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I am learning for a regression course. For my homework, I was given the residual plot below and I have to analyze it. This is how I interpreted it. I want to know if there are any wrong interpretations.

  1. Since the variability at each $x$ is not the same, there is non constant variance.
  2. The residuals are not randomly scattered as a grouping of some kind can be seen and therefore the assumption on residuals are independent is violated.
  3. Since the residuals are scattered around 0, it can be concluded that the mean of the residuals is 0.

enter image description here

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    $\begingroup$ If the x axis represenst the corresponding fitted values, then you need to rethink at least two (rather three) of your answers. First hint: The average residual is always zero, no matter how bad the regression model is. $\endgroup$ – Michael M May 6 '14 at 16:34
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    $\begingroup$ The conditional variance looks pretty close to constant to me. The problem lies elsewhere. The dependence issue is certainly a clue to the problem, but it's not likely to be the independence assumption that's causing that - a far simpler explanation suffices. Try thinking in terms of the model for the mean. $\endgroup$ – Glen_b May 6 '14 at 23:56
  • $\begingroup$ Why is the variance constant?Is it because data points lie between $(-1,+1)$?Does the y axis represent how far away the data points are from mean 0?There's a lack of fit error because for same value of x there are multiple y values.So the independence of errors is not also violated? $\endgroup$ – clarkson May 7 '14 at 1:27
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    $\begingroup$ If you want to get the attention of someone who has commented you need to put @ in front of their name, like so: @Glen_b Why is the variance constant? ... -- then I'll see you responded. The variance looks close to constant because the spread of values at each $x$ is about the same, roughly 1-1.3 (though it's a bit hard to judge because your image has been taking at a slant). There's certainly lack of fit in the mean, as you say. We have no basis to say "the independence of errors in not violated". Rather, we don't have any indication that it is violated (except that caused by lack of fit) $\endgroup$ – Glen_b May 9 '14 at 2:17
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Since the OP already seems to have arrived at what I regard as the critical insight (the lack of fit issue), I'll expand my comments into an answer:

The conditional variance looks pretty close to constant to me - the spread of values at each $x$ is about the same, roughly around 1 or so (though it's a bit hard to judge because your image has been taken at a slant).

enter image description here

The green braces represent a roughly constant spread (adjusted for the slant).

The seeming dependence is a clue to the problem, but a far simpler explanation suffices - there's a lack of fit in the mean:

enter image description here

... as you suggested.

However, we really have no basis to say "the independence of errors is not violated". Rather, we don't have any indication that it is violated; there's nothing that needs dependence of conditional y's to explain it.

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  • $\begingroup$ Thanks. With your graph I now understand what constant variance is.Also if dependence of y's have nothing to do with independence of errors, say if for different x values the y values are same and they are scattered according to some grouping.Then can i say that since the errors are not randomly distributed the assumption om errors are independent is violated.Also in your early comment you say that " The variance looks close to constant because the spread of values at each x is about the same, roughly 1-1.3".Here 1-1.3 indicate the height of green braces right? $\endgroup$ – clarkson May 10 '14 at 5:51
  • $\begingroup$ The height of the green brace is roughly 1.05 each time (give or take, it's only rough). The range of values (distance between high point to low point at each x-value) looks more like 0.9-1.2 now that I've marked the braces in. $\endgroup$ – Glen_b May 10 '14 at 7:56

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