2
$\begingroup$

Let $Y_1,\ldots,Y_n$ be independent random variables with $Y_k$ distributed as $N\sim(a_k,\sigma^2)$, and $\bar Y=\sum_{k=1}^{n}\frac{Y_k}{n}$ denote the sample mean, $S^2$ denotes the sample variance.

Then show that $$\frac{(n-1)S^2}{\sigma^2}=\frac{1}{\sigma^2}\sum_{k=1}^{n}(Y_k-\bar Y)^2$$

is distributed as noncentral $\chi^2$ with $(n-1)$ degrees of freedom and noncentrality parameter $\lambda=\sum_{k=1}^n\frac{(a_k-\bar a)^2}{\sigma^2}$, where $\bar a=\sum_{k=1}^{n}\frac{a_k}{n}$.

I have only the idea of Moment Generating Function(MGF) Technique to prove this type of proof based on transformation. But this is almost impossible to me use the MGF technique for the given exercise.

How can i prove it easily?

$\endgroup$
2
$\begingroup$

Since the $Y$s are independent and the distributions are normal and the variances are all equal, the distribution is spherically symmetric, so the distribution of sum of squares will still be the same if we look at a rotated coordinate system.

Let $P$ be the orthogonal projection of the column vector $(Y_1,\ldots, Y_n)^T$ onto the one-dimensional space $x_1=\cdots=x_n$, and let $Q=I-P$ be the complementary orthogonal projection onto the $(n-1)$-dimensional space $x_1+\cdots+x_n=0$. Then $$ Q\begin{bmatrix} Y_1 \\ \vdots \\ Y_n \end{bmatrix} = \begin{bmatrix} Y_1-\bar Y \\ \vdots \\ Y_n-\bar Y \end{bmatrix} \text{ and }Q\begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix} = \begin{bmatrix} a_1-\bar a \\ \vdots \\ a_n-\bar a \end{bmatrix}. $$ If we express everything in a rotated coordinate system with one axis in the direction of the space $x_1=\cdots=x_n$ and the others orthogonal to that, this is expressed as $$ Q\begin{bmatrix} U_1 \\ U_2 \\ \vdots \\ U_n \end{bmatrix} = \begin{bmatrix} 0 \\ U_2 \\ \vdots \\ U_n \end{bmatrix} \text{ and }Q\begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\ \vdots \\ b_n \end{bmatrix}. $$ Then $$ \frac{1}{\sigma^2}(U_2^2+\cdots +U_n^2) \sim \chi^2_{n-1,b_2^2+\cdots+b_n^2}. $$ But $$ (Y_1-\bar Y)^2 + \cdots+(Y_n-\bar Y)^2 = U_2^2+\cdots+U_n^2 $$ and $$ (a_1-\bar a)^2+\cdots+(a_n-\bar a)^2 = b_2^2+\cdots+b_n^2. $$

$\endgroup$
  • $\begingroup$ Please correct me if I am wrong, but $Q=I-P$ is not an orthogonal matrix, is it? $\endgroup$ – Samrat Mukhopadhyay Jun 14 '15 at 14:07
  • $\begingroup$ I think it is better to use $Q=I-2P$ where $P$ is the orthogonal projection matrix on $span(\overline{Y})$. $\endgroup$ – Samrat Mukhopadhyay Jun 14 '15 at 14:17
  • $\begingroup$ @SamratMukhopadhyay : Certainly $Q=I-P$ is not an orthogonal matrix. It is the matrix of an orthogonal projection onto a proper subspace, so it is singular. $I-2P$, on the other hand, is an orthogonal matri. I don't know how $I-2P$ could be used to answer this question. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 14 '15 at 15:09
  • $\begingroup$ Yes, @Michael Hardy, I was wrong to propose $I-2P$ here, but I am having some problem understanding how $Q$ is going to perform a rotation operation, i.e. we cannot say that $\|Y\|_2^2=\|QY\|_2^2$, then how can we say that the distribution of the sum of squares is the same as in a rotated coordinate system. maybe I am missing something. It'll be so kind of you to put some light on it. $\endgroup$ – Samrat Mukhopadhyay Jun 14 '15 at 15:25
  • $\begingroup$ $Q$ doesn't perform a rotation. But there is a rotation that transforms the standard basis to the union of a basis of the column space of $Q$ and a basis of the orthogonal complement of that column space. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 14 '15 at 15:56

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.