Right-skewed distribution with mean equals to mode?
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Is it possible to have a right-skewed distribution with mean equal to mode? If so, could you give me some example?
distributions mean skewness mode
edited yesterday
Nick Cox
39.4k588132
asked yesterday
Don TawanpitakDon Tawanpitak
363
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Don Tawanpitak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
Is it possible to have a right-skewed distribution with mean equal to mode? If so, could you give me some example?
distributions mean skewness mode
edited yesterday
Nick Cox
39.4k588132
asked yesterday
Don TawanpitakDon Tawanpitak
363
New contributor
Don Tawanpitak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
Take a suitable mixture of any finite-mean skewed distribution and any finite-mean unimodal symmetric distribution with the same mean. All continuous examples and all discrete examples arise in this way.
$endgroup$
– whuber♦
yesterday
$begingroup$
@whuber, That's a great idea. If you have time, it would be terrific if you made a slightly more detailed answer out of that.
$endgroup$
– beta1_equals_beta2
yesterday
add a comment |
$begingroup$
Is it possible to have a right-skewed distribution with mean equal to mode? If so, could you give me some example?
distributions mean skewness mode
edited yesterday
Nick Cox
39.4k588132
asked yesterday
Don TawanpitakDon Tawanpitak
363
New contributor
Don Tawanpitak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Is it possible to have a right-skewed distribution with mean equal to mode? If so, could you give me some example?
distributions mean skewness mode
distributions mean skewness mode
edited yesterday
Nick Cox
39.4k588132
asked yesterday
Don TawanpitakDon Tawanpitak
363
New contributor
Don Tawanpitak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Nick Cox
39.4k588132
asked yesterday
Don TawanpitakDon Tawanpitak
363
New contributor
Don Tawanpitak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Nick Cox
39.4k588132
edited yesterday
Nick Cox
39.4k588132
edited yesterday
Nick Cox
39.4k588132
39.4k588132
asked yesterday
Don TawanpitakDon Tawanpitak
363
New contributor
Don Tawanpitak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Don TawanpitakDon Tawanpitak
363
asked yesterday
Don TawanpitakDon Tawanpitak
363
363
New contributor
Don Tawanpitak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Don Tawanpitak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Don Tawanpitak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
Take a suitable mixture of any finite-mean skewed distribution and any finite-mean unimodal symmetric distribution with the same mean. All continuous examples and all discrete examples arise in this way.
$endgroup$
– whuber♦
yesterday
$begingroup$
@whuber, That's a great idea. If you have time, it would be terrific if you made a slightly more detailed answer out of that.
$endgroup$
– beta1_equals_beta2
yesterday
add a comment |
3
$begingroup$
Take a suitable mixture of any finite-mean skewed distribution and any finite-mean unimodal symmetric distribution with the same mean. All continuous examples and all discrete examples arise in this way.
$endgroup$
– whuber♦
yesterday
$begingroup$
@whuber, That's a great idea. If you have time, it would be terrific if you made a slightly more detailed answer out of that.
$endgroup$
– beta1_equals_beta2
yesterday
3
3
$begingroup$
Take a suitable mixture of any finite-mean skewed distribution and any finite-mean unimodal symmetric distribution with the same mean. All continuous examples and all discrete examples arise in this way.
$endgroup$
– whuber♦
yesterday
$begingroup$
Take a suitable mixture of any finite-mean skewed distribution and any finite-mean unimodal symmetric distribution with the same mean. All continuous examples and all discrete examples arise in this way.
$endgroup$
– whuber♦
yesterday
$begingroup$
@whuber, That's a great idea. If you have time, it would be terrific if you made a slightly more detailed answer out of that.
$endgroup$
– beta1_equals_beta2
yesterday
$begingroup$
@whuber, That's a great idea. If you have time, it would be terrific if you made a slightly more detailed answer out of that.
$endgroup$
– beta1_equals_beta2
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success 0.1. Then the mean is 10 $times$ 0.1 = 1, and 1 also is the mode (and for a bonus the median too), but the distribution is manifestly right skewed.
The code giving the number of successes 0 to 10 and their probabilities 0.348678... and so forth is Mata code from Stata, but your favourite statistical platform should be able to do it. (If not, you need a new favourite.)
