5
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.