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It seems these terms are confusing. Some experts think that these terms have a contrasting meaning which is incorrect. Is there someone who can justify the interpretation.

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    $\begingroup$ They are opposites. $\endgroup$
    – Michael R. Chernick
    May 4, 2019 at 18:24
  • $\begingroup$ Homoscedasticity term is used to represent dispersion in continuous data. The term heteroscedasticity measures dispersion of binomial-effects (here in terms of extent of skewness) e.g. treatment of patient results in success i.e. 1 or failure I.e. 0. I have stated in my answer 0, 1 type of data. In case of meta-analysis, we have data of this type and proceed with moderated regression using one or more possible moderator variables as independent variables. thanks for comment. $\endgroup$
    – user10619
    May 5, 2019 at 2:41
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    $\begingroup$ I don't buy your explanation. "Homo" means the same or similar . "hetero" means diverse.. $\endgroup$
    – Michael R. Chernick
    May 5, 2019 at 14:43
  • $\begingroup$ thanks a lot for your comment. I agree with you (upvote). Howevever, I disagree with the classfication - opposites. $\endgroup$
    – user10619
    May 5, 2019 at 15:52
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    $\begingroup$ The experts are correct and there is no confusion at all. $\endgroup$
    – Peter Flom
    May 6, 2019 at 10:57

2 Answers 2

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They are opposites. skedasis means “dispersion”, so hetero mean different variances and homo indicates same/constant variances of the distribution where the shocks/errors/disturbances come from.

For example, if some observations get their errors from the blue distribution (lower variance), while others are drawn from the red (higher variance), you have heteroskedasticity.

enter image description here

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    $\begingroup$ @SubhashC.Davar Skew is about the asymmetry of the distribution about the mean, variance is about dispersion around the mean. For a unimodal distribution, negative skew means there is a long tail is on the left side of the distribution, and positive skew indicates that the long tail is on the right. $\endgroup$
    – dimitriy
    May 4, 2019 at 1:52
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    $\begingroup$ @SubhashC.Davar I don't see anything at that link that contradicts what I wrote, and I have nothing more to add to the discussion here. Best of luck! $\endgroup$
    – dimitriy
    May 6, 2019 at 16:00
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    $\begingroup$ @SubhashC.Davar To what has been said to you many times, I add just two further points. I can't see any useful meaning at all to the word "scedastic" without prefix. If it means anything it just means variable, which is useless, because the word variable is already in use and simpler for most people not Greek. Also homo- or heteroscedasticity has nothing whatsoever to do with skewness. If you can cite literature that says so, that would be a public service, as you would be flagging a paper or textbook to avoid. $\endgroup$
    – Nick Cox
    May 7, 2019 at 10:13
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    $\begingroup$ Subhash, you had to look hard for an inferior definition of "homoscedastic" and even then you chose to cite the least relevant of the definitions offered. Using selective quotations from unauthoritative sources to confront people (who are trying to help you out of your confusion) and to call them wrong is likely to offend anybody. Even when people are not affronted by your attitude, I'm sure they will learn not to answer any more questions you might post. $\endgroup$
    – whuber ♦
    May 8, 2019 at 3:35
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    $\begingroup$ Any good English dictionary explains that homo and hetero are standard Greek roots, from words meaning the same or different respectively. They are in use for many words more common than homoscedastic or heteroscedastic. The thread stats.stackexchange.com/questions/153526/… comments on their origin and use in statistics. I doubt that I can add anything else that will be helpful. There are many problematic terms in statistics, but these aren't among them: the definitions are agreed and simple. $\endgroup$
    – Nick Cox
    May 8, 2019 at 7:52
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They are actually opposite!

Think, for instance, of a linear model where $Y=\beta_0 + \beta_1 x + \epsilon$ where $\epsilon$ is constant. Here you have homoskedasticity, since variance will always be the same, regardless of $x$

If $\epsilon$ depended on the values of $x$, you would have heteroskedasticity, with the variance being different depending on the value of the regressor

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    $\begingroup$ I don't think this is quite right. If the epsilons come from distributions with different variances, you have heteroskedasticity. This does not require dependence on $x$ (though that is common) and epsilon is definitely not constant. It is a random variable, with varying or constant variance. $\endgroup$
    – dimitriy
    May 3, 2019 at 23:53
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    $\begingroup$ @DimitriyV.Masterov Right. Your point about the variance of $\epsilon$ is especially apt. David I think means that the variance of the distribution of $\epsilon$ is constant a la $\epsilon \sim \mathcal{N}(0,\sigma_{\epsilon})$, where $\sigma_{\epsilon}$ is some constant value. $\endgroup$
    – Alexis
    May 4, 2019 at 4:09
  • $\begingroup$ I was just giving an example. You can have heteroskedascity in many different situations, not even necessarely in linear regression $\endgroup$
    – David
    May 4, 2019 at 8:25
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    $\begingroup$ You probably mean that the variance of e is constant, not e itself. $\endgroup$
    – Michael M
    May 6, 2019 at 8:37

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