Confusion population variance

Mathstudent7

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The formula for the population variance is 1601480358373.png


Is this formula valid just for a certain population distribution or for any?

The confusion for me occure when I look at the variance for the exponential distribution:

1/Λ2 (1 divided by lambda squared). I know how to get this using integral calculation:
But I am struck by the difference between the two ways of calculating variance.

Assume the mean (my) for a population with a exponential distribution is 2, i.e lambda is then 0.5 (1/2). If i plug in the values my = 2 in the population variance formula and lambda = 0.5 i the variance formula for exponential distribution. Will they give the same answer?
 

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The definition you show for variance applies to any discrete distribution. They are using the analogous definition for continuous distributions, where a summation is replaced by an integral. The exponential distribution is not discrete.
 
Thank you. This was somewhat surprising, nowhere in my textbook it had been mentioned that the formula for population variance should not be applied on continous distributions. This make me wonder:
- https://davegiles.blogspot.com/2013/12/unbiased-estimation-of-standard.html?m=1
This professor states that the unbiasedness of S^2 or (s^2) as an estimator of σ^2 doesn't require that the population follows any particular distribution. How do this add upp to that it must be a discrete distribution?
- how then estimate a population variance/standard deviation in the case of a sample from an exponential distribution? If its not possible to use bessels correction, what then to use?
- https://en.m.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation
"For non-normal distributions an approximate (up to O(n−1) terms) formula for the unbiased estimator of the standard deviation is similiar to the bessels correction but with a smaller denominator (n - 1.5 - skewness)
"
If i have understood you correctly, this formula to estimate standard deviation, can not be used in the case of a sample from a continous distribution?
 
If i have understood you correctly, this formula to estimate standard deviation, can not be used in the case of a sample from a continuous distribution?

You asked about population variance. Nothing was said about a sample.

Now you are talking about the variance of a sample, used as an estimate of the variance of the population from which it comes. Those are different things!

But I misspoke in relating the distinction to discrete vs. continuous distributions, due to your juxtaposing two formulas as if they should be related. The summation formula you gave applies to a specific finite population, not to the distribution of the population; I misread it as the definition of variance of a discrete distribution, which looks similar. The integral formula applies to a continuous distribution, not to a population. Frankly, I'm still a little confused about what you were asking.
 
Sorry, let me put it a bit clearer.

Assume we have a finite population which is exponentially distributed. If I have understood it correctly: 1602783172309.png can be applied in order to find the variance for the population.

But let say that we already know the lambda of the population. We know that the variance for an exponential distribution is: 1/lambda^2 will then:
1/lambda^2 equal to 1602783477225.png i.e. can both formulas be used in order to find the population variance?
 
I think you are comparing two different things, though they should be related. (I'm not a statistician, and haven't done this in a long time, so I may not be thinking quite right.)

The summation will give the variance of a particular population, which may differ from the variance of the distribution itself. You can almost think of the population as a sample from the (infinite) distribution; no one population will have exactly the theoretical variance.

I hope someone else will answer who can be more definite.
 
Sorry, let me put it a bit clearer.

Assume we have a finite population which is exponentially distributed. If I have understood it correctly: View attachment 22336 can be applied in order to find the variance for the population.

But let say that we already know the lambda of the population. We know that the variance for an exponential distribution is: 1/lambda^2 will then:
1/lambda^2 equal to View attachment 22337 i.e. can both formulas be used in order to find the population variance?
It is not a sensible question. If you have the entire population, then that population is used to calculate the moments of the population. If you also have a model of the population, perhaps an exponential distribution, that should be useful for calculating estimates of the population moments, mean, variance, etc. You may be able to pick a model distribution to match more than one or two moments of the actual distribution, but this model will not have any authority over the population, itself. There is no point to a model when you have the entire population - excepting perhaps the ease of calculation of some things. When you build a model, DO NOT forget that it is a model. As soon as one member of the population moves in or out, your model is now invalid, at least slightly.
 
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