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Right-skewed distribution with mean equals to mode?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

7

$begingroup$

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

share|cite|improve this question

edited 7 hours ago

Nick Cox

39.4k588131

asked 8 hours ago

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♦
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  3 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

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

share|cite|improve this question

edited 7 hours ago

Nick Cox

39.4k588131

asked 8 hours ago

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♦
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  3 hours ago
  

  
  
  
  

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

share|cite|improve this question

edited 7 hours ago

Nick Cox

39.4k588131

asked 8 hours ago

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.

$endgroup$

share|cite|improve this question

edited 7 hours ago

Nick Cox

39.4k588131

asked 8 hours ago

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.

share|cite|improve this question

edited 7 hours ago

Nick Cox

39.4k588131

asked 8 hours ago

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 7 hours ago

Nick Cox

39.4k588131

edited 7 hours ago

Nick Cox

39.4k588131

asked 8 hours ago

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 8 hours ago

Don TawanpitakDon Tawanpitak

363

2 Answers
2

active

oldest

votes

8

$begingroup$

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.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Nick CoxNick Cox

39.4k588131

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
  $endgroup$
  – Don Tawanpitak
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
  $endgroup$
  – Don Tawanpitak
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  5 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered 7 hours ago

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
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  6 hours ago
  

  
  
  
  

add a comment | 

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8

$begingroup$

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.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Nick CoxNick Cox

39.4k588131

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
  $endgroup$
  – Don Tawanpitak
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
  $endgroup$
  – Don Tawanpitak
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  5 hours ago
  

  
  
  
  

add a comment | 

8

$begingroup$

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.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Nick CoxNick Cox

39.4k588131

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
  $endgroup$
  – Don Tawanpitak
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
  $endgroup$
  – Don Tawanpitak
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  5 hours ago
  

  
  
  
  

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.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Nick CoxNick Cox

39.4k588131

$endgroup$

: (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 |
 +-----------------------------+

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Nick CoxNick Cox

39.4k588131

answered 7 hours ago

Nick CoxNick Cox

39.4k588131

answered 7 hours ago

Nick CoxNick Cox

39.4k588131

4

$begingroup$

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.

share|cite|improve this answer

answered 7 hours ago

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
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  6 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered 7 hours ago

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
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  6 hours ago
  

  
  
  
  

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.

share|cite|improve this answer

answered 7 hours ago

beta1_equals_beta2beta1_equals_beta2

1063

$endgroup$

share|cite|improve this answer

answered 7 hours ago

beta1_equals_beta2beta1_equals_beta2

1063

answered 7 hours ago

beta1_equals_beta2beta1_equals_beta2

1063

answered 7 hours ago

beta1_equals_beta2beta1_equals_beta2

1063