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What is the expected value in a non-normal distribution?
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I'm really confused about an elementary aspect of distributions, and I can't quite seem to find an answer to this in my old stats texts.
I'm having a hard time understanding how the expected value or mean of a continuous distribution could be different from the median or the point where the probability dist function integrates to .5. And yet non-normal distributions like the log normal dist have different means and medians:
I think my whole intuition about the expected value was that it was the point where the pdf evaluates to .5. Can someone explain expected values more clearly than my stats text does?
probability normal-distribution
asked 5 hours ago
Ted Middleton
1062
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up vote
1
down vote
favorite
I'm really confused about an elementary aspect of distributions, and I can't quite seem to find an answer to this in my old stats texts.
I'm having a hard time understanding how the expected value or mean of a continuous distribution could be different from the median or the point where the probability dist function integrates to .5. And yet non-normal distributions like the log normal dist have different means and medians:
I think my whole intuition about the expected value was that it was the point where the pdf evaluates to .5. Can someone explain expected values more clearly than my stats text does?
probability normal-distribution
asked 5 hours ago
Ted Middleton
1062
The median is the point where the CDF evaluates to $1/2$ (really a point since this may not be unique, but usually is for continuous distributions). Points where the PDF evaluates to $1/2$ have no special meaning.
– spaceisdarkgreen
3 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm really confused about an elementary aspect of distributions, and I can't quite seem to find an answer to this in my old stats texts.
I'm having a hard time understanding how the expected value or mean of a continuous distribution could be different from the median or the point where the probability dist function integrates to .5. And yet non-normal distributions like the log normal dist have different means and medians:
I think my whole intuition about the expected value was that it was the point where the pdf evaluates to .5. Can someone explain expected values more clearly than my stats text does?
probability normal-distribution
asked 5 hours ago
Ted Middleton
1062
I'm really confused about an elementary aspect of distributions, and I can't quite seem to find an answer to this in my old stats texts.
I'm having a hard time understanding how the expected value or mean of a continuous distribution could be different from the median or the point where the probability dist function integrates to .5. And yet non-normal distributions like the log normal dist have different means and medians:
I think my whole intuition about the expected value was that it was the point where the pdf evaluates to .5. Can someone explain expected values more clearly than my stats text does?
probability normal-distribution
asked 5 hours ago
Ted Middleton
1062
asked 5 hours ago
Ted Middleton
1062
asked 5 hours ago
Ted Middleton
1062
asked 5 hours ago
Ted Middleton
1062
1062
The median is the point where the CDF evaluates to $1/2$ (really a point since this may not be unique, but usually is for continuous distributions). Points where the PDF evaluates to $1/2$ have no special meaning.
– spaceisdarkgreen
3 hours ago
add a comment |Â
The median is the point where the CDF evaluates to $1/2$ (really a point since this may not be unique, but usually is for continuous distributions). Points where the PDF evaluates to $1/2$ have no special meaning.
– spaceisdarkgreen
3 hours ago
The median is the point where the CDF evaluates to $1/2$ (really a point since this may not be unique, but usually is for continuous distributions). Points where the PDF evaluates to $1/2$ have no special meaning.
– spaceisdarkgreen
3 hours ago
The median is the point where the CDF evaluates to $1/2$ (really a point since this may not be unique, but usually is for continuous distributions). Points where the PDF evaluates to $1/2$ have no special meaning.
– spaceisdarkgreen
3 hours ago
add a comment |Â
2 Answers
2
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The mean $mu$ and the median $tildemu$ are equal if and only if the pdf $f$ is symmetric with respect to $x=mu$, that is, if $f(mu-x) =f(mu+x)$ for every $x>0$.
answered 5 hours ago
Rócherz
2,1811417
1
No, symmetry will make them equal but there are asymmetric distributions where the mean and median are equal.
– Ross Millikan
5 hours ago
Plus the mean has to exist. Take the Cauchy distribution for example.
– StubbornAtom
5 hours ago
add a comment |Â
up vote
0
down vote
Maybe it is easier to see with a finite distribution. Suppose the values are $1,1,1,2,2,2,2,2,5,5,100$. The median is $2$ because half the values are above and half below. The mean is $frac 111(1+1+1+2+2+2+2+2+5+5+100)approx 12.2$. The mode is again $2$ because there are more of them than anything else.
Which is useful depends on what you are using it for. The mean will give the average value over many samples. It gets pulled most by one outlier, here the $100$. The median might be "what you are close to" most of the time. The mode is the most likely value of a single sample.
answered 5 hours ago
Ross Millikan
275k21183348
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The mean $mu$ and the median $tildemu$ are equal if and only if the pdf $f$ is symmetric with respect to $x=mu$, that is, if $f(mu-x) =f(mu+x)$ for every $x>0$.
answered 5 hours ago
Rócherz
2,1811417
1
No, symmetry will make them equal but there are asymmetric distributions where the mean and median are equal.
– Ross Millikan
5 hours ago
Plus the mean has to exist. Take the Cauchy distribution for example.
