Degree of freedom and corrected standard deviation. It is often said that degree of freedom causes the need for standard deviation formula to be corrected. When explaining degree of freedom, it is often said that when one knows the mean of the formula, only n-1 data are actually needed, as the last data can be determined using mean and n-1 data. However, I see the same thing occuring in population - not just in sample. So what's going on here, and how is this justification really working?

Filloltarninsv9p 2022-11-19 Answered
Degree of freedom and corrected standard deviation
It is often said that degree of freedom causes the need for standard deviation formula to be corrected. When explaining degree of freedom, it is often said that when one knows the mean of the formula, only n 1 data are actually needed, as the last data can be determined using mean and n 1 data. However, I see the same thing occuring in population - not just in sample. So what's going on here, and how is this justification really working?
For example, in simple linear regression model, variance of error terms are often sum of variance of each data divided by n 2. This is justified as said above. But if this justification is also true for population, not just sample, how is this really working?
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Answers (1)

mignonechatte00f
Answered 2022-11-20 Author has 13 answers
We can write
ln ( 1 + 1 s 2 ) = ln ( s + i ) + ln ( s i ) 2 ln ( s ) = ( s + i ) ln ( s + i ) + γ s + i + ( s i ) ln ( s i ) + γ s i 2 s ln ( s ) + γ s
We know that
L ( ln ( t ) u ( t ) ) = ln ( s ) + γ s
Therefore
L ( D ( ln ( t ) u ( t ) ) ) = s ln ( s ) + γ s
where D is derivative and
L ( e ± i t D ( ln ( t ) u ( t ) ) ) = ( s ± i ) ln ( s ± i ) + γ s ± i
Now you can complete it.
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