# For $Y \sim N_n(X\theta,\sigma^2I_n)$ show that an unbiased linear estimator, $b'Y$, of $c'\theta$ is independent of $Y'(I-P_x)Y$

For $Y \sim N_n(X\theta,\sigma^2I_n)$ show that an unbiased linear estimator, $b'Y$, of $c'\theta$ is independent of $Y'(I-P_x)Y$

Ok...first some notation: $'$ means transpose and $P_x$ is the projection matrix $X(X'X)^-X'$

Given: the distribution of $Y$, $X$ is an $n$ by $m$ matrix of rank $r \leq m < n$, $\sigma^2$ is unknown, and $c'\theta$ is estimable (by $b'Y$ which need not be the BLUE). I have to find the denominator $D$ such that $\frac{b'Y-c'\theta}{D}$ follows a t distribution.

I've actually pieced this together for the most part. We known that if $U \sim N(0,1)$ and $V \sim \chi^2(r)$ are independent, $\frac{U}{\sqrt{V/r}}$ follows a t distribution.

Well, $b'Y-c'\theta \sim N(0, \sigma^2b'b)$ so $\frac{b'Y-c'\theta}{\sqrt{\sigma^2b'b}} \sim N(0, 1)$

I also know that $(1/\sigma^2)Y'(I-P_x)Y \sim \chi^2(