Given that $Y$ follows multivariate normal distribution ,i.e, $N_n (0, \sigma^2 I_n)$, we want to find the distribution of $Y'Y$ given that $a'Y=0$ where $a$ is a non zero constant vector.

I know that the distribution of $Y'Y$ would be $\sigma^2 \chi_n^2$ if the condition is not given. How to approach for the given condition?

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    $\begingroup$ Welcome to the site, @Shyam. I have added the 'homework' tag, because your Q looks like it is. If it's not, you can edit your Q to remove the tag w/ my apologies. Note that homework gets special treatment here: we provide hints to help you work through the problem, instead of a full solution, so that you can learn from the process (see our FAQ). $\endgroup$ – gung - Reinstate Monica Dec 2 '12 at 16:55
  • $\begingroup$ Note that $a^{\prime} Y = 0$ part implies that $Y$ is not linearly independent so I don't think that the covariance matrix of $Y$ can be $\sigma^2 I_{n}$. Is it possible that the question has a typo ? $\endgroup$ – mlofton Aug 15 '18 at 15:38
  • $\begingroup$ @mlofton The question asks for a conditional distribution. $\endgroup$ – whuber Aug 15 '18 at 15:40
  • $\begingroup$ gotcha. my mistake. $\endgroup$ – mlofton Aug 16 '18 at 17:13

This distribution has spherical symmetry: that is, any rotation of $Y$ about the origin has the same distribution. Exploit this by applying a rotation that puts $a$ parallel to the last coordinate. The condition $a^\prime Y=0$ is tantamount to $Y_n=0.$ Since $Y_n$ is independent of $Y_1, \ldots, Y_{n-1},$ the question asks for the distribution of $Y_1^2 + Y_2^2 + \cdots + Y_{n-1}^2.$ As stated in the question (replacing $n$ by $n-1$), this will be the distribution of $\sigma^2$ times a $\chi^2(n-1)$ variable.

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  • $\begingroup$ Thanks whuber but I don't see why $a^\prime Y = 0$ implies that $Y_{n} = 0$. .Could you explain that in more detail ? $\endgroup$ – mlofton Aug 16 '18 at 17:12
  • $\begingroup$ @mlofton In the rotated coordinate system the coordinates for $a$ are $(0,0,\ldots, 0, |a|),$ whence the equation is $$0 = a^\prime Y = 0(Y_1)+0(Y_2)+\cdots+0(Y_{n-1})+|a|Y_n = |a|Y_n.$$ $\endgroup$ – whuber Aug 16 '18 at 18:06
  • $\begingroup$ thanks. I'm pretty bad once I go past three dimensions so I'll think about it some. $\endgroup$ – mlofton Aug 16 '18 at 20:40
  • $\begingroup$ @mlofton Fortunately, this situation can be visualized in just two dimensions. The two dimensions needed are those spanned by $a$ and the vector $(0,0,\ldots, 1).$ $\endgroup$ – whuber Aug 16 '18 at 21:10
  • $\begingroup$ Unfortunately, I'm still not following. So, if you don't mind, going back to your statement, suppose that we just have two dimensions so $Y_{1}$ and $Y_{2}$. Now, it is assumed that $a_{1} Y_{1} + a_{2} Y_{2} = 0$. So, are you saying that $a_{1} Y_{1} + a_{2} Y_{2}$ then has the same distribution as $\sqrt{a_{1}^2 + a_{2}^2} \times Y_{1}$ which has the same distribution as $\sqrt{a_{1}^2 + a_{2}^2} \times Y_{2}$. If so, I believe you but I don't have the intuition for why ? Thanks a lot for your help. $\endgroup$ – mlofton Aug 17 '18 at 7:34

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