Ask HN: Regression to the mean math question
8 comments
Don't do a least squares line. That doesn't help. In the first plot, you'll see that in general:
x < mean => y > x
x > mean => y < x
If the scores are normalized. Regression to the mean is that most people move towards the mean in subsequent games/attempts/whatever.
But I fail to see what has changed in the analysis -- a and b are both just supposed to be samples from the same distribution, right?
Not at all. b is not independent of a, thats the whole point of regression to the mean. If you take ordered pairs where there is no connection between a and b, then you won't get any regression to the mean, you'll get points essentially randomly placed on the plane.
x < mean => y > x
x > mean => y < x
If the scores are normalized. Regression to the mean is that most people move towards the mean in subsequent games/attempts/whatever.
But I fail to see what has changed in the analysis -- a and b are both just supposed to be samples from the same distribution, right?
Not at all. b is not independent of a, thats the whole point of regression to the mean. If you take ordered pairs where there is no connection between a and b, then you won't get any regression to the mean, you'll get points essentially randomly placed on the plane.
Unfortunately what is really tripping me up is when we draw that line. I get the principle of regression to the mean, and why it occurs. What I don't get is this particular manifestation of it.
Fair point about "no connection between a and b".
What I should have said was something like: Why is it important that a come before b chronologically? If we were mistaken, and we thought that b came first, then what we would be seeing is "progression from the mean".
Does the concept of regression to the mean depend on the chronology of events? That would be weird -- most probability doesn't, right?
Fair point about "no connection between a and b".
What I should have said was something like: Why is it important that a come before b chronologically? If we were mistaken, and we thought that b came first, then what we would be seeing is "progression from the mean".
Does the concept of regression to the mean depend on the chronology of events? That would be weird -- most probability doesn't, right?
Hmm, I read this:
http://en.wikipedia.org/wiki/Regression_toward_the_mean
And I guess I made an assumption about the situation you described. If the students were all answering in identical random ways, then you'd see what I describe. I think this part of the wikipedia article describes it well:
"Consider a simple example: a class of students takes a 100-item true/false test on a subject. Suppose that all students choose randomly on all questions. Then, each student’s score would be a realization of one of a set of i.i.d. random variables, with a mean of 50. Naturally, some students will score substantially above 50 and some substantially below 50 just by chance. If one takes only the top scoring 10% of the students and gives them a second test on which they again guess on all items, the mean score would again be expected to be close to 50. Thus the mean of these students would “regress” all the way back to the mean of all students who took the original test. No matter what a student scores on the original test, the best prediction of their score on second test is 50."
So, to your question, why is time important? Its important in the sense that you need the first test to determine who your "high flyers" are for the second experiment.
http://en.wikipedia.org/wiki/Regression_toward_the_mean
And I guess I made an assumption about the situation you described. If the students were all answering in identical random ways, then you'd see what I describe. I think this part of the wikipedia article describes it well:
"Consider a simple example: a class of students takes a 100-item true/false test on a subject. Suppose that all students choose randomly on all questions. Then, each student’s score would be a realization of one of a set of i.i.d. random variables, with a mean of 50. Naturally, some students will score substantially above 50 and some substantially below 50 just by chance. If one takes only the top scoring 10% of the students and gives them a second test on which they again guess on all items, the mean score would again be expected to be close to 50. Thus the mean of these students would “regress” all the way back to the mean of all students who took the original test. No matter what a student scores on the original test, the best prediction of their score on second test is 50."
So, to your question, why is time important? Its important in the sense that you need the first test to determine who your "high flyers" are for the second experiment.
"If you take ordered pairs where there is no connection between a and b, then you won't get any regression to the mean, you'll get points essentially randomly placed on the plane."
This is exactly reversed. If A and B are perfectly correlated, then you will have no regression to the mean. If they are perfectly independent, then you will have full regression to the mean. If they are only partially correlated, then you will have only partial regression to the mean.
(This is easy to see if you run a simulation of each case.)
This is exactly reversed. If A and B are perfectly correlated, then you will have no regression to the mean. If they are perfectly independent, then you will have full regression to the mean. If they are only partially correlated, then you will have only partial regression to the mean.
(This is easy to see if you run a simulation of each case.)
Regression to the mean does not imply that the first slope should be less than one.
If for some reason only the above-average students regressed, then the slope would be <1. But regression to the mean also affects the scores of students who started below average; as a group we should expect them to regress upward toward the mean. Combine the two groups, and the effects exactly cancel out, leaving a slope of 1.
(Since you say the slope "should be" one, I assume the scores are normalized somehow so that the mean score for exam A is the same as the mean for exam B.)
