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Understanding skewness

  1. Nov 25, 2011 #1
    Hey

    In my textbook it says if E[(x-u)^3] is positive the skew is to the right, and negative to the left. u=expected value.

    This is ok and I believe it, but I can't see why it is like this. Is there an easy explanation of why this has to hold?
     
  2. jcsd
  3. Nov 25, 2011 #2

    D H

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    The first thing to note is that skewness is defined as E[((x-μ)/σ)3], so your E[(x-μ)3] is just the skewness scaled by the cube of the standard deviation. If a distribution has positive (negative) skewness then E[(x-μ)3] is positive (negative) by definition.

    The next thing to note is that skewness is an odd moment (e.g., E[(x-μ)1], E[(x-μ)3], E[(x-μ)5], ...) about the mean. All of these odd moments would be zero if the distribution was symmetric. So a non-zero value skewness necessarily means the distribution is not symmetric. (The converse is not true. A zero skewness does not necessarily mean the distribution is symmetric.)

    So when talking about a distribution with a positive (negative) skewness we are talking about a distribution that is not symmetric. Think about what E[((x-μ)/σ)3] is measuring. Values far from the mean (|x-μ|/σ > 1) will weigh much more heavily than those near the mean (|x-μ|/σ < 1). It is those values that are far from the mean that determine the skewness. For the skewness to be negative, there must be more such values far from the mean for x<μ than for x>μ. The opposite is the case when the skewness is positive.
     
  4. Nov 26, 2011 #3
    Thanks DH you obviously know what you are talking about. And sorry for not scaling it as you mentioned, I didn't do it because I only wanted to know why it was negative and why it was positive at times.

    However I hope you can give me some more of your time, cause there is one thing that is still unclear:

    Regarding the last paragraph of yours, I have had the same thought as you. But lets for argument sake look at the first moment E[(x-u)], with this moment you can use the same arguments as you use. If the distribution is skewed to the right x-u will tend to be bigger than |x-u| to the left. But the moment is still zero, because the p(x)-values to the left will tend to be bigger, so this will make the moment zero.

    My question is: how do you disregard this when looking at the third moment, cause even though |x-u| will tend to be bigger at the skewed side. p(x) will tend to be bigger at the non-skewed side. So why does the effect of |x-u|^3 beat the p(x) "effect"? That is, if we have right skew, why does
    (x-u)^3*p(x)dx tend to be more positive on the right, than (x-u)^3*p(x)dx is negative on the left side. Because p(x) will be bigger on the left, ex: chi-squared distributions.
     
    Last edited: Nov 26, 2011
  5. Nov 26, 2011 #4

    chiro

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    Consider each term in the expectation: If there are more terms left of the mean than the right when considered with respect to the frequency of each value in the domain, it would make sense that you get a negative value.

    If the distribution was skewed to the left (i.e. had a right tail), you would expect a negative result since the proportion of values of (x-mu)^3 would be negative and we would have a higher frequency of weight being negative.

    Think of a scale where you have an object on the left scale, and weights on the right hand scale. You have to balance the weights with the object to get an exact reading: it works in the same sort of way for the skewness.
     
    Last edited: Nov 26, 2011
  6. Nov 26, 2011 #5

    D H

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    That's a bit counter to the more or less standard mean of skewed to the left versus skewed to the right.

    For example, this distribution with a long tail to the right is typically denoted as skewed right.
    histogr6.gif

    This nomenclature (skewed right means a longer tail to the right) is widespread but is not universal. Some would call the above distribution as being skewed to the left, evidenced by the mode and median being to the left of the mean.
     
  7. Nov 26, 2011 #6

    chiro

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    If my terminology was wrong I apologize to the OP (and think D H for the correction). In saying this the premise of my general message still stands.
     
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