# Understanding sum of square deviations [duplicate]

Given $$X_1...X_n\stackrel{iid}{\sim} N(\mu,\sigma^2)$$ and $$U=\sum_{i=1}^n (X_i-\overline{X})^2$$, why is $$U\sim\sigma^2 \chi_{n-1}^2$$ ?

And what would be the distribution of $$V=\sum_{i=1}^n (X_i-\mu)^2$$ ?

• $X_i$ are independent right? And also, is this a self-study question? If so, please add a tag. – psarka Jan 3 '19 at 11:57
• yeah, sorry. fixed it – therealcode Jan 3 '19 at 12:35
• Hint: $X_1-\mu, X_2-\mu, \ldots, X_n-\mu$ are iid $N(0,\sigma^2)$ random variables. What does your book have to say about this distribution? Nothing? How about considering $(X_1-\mu)/\sigma, (X_2-\mu)/\sigma, \ldots, (X_n-\mu)/\sigma$ which are iid $N(0,1)$ random variables? – Dilip Sarwate Jan 3 '19 at 14:43
• See stats.stackexchange.com/questions/385032/… and the links posted there. – Christoph Hanck Jan 3 '19 at 15:00