If for some reason only the above-average students regressed, then the slope would be <1. But regression to the mean also affects the scores of students who started below average; as a group we should expect them to regress upward toward the mean. Combine the two groups, and the effects exactly cancel out, leaving a slope of 1.
(Since you say the slope "should be" one, I assume the scores are normalized somehow so that the mean score for exam A is the same as the mean for exam B.)
Not only is it true that the first slope doesn't have to be less than one, it could plausibly be greater than one.
Suppose the course material is really cumulative, so that some students "get it" and take off, while other students fall by the wayside. Then scoring well on the first test predicts scoring well on the second midterm, while scoring poorly on the first predicts scoring really badly on the second. Then the slope of your least-squares-fit line could easily be greater than 1.
In other words, the mean could stay the same, due to above-average students (on the first test) getting better, and below-average students getting worse. There's no reason to suppose that below-average students will magically get better.
Suppose the course material is really cumulative, so that some students "get it" and take off, while other students fall by the wayside. Then scoring well on the first test predicts scoring well on the second midterm, while scoring poorly on the first predicts scoring really badly on the second. Then the slope of your least-squares-fit line could easily be greater than 1.
In other words, the mean could stay the same, due to above-average students (on the first test) getting better, and below-average students getting worse. There's no reason to suppose that below-average students will magically get better.
Yep, I assumed they were normalized. I don't see the canceling out, though.
If you do poorly on the first test, the x-coordinate is low/close to y-axis. You're then expected to do better on the second test, so the y-coordinate is high. This will flatten the left half of the line.
As you said, if you do well on the first test, the x-coordinate is high, and the y-coordinate is low. This will flatten the right half of the line.
Right?
If you do poorly on the first test, the x-coordinate is low/close to y-axis. You're then expected to do better on the second test, so the y-coordinate is high. This will flatten the left half of the line.
As you said, if you do well on the first test, the x-coordinate is high, and the y-coordinate is low. This will flatten the right half of the line.
Right?
I'll assume up front that thee scores on the two tests are independent, since that's the scenario in which regression to the mean applies. (It also applies if they are correlated but have some independent "noise" component, but that complicates things.)
Your mistake is here:
"If you do poorly on the first test... you're then expected to do better on the second test, so the y-coordinate is high."
Actually, under my assumption, a student who does poorly on the first test is no more or less likely than anyone else to do well on the second test. Their y-coordinate will not be "high" in absolute terms; on average, it will be the same as the mean for the whole class. The regression exists because the group has a low starting point, not because it has a high ending point. As a group the high scorers will regress to the mean, not past it. (In the case where scores are partially correlated, the group will regress toward the mean.)
For example, suppose scores are independently, uniformly distributed on both tests. Then your scatterplot will have dots distributed uniformly over its entire area - obviously this does not change if you switch axes. And yet there is regression to the mean. Divide the graph into quadrants. If you look at the right half of the plot (high scorers on first exam), you'll see that there are as many in the upper quadrant as the lower (their mean on the second exam is the class mean). Same for the left side of the plot. This isn't order-dependent; you'll find that the high-scorers on B also regress to the mean when you look at their A scores.
Your mistake is here:
"If you do poorly on the first test... you're then expected to do better on the second test, so the y-coordinate is high."
Actually, under my assumption, a student who does poorly on the first test is no more or less likely than anyone else to do well on the second test. Their y-coordinate will not be "high" in absolute terms; on average, it will be the same as the mean for the whole class. The regression exists because the group has a low starting point, not because it has a high ending point. As a group the high scorers will regress to the mean, not past it. (In the case where scores are partially correlated, the group will regress toward the mean.)
For example, suppose scores are independently, uniformly distributed on both tests. Then your scatterplot will have dots distributed uniformly over its entire area - obviously this does not change if you switch axes. And yet there is regression to the mean. Divide the graph into quadrants. If you look at the right half of the plot (high scorers on first exam), you'll see that there are as many in the upper quadrant as the lower (their mean on the second exam is the class mean). Same for the left side of the plot. This isn't order-dependent; you'll find that the high-scorers on B also regress to the mean when you look at their A scores.
I have a question about regression to the mean.
Suppose you have a set of pairs (a,b) corresponding to students in a class. a = the student's score on the first midterm, b = score on second midterm.
If you plot the pairs with a on x-axis, b on y-axis, then get the least-squares line, you have an upward sloping line.
The line slope should be less than 1, indicating regression to the mean.
If you plot b on x-axis, a on y-axis, the slope is necessarily now greater than 1. But I fail to see what has changed in the analysis -- a and b are both just supposed to be samples from the same distribution, right?
This has been driving me crazy, so I'd love some help.
Thank you!