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 $$V=\sum_{i=1}^n (X_i-\mu)^2$$ ?

marked as duplicate by whuber♦ self-study StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 3 at 16:26

• $X_i$ are independent right? And also, is this a self-study question? If so, please add a tag. – psarka Jan 3 at 11:57
• 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 at 14:43