Suppose there are independent samples of $\{x_1,x_2,\cdots,x_n\}$, where each sample is bounded in $[a,b]$. The sample mean is designed as $\hat{\mu}=\frac{\sum_i x_i}{n}$ and the sample variance is designed as $s^2=\frac{\sum_i(x_i-\hat{\mu})^2}{n-1}$.

Now I am trying to prove the independence between $\hat{\mu}$ and $s^2$. I have checked the previous result on independence of sample mean and variance for variables following Gaussian distributions (e.g., Proof that $\frac{(\bar X-\mu)}{\sigma}$ and $\sum_{i=1}^n\frac{(X_i-\bar X)^2}{\sigma^2}$ are independent and Variance of the mean of independent (not IID) normal random variables).

However, it seems a bit tricky to prove the independence result for bounded variable. I am not sure whether the claim is right or not. So, how to prove the independence between sample mean and variance for bounded variable (or disprove)? Could you please give some hints? Thanks in advance.

  • $\begingroup$ If the sample mean and the sample variance are independent, then samples are drawn from Gaussian distribution. See stats.stackexchange.com/a/4359 for details. $\endgroup$ – NCh Mar 22 '17 at 18:58
  • $\begingroup$ @NCh Thanks. Got it now. $\endgroup$ – aaronyxt Mar 23 '17 at 3:34

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