Joint distribution of $Y$ and $S^2-Y^2$

Question

Let $\{X_i\}_{i=1}^n\overset{iid}{\sim}\mathcal{N}(\mu,\sigma^2)$. Let $\{b_i\}_{i=1}^n$ be a sequence of numbers so that $\sum_{i=1}^nb_i=0$ and $\sum_{i=1}^nb^2_i=1$. Define $$S^2=\sum_{i=1}^n(X_i-\bar{X})^2 \text{ and } Y=\sum_{i=1}^nb_iX_i$$.

What can you say about the joint distribution of $Y$ and $S^2-Y^2$?

I only can get the marginal distribution of $Y\sim\mathcal{N}(0,\sigma^2)$, $S^2\sim\sigma^2\chi_{n-1}^2$ and $Y^2/\sigma^2\sim\chi^2_{1}$. What can we say about the joint distribution of $Y$ and $S^2-Y^2$? I tried Basu's theorem, based on the fact that $S^2$ is complete and sufficient for $\sigma^2$ but I cannot show anything using this...

Answer 1

Denote $n$-by-$1$ vectors $(b_1, b_2, \ldots, b_n)^\top$ and $(1, 1, \ldots, 1)^\top$ by $b$ and $e$ respectively. By assumption, we have \begin{align*} e^\top b = 0, \quad b^\top b = 1. \tag{1}\label{1} \end{align*}

Without loss of generality, assume $\sigma^2 = 1$. In addition, since both $Y$ and $S^2$ are invariant under the translation transformation, we can answer the question under the simplified condition $X_1, \ldots, X_n \text{ i.i.d. } \sim N(0, 1)$. In other words, $X := (X_1, \ldots, X_n)^\top \sim N_n(0, I_{(n)})$.

As a standard trick in multivariate analysis, write $S^2 = X^\top(I_{(n)} - n^{-1}ee^\top)X$, $Y = b^\top X$. It then follows that \begin{align*} S^2 - Y^2 = X^\top(I_{(n)} - n^{-1}ee^\top - bb^\top)X := X^\top P X. \end{align*} Due to the condition $\eqref{1}$, it is straightforward to verify that the matrix $P$ is symmetric and idempotent, whence \begin{align*} \operatorname{rank}(P) = \operatorname{tr}(P) = \operatorname{tr}(I_{(n)}) - n^{-1}\operatorname{tr}(ee^\top) - \operatorname{tr}(bb^\top) = n - 1 - 1 = n - 2. \end{align*} It thus follows by the distribution of multivariate normal quadratic form that $S^2 - Y^2 \sim \chi^2_{n - 2}$. On the other hand, clearly we have $Y = b^\top X \sim N(0, 1)$.

Last, let's show $Y$ and $S^2 - Y^2$ are independent, which is true if one can show $PX$ and $b^\top X$ are independent in view of $S^2 - Y^2 = X^\top P X = (PX)^\top(PX)$ is a function of $PX$. This can be further reduced to show $PX$ and $b^\top X$ are uncorrelated, because both of them are normally distributed as linear transformations of $X \sim N(0, I_{(n)})$. This is also straightforward to verify by $\eqref{1}$: \begin{align*} \operatorname{Cov}(PX, b^\top X) = PI_{(n)}b = (I_{(n)} - n^{-1}ee^\top - bb^\top)b = b - b = 0. \end{align*}

In summary, for the joint distribution of $Y$ and $S^2 - Y^2$, we can "say" that

1. $Y$ and $S^2 - Y^2$ are independent.
2. The marginal distribution of $Y$ is $N(0, 1)$, the marginal distribution of $S^2 - Y^2$ is $\chi^2_{n - 2}$.

Answer 2

we know, $$\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}$$ $$Y = \sum_{i=1}^{n} b_iX_i$$ notice, this is a linear combination of normal random variables so, this should follow normal distribution with $$ E(Y) = E(\sum_{i=1}^{n} b_iX_i)$$ $$ = \sum_{i=1}^{n} b_i E(X_i)$$ $$ = \mu.0$$ $$ = 0$$ similarly check $var(Y)=\sigma^2$ i.e. $Y \sim N(0,\sigma^2)$ $$\frac{Y}{\sigma} \sim N(0,1)$$ $$\frac{Y^2}{\sigma^2} \sim \chi^2_1$$ now, $$\frac{(n-1)S^2}{\sigma^2}-\frac{Y^2}{\sigma^2} \sim \chi^2_{(n-1)-1}$$ now to check independence, you can check that the covariance of $Y$ and $S^2-Y^2$ is zero. my doubt is that, cov = 0 does not always implies independence.. :!