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I have learnt from solving different statistical problems that if we have a sample of size $n$ picked from the population with normal distribution, the degree of freedom for, say, Student distribution or Chi-square distribution distribution will be always $n-1$. Recently i have discovered the following problem:

Let ${X_1},{X_2}, \cdots ,{X_{10}}$ be independent $N(0,1)$-distributed (standard normal distribution) random variables. Define $$X = X_1^2 + X_2^2 + \cdots + X_{10}^2$$ What is the distribution, expectation and variance of $X$?

I understand that this is a Chi-square distribution. What i don't get is why the degree of freedom is $10$ instead of $9$? If $X$ is a sum of $10$ squared $RV$ shouldn't we treat it as a sample with size $n=10$ and hence $X \sim {\chi}^2(n - 1)$?

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In your sum-of-squares expression, all 10 variables are free to vary. Your reference to a sample of size 10 refers to the case when we average the squares of 10 deviations of 10 scores from the mean of those scores. In that case, 9 of the scores can be arbitrary, but the 10th one is determined since all 10 must now have produce the given mean value. T. Kasper

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