– StubbornAtom
5 hours ago
add a comment |Â
up vote
0
down vote
The mean $mu$ and the median $tildemu$ are equal if and only if the pdf $f$ is symmetric with respect to $x=mu$, that is, if $f(mu-x) =f(mu+x)$ for every $x>0$.
answered 5 hours ago
Rócherz
2,1811417
1
No, symmetry will make them equal but there are asymmetric distributions where the mean and median are equal.
– Ross Millikan
5 hours ago
Plus the mean has to exist. Take the Cauchy distribution for example.
– StubbornAtom
5 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The mean $mu$ and the median $tildemu$ are equal if and only if the pdf $f$ is symmetric with respect to $x=mu$, that is, if $f(mu-x) =f(mu+x)$ for every $x>0$.
answered 5 hours ago
Rócherz
2,1811417
The mean $mu$ and the median $tildemu$ are equal if and only if the pdf $f$ is symmetric with respect to $x=mu$, that is, if $f(mu-x) =f(mu+x)$ for every $x>0$.
answered 5 hours ago
Rócherz
2,1811417
answered 5 hours ago
Rócherz
2,1811417
answered 5 hours ago
Rócherz
2,1811417
answered 5 hours ago
Rócherz
2,1811417
2,1811417
1
No, symmetry will make them equal but there are asymmetric distributions where the mean and median are equal.
– Ross Millikan
5 hours ago
Plus the mean has to exist. Take the Cauchy distribution for example.
– StubbornAtom
5 hours ago
add a comment |Â
1
No, symmetry will make them equal but there are asymmetric distributions where the mean and median are equal.
– Ross Millikan
5 hours ago
Plus the mean has to exist. Take the Cauchy distribution for example.
– StubbornAtom
5 hours ago
1
1
No, symmetry will make them equal but there are asymmetric distributions where the mean and median are equal.
– Ross Millikan
5 hours ago
No, symmetry will make them equal but there are asymmetric distributions where the mean and median are equal.
– Ross Millikan
5 hours ago
Plus the mean has to exist. Take the Cauchy distribution for example.
– StubbornAtom
5 hours ago
Plus the mean has to exist. Take the Cauchy distribution for example.
– StubbornAtom
5 hours ago
add a comment |Â
up vote
0
down vote
Maybe it is easier to see with a finite distribution. Suppose the values are $1,1,1,2,2,2,2,2,5,5,100$. The median is $2$ because half the values are above and half below. The mean is $frac 111(1+1+1+2+2+2+2+2+5+5+100)approx 12.2$. The mode is again $2$ because there are more of them than anything else.
Which is useful depends on what you are using it for. The mean will give the average value over many samples. It gets pulled most by one outlier, here the $100$. The median might be "what you are close to" most of the time. The mode is the most likely value of a single sample.
answered 5 hours ago
Ross Millikan
275k21183348
add a comment |Â
up vote
0
down vote
Maybe it is easier to see with a finite distribution. Suppose the values are $1,1,1,2,2,2,2,2,5,5,100$. The median is $2$ because half the values are above and half below. The mean is $frac 111(1+1+1+2+2+2+2+2+5+5+100)approx 12.2$. The mode is again $2$ because there are more of them than anything else.
Which is useful depends on what you are using it for. The mean will give the average value over many samples. It gets pulled most by one outlier, here the $100$. The median might be "what you are close to" most of the time. The mode is the most likely value of a single sample.
answered 5 hours ago
Ross Millikan
275k21183348
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Maybe it is easier to see with a finite distribution. Suppose the values are $1,1,1,2,2,2,2,2,5,5,100$. The median is $2$ because half the values are above and half below. The mean is $frac 111(1+1+1+2+2+2+2+2+5+5+100)approx 12.2$. The mode is again $2$ because there are more of them than anything else.
Which is useful depends on what you are using it for. The mean will give the average value over many samples. It gets pulled most by one outlier, here the $100$. The median might be "what you are close to" most of the time. The mode is the most likely value of a single sample.
answered 5 hours ago
Ross Millikan
275k21183348
Maybe it is easier to see with a finite distribution. Suppose the values are $1,1,1,2,2,2,2,2,5,5,100$. The median is $2$ because half the values are above and half below. The mean is $frac 111(1+1+1+2+2+2+2+2+5+5+100)approx 12.2$. The mode is again $2$ because there are more of them than anything else.
Which is useful depends on what you are using it for. The mean will give the average value over many samples. It gets pulled most by one outlier, here the $100$. The median might be "what you are close to" most of the time. The mode is the most likely value of a single sample.
answered 5 hours ago
Ross Millikan
275k21183348
answered 5 hours ago
Ross Millikan
275k21183348
answered 5 hours ago
Ross Millikan
275k21183348
answered 5 hours ago
Ross Millikan
275k21183348
275k21183348
add a comment |Â
add a comment |Â
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The median is the point where the CDF evaluates to $1/2$ (really a point since this may not be unique, but usually is for continuous distributions). Points where the PDF evaluates to $1/2$ have no special meaning.
– spaceisdarkgreen
3 hours ago