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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?

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- Thread starter skwey
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In summary, skewness is defined as E[((x-μ)/σ)^3] and is an odd moment about the mean. A non-zero skewness indicates that the distribution is not symmetric. When a distribution is skewed to the right, E[(x-μ)^3] is positive, while it is negative when skewed to the left. This is because values far from the mean have a greater impact on skewness and when the distribution is skewed to the right, there are more values on the right side of the mean. This can be compared to balancing weights on a scale. While there may be some variation in the nomenclature, the general idea is the same.

- #1

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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?

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The next thing to note is that skewness is an odd moment (e.g., E[(x-μ)

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

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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 let's 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.

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 let's 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.

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- #4

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skwey said:

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 let's 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.

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.

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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.chiro said: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..

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

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.

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