: (0::10), binomialp(10, (0::10), 0.1)
1 2
+-----------------------------+
1 | 0 .3486784401 |
2 | 1 .387420489 |
3 | 2 .1937102445 |
4 | 3 .057395628 |
5 | 4 .011160261 |
6 | 5 .0014880348 |
7 | 6 .000137781 |
8 | 7 8.74800e-06 |
9 | 8 3.64500e-07 |
10 | 9 9.00000e-09 |
11 | 10 1.00000e-10 |
+-----------------------------+
Among continuous distributions, the Weibull distribution can show equal mean and mode yet be right-skewed.
answered yesterday
Nick CoxNick Cox
39.4k588132
$endgroup$
$begingroup$
Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
$endgroup$
– Don Tawanpitak
yesterday
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
yesterday
2
$begingroup$
@DonTawanpitak A quick numerical search for the Weibull only revealed one solution: $alpha = 3.3125, beta = 1$ where $alpha$ is the shape and $beta$ is the scale. The mode and mean are then $0.897186$. But this Weibull isn't terribly right skewed (its skewness is $0.074$).
$endgroup$
– COOLSerdash
yesterday
add a comment |
$begingroup$
If the distribution is discrete, sure. It's easy. For example, a distribution with probability mass function
$P(X=0) = 0.36$
$P(X=1) = 0.40$
$P(X=2) = 0.13$
$P(X=3) = 0.10$
$P(X=4) = 0.01$
is right (i.e. positively) skewed and has both a mean and a mode of 1.
answered yesterday
beta1_equals_beta2beta1_equals_beta2
1063
$endgroup$
$begingroup$
Is this suggesting that discrete distributions, as opposed to continuous distributions are more likely to exhibit this property? There seems to be a strong argument for that.
$endgroup$
– Thomas Cleberg
yesterday
1
$begingroup$
No, I'm not suggesting that. I'm just saying it's easy to come up with an example (which is all the OP asked for) of a discrete distribution with that property.
$endgroup$
– beta1_equals_beta2
yesterday
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success 0.1. Then the mean is 10 $times$ 0.1 = 1, and 1 also is the mode (and for a bonus the median too), but the distribution is manifestly right skewed.
The code giving the number of successes 0 to 10 and their probabilities 0.348678... and so forth is Mata code from Stata, but your favourite statistical platform should be able to do it. (If not, you need a new favourite.)
: (0::10), binomialp(10, (0::10), 0.1)
1 2
+-----------------------------+
1 | 0 .3486784401 |
2 | 1 .387420489 |
3 | 2 .1937102445 |
4 | 3 .057395628 |
5 | 4 .011160261 |
6 | 5 .0014880348 |
7 | 6 .000137781 |
8 | 7 8.74800e-06 |
9 | 8 3.64500e-07 |
10 | 9 9.00000e-09 |
11 | 10 1.00000e-10 |
+-----------------------------+
Among continuous distributions, the Weibull distribution can show equal mean and mode yet be right-skewed.
answered yesterday
Nick CoxNick Cox
39.4k588132
$endgroup$
$begingroup$
Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
$endgroup$
– Don Tawanpitak
yesterday
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
yesterday
2
$begingroup$
@DonTawanpitak A quick numerical search for the Weibull only revealed one solution: $alpha = 3.3125, beta = 1$ where $alpha$ is the shape and $beta$ is the scale. The mode and mean are then $0.897186$. But this Weibull isn't terribly right skewed (its skewness is $0.074$).
$endgroup$
– COOLSerdash
yesterday
add a comment |
$begingroup$
Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success 0.1. Then the mean is 10 $times$ 0.1 = 1, and 1 also is the mode (and for a bonus the median too), but the distribution is manifestly right skewed.
The code giving the number of successes 0 to 10 and their probabilities 0.348678... and so forth is Mata code from Stata, but your favourite statistical platform should be able to do it. (If not, you need a new favourite.)
: (0::10), binomialp(10, (0::10), 0.1)
1 2
+-----------------------------+
1 | 0 .3486784401 |
2 | 1 .387420489 |
3 | 2 .1937102445 |
4 | 3 .057395628 |
5 | 4 .011160261 |
6 | 5 .0014880348 |
7 | 6 .000137781 |
8 | 7 8.74800e-06 |
9 | 8 3.64500e-07 |
10 | 9 9.00000e-09 |
11 | 10 1.00000e-10 |
+-----------------------------+
Among continuous distributions, the Weibull distribution can show equal mean and mode yet be right-skewed.
answered yesterday
Nick CoxNick Cox
39.4k588132
$endgroup$
$begingroup$
Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
$endgroup$
– Don Tawanpitak
yesterday
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
yesterday
2
$begingroup$
@DonTawanpitak A quick numerical search for the Weibull only revealed one solution: $alpha = 3.3125, beta = 1$ where $alpha$ is the shape and $beta$ is the scale. The mode and mean are then $0.897186$. But this Weibull isn't terribly right skewed (its skewness is $0.074$).
$endgroup$
– COOLSerdash
yesterday
add a comment |
$begingroup$
Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success 0.1. Then the mean is 10 $times$ 0.1 = 1, and 1 also is the mode (and for a bonus the median too), but the distribution is manifestly right skewed.
