, then draw points uniformly along them, you don’t obtain a uniform distribution in the disk, and this biases the estimate in favor of shorter radii.

        share|cite|improve this answer

        If you draw radii with uniform direction, then draw points uniformly along them, you don’t obtain a uniform distribution in the disk, and this biases the estimate in favor of shorter radii.

        share|cite|improve this answer

        share|cite|improve this answer

        share|cite|improve this answer

        answered Nov 29 at 21:17

        Yves Daoust

        123k668218

        123k668218

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