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Right-skewed distribution with mean equals to mode?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)CDFs for Right-Skewed DistributionsDistributions with Median=Mode=Average?Skewness - why is this distribution right skewed?Compute mode of distributionWhy not take the mode of a bootstrap distribution?Does mean=mode imply a symmetric distribution?Bounding the distance between the mean and the mode of a unimodal distributionCounterexamples where Median is outside [Mode-Mean]R glmer: distribution for strongly right skewed dataHow to generate a skewed bell curve distribution function with known min, max, and mode
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
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 Apr 15 at 17:14
Nick Cox
39.4k588132
asked Apr 15 at 16:03
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.
$endgroup$
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 Apr 15 at 17:14
Nick Cox
39.4k588132
asked Apr 15 at 16:03
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♦
Apr 15 at 20:27
$begingroup$
@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
Apr 15 at 21:25
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 Apr 15 at 17:14
Nick Cox
39.4k588132
asked Apr 15 at 16:03
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 Apr 15 at 17:14
Nick Cox
39.4k588132
asked Apr 15 at 16:03
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 Apr 15 at 17:14
Nick Cox
39.4k588132
asked Apr 15 at 16:03
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 Apr 15 at 17:14
Nick Cox
39.4k588132
edited Apr 15 at 17:14
Nick Cox
39.4k588132
edited Apr 15 at 17:14
Nick Cox
39.4k588132
39.4k588132
asked Apr 15 at 16:03
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 Apr 15 at 16:03
Don TawanpitakDon Tawanpitak
363
asked Apr 15 at 16:03
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♦
Apr 15 at 20:27
$begingroup$
@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
Apr 15 at 21:25
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♦
Apr 15 at 20:27
$begingroup$
@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
Apr 15 at 21:25
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♦
Apr 15 at 20:27
$begingroup$
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♦
Apr 15 at 20:27
$begingroup$
@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
Apr 15 at 21:25
$begingroup$
@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
Apr 15 at 21:25
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 Apr 15 at 17:20
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
Apr 15 at 19:00
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
Apr 15 at 19:04
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
Apr 15 at 19:12
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 Apr 15 at 17:00
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
Apr 15 at 18:47
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
Apr 15 at 18:53
add a comment |
Your Answer
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
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 Apr 15 at 17:20
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
Apr 15 at 19:00
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
Apr 15 at 19:04
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
Apr 15 at 19:12
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 Apr 15 at 17:20
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
Apr 15 at 19:00
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
Apr 15 at 19:04
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
Apr 15 at 19:12
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 Apr 15 at 17:20
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 Apr 15 at 17:20
Nick CoxNick Cox
39.4k588132
edited Apr 15 at 17:33
answered Apr 15 at 17:20
Nick CoxNick Cox
39.4k588132
answered Apr 15 at 17:20
Nick CoxNick Cox
39.4k588132
answered Apr 15 at 17:20
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
Apr 15 at 19:00
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
Apr 15 at 19:04
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
Apr 15 at 19:12
add a comment |
$begingroup$
Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
$endgroup$
– Don Tawanpitak
Apr 15 at 19:00
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
Apr 15 at 19:04
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
Apr 15 at 19:12
$begingroup$
Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
$endgroup$
– Don Tawanpitak
Apr 15 at 19:00
$begingroup$
Thank you! This is absolutely helpful! I will also look into the Weibull distribution you mentioned.
$endgroup$
– Don Tawanpitak
Apr 15 at 19:00
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
Apr 15 at 19:04
$begingroup$
By the way, do you know some other continuous distribution with finite support that can exhibit the same property?
$endgroup$
– Don Tawanpitak
Apr 15 at 19:04
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
Apr 15 at 19:12
$begingroup$
@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
Apr 15 at 19:12
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 Apr 15 at 17:00
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
Apr 15 at 18:47
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
Apr 15 at 18:53
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 Apr 15 at 17:00
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
Apr 15 at 18:47
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
Apr 15 at 18:53
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 Apr 15 at 17:00
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 Apr 15 at 17:00
beta1_equals_beta2beta1_equals_beta2
1063
answered Apr 15 at 17:00
beta1_equals_beta2beta1_equals_beta2
1063
answered Apr 15 at 17:00
beta1_equals_beta2beta1_equals_beta2
1063
answered Apr 15 at 17:00
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
Apr 15 at 18:47
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
Apr 15 at 18:53
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
Apr 15 at 18:47
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
Apr 15 at 18:53
$begingroup$
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
Apr 15 at 18:47
$begingroup$
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
Apr 15 at 18:47
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
Apr 15 at 18:53
$begingroup$
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
Apr 15 at 18:53
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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$begingroup$
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♦
Apr 15 at 20:27
$begingroup$
@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
Apr 15 at 21:25