The code giving the number of successes 0 to 10 and their probabilities 0.348678... and so forth is Mata code from Stata, but your favourite statistical platform should be able to do it. (If not, you need a new favourite.)
: (0::10), binomialp(10, (0::10), 0.1)
1 2
+-----------------------------+
1 | 0 .3486784401 |
2 | 1 .387420489 |
3 | 2 .1937102445 |
4 | 3 .057395628 |
5 | 4 .011160261 |
6 | 5 .0014880348 |
7 | 6 .000137781 |
8 | 7 8.74800e-06 |
9 | 8 3.64500e-07 |
10 | 9 9.00000e-09 |
11 | 10 1.00000e-10 |
+-----------------------------+
Among continuous distributions, the Weibull distribution can show equal mean and mode yet be right-skewed.
answered yesterday
Nick CoxNick Cox
39.4k588132
$endgroup$
Easy examples come from binomial distributions -- which can hardly be dismissed as pathological or as bizarre counter-examples constructed ad hoc. Here is one for 10 trials and probability of success 0.1. Then the mean is 10 $times$ 0.1 = 1, and 1 also is the mode (and for a bonus the median too), but the distribution is manifestly right skewed.
The code giving the number of successes 0 to 10 and their probabilities 0.348678... and so forth is Mata code from Stata, but your favourite statistical platform should be able to do it. (If not, you need a new favourite.)
: (0::10), binomialp(10, (0::10), 0.1)
1 2
+-----------------------------+
1 | 0 .3486784401 |
2 | 1 .387420489 |
3 | 2 .1937102445 |
4 | 3 .057395628 |
5 | 4 .011160261 |
6 | 5 .0014880348 |
7 | 6 .000137781 |
8 | 7 8.74800e-06 |
9 | 8 3.64500e-07 |
10 | 9 9.00000e-09 |
11 | 10 1.00000e-10 |
+-----------------------------+
Among continuous distributions, the Weibull distribution can show equal mean and mode yet be right-skewed.
answered yesterday
Nick CoxNick Cox
39.4k588132
edited yesterday
answered yesterday
Nick CoxNick Cox
39.4k588132
answered yesterday
Nick CoxNick Cox
39.4k588132
answered yesterday
Nick CoxNick Cox
39.4k588132
39.4k588132
$begingroup$
Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
$endgroup$
– Don Tawanpitak
yesterday
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
yesterday
2
$begingroup$
@DonTawanpitak A quick numerical search for the Weibull only revealed one solution: $alpha = 3.3125, beta = 1$ where $alpha$ is the shape and $beta$ is the scale. The mode and mean are then $0.897186$. But this Weibull isn't terribly right skewed (its skewness is $0.074$).
$endgroup$
– COOLSerdash
yesterday
add a comment |
$begingroup$
Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
$endgroup$
– Don Tawanpitak
yesterday
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
yesterday
2
$begingroup$
@DonTawanpitak A quick numerical search for the Weibull only revealed one solution: $alpha = 3.3125, beta = 1$ where $alpha$ is the shape and $beta$ is the scale. The mode and mean are then $0.897186$. But this Weibull isn't terribly right skewed (its skewness is $0.074$).
$endgroup$
– COOLSerdash
yesterday
$begingroup$
Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
$endgroup$
– Don Tawanpitak
yesterday
$begingroup$
Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
$endgroup$
– Don Tawanpitak
yesterday
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
yesterday
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
yesterday
2
2
$begingroup$
@DonTawanpitak A quick numerical search for the Weibull only revealed one solution: $alpha = 3.3125, beta = 1$ where $alpha$ is the shape and $beta$ is the scale. The mode and mean are then $0.897186$. But this Weibull isn't terribly right skewed (its skewness is $0.074$).
$endgroup$
– COOLSerdash
yesterday
$begingroup$
@DonTawanpitak A quick numerical search for the Weibull only revealed one solution: $alpha = 3.3125, beta = 1$ where $alpha$ is the shape and $beta$ is the scale. The mode and mean are then $0.897186$. But this Weibull isn't terribly right skewed (its skewness is $0.074$).
$endgroup$
– COOLSerdash
yesterday
add a comment |
$begingroup$
If the distribution is discrete, sure. It's easy. For example, a distribution with probability mass function
$P(X=0) = 0.36$
$P(X=1) = 0.40$
$P(X=2) = 0.13$
$P(X=3) = 0.10$
$P(X=4) = 0.01$
is right (i.e. positively) skewed and has both a mean and a mode of 1.
answered yesterday
beta1_equals_beta2beta1_equals_beta2
1063
$endgroup$
$begingroup$
Is this suggesting that discrete distributions, as opposed to continuous distributions are more likely to exhibit this property? There seems to be a strong argument for that.
$endgroup$
– Thomas Cleberg
yesterday
1
$begingroup$
No, I'm not suggesting that. I'm just saying it's easy to come up with an example (which is all the OP asked for) of a discrete distribution with that property.
$endgroup$
– beta1_equals_beta2
yesterday
add a comment |
$begingroup$
If the distribution is discrete, sure. It's easy. For example, a distribution with probability mass function
$P(X=0) = 0.36$
$P(X=1) = 0.40$
$P(X=2) = 0.13$
$P(X=3) = 0.10$
$P(X=4) = 0.01$
is right (i.e. positively) skewed and has both a mean and a mode of 1.
answered yesterday
beta1_equals_beta2beta1_equals_beta2
1063
$endgroup$
$begingroup$
Is this suggesting that discrete distributions, as opposed to continuous distributions are more likely to exhibit this property? There seems to be a strong argument for that.
$endgroup$
– Thomas Cleberg
yesterday
1
$begingroup$
No, I'm not suggesting that. I'm just saying it's easy to come up with an example (which is all the OP asked for) of a discrete distribution with that property.
$endgroup$
– beta1_equals_beta2
yesterday
add a comment |
$begingroup$
If the distribution is discrete, sure. It's easy. For example, a distribution with probability mass function
$P(X=0) = 0.36$
$P(X=1) = 0.40$
$P(X=2) = 0.13$
$P(X=3) = 0.10$
$P(X=4) = 0.01$
is right (i.e. positively) skewed and has both a mean and a mode of 1.
answered yesterday
beta1_equals_beta2beta1_equals_beta2
1063
$endgroup$
If the distribution is discrete, sure. It's easy. For example, a distribution with probability mass function
$P(X=0) = 0.36$
$P(X=1) = 0.40$
$P(X=2) = 0.13$
$P(X=3) = 0.10$
$P(X=4) = 0.01$
is right (i.e. positively) skewed and has both a mean and a mode of 1.
answered yesterday
beta1_equals_beta2beta1_equals_beta2
1063
answered yesterday
beta1_equals_beta2beta1_equals_beta2
1063
answered yesterday
beta1_equals_beta2beta1_equals_beta2
1063
answered yesterday
beta1_equals_beta2beta1_equals_beta2
1063
1063
$begingroup$
Is this suggesting that discrete distributions, as opposed to continuous distributions are more likely to exhibit this property? There seems to be a strong argument for that.
$endgroup$
– Thomas Cleberg
yesterday
1
$begingroup$
No, I'm not suggesting that. I'm just saying it's easy to come up with an example (which is all the OP asked for) of a discrete distribution with that property.
$endgroup$
– beta1_equals_beta2
yesterday
add a comment |
$begingroup$
Is this suggesting that discrete distributions, as opposed to continuous distributions are more likely to exhibit this property? There seems to be a strong argument for that.
$endgroup$
– Thomas Cleberg
yesterday
1
$begingroup$
No, I'm not suggesting that. I'm just saying it's easy to come up with an example (which is all the OP asked for) of a discrete distribution with that property.
$endgroup$
– beta1_equals_beta2
yesterday
$begingroup$
Is this suggesting that discrete distributions, as opposed to continuous distributions are more likely to exhibit this property? There seems to be a strong argument for that.
$endgroup$
– Thomas Cleberg
yesterday
$begingroup$
Is this suggesting that discrete distributions, as opposed to continuous distributions are more likely to exhibit this property? There seems to be a strong argument for that.
$endgroup$
– Thomas Cleberg
yesterday
1
1
$begingroup$
No, I'm not suggesting that. I'm just saying it's easy to come up with an example (which is all the OP asked for) of a discrete distribution with that property.
$endgroup$
– beta1_equals_beta2
yesterday
$begingroup$
No, I'm not suggesting that. I'm just saying it's easy to come up with an example (which is all the OP asked for) of a discrete distribution with that property.
$endgroup$
– beta1_equals_beta2
yesterday
add a comment |
Don Tawanpitak is a new contributor. Be nice, and check out our Code of Conduct.
Don Tawanpitak is a new contributor. Be nice, and check out our Code of Conduct.
Don Tawanpitak is a new contributor. Be nice, and check out our Code of Conduct.
Don Tawanpitak is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
Take a suitable mixture of any finite-mean skewed distribution and any finite-mean unimodal symmetric distribution with the same mean. All continuous examples and all discrete examples arise in this way.
$endgroup$
– whuber♦
yesterday
$begingroup$
@whuber, That's a great idea. If you have time, it would be terrific if you made a slightly more detailed answer out of that.
$endgroup$
– beta1_equals_beta2
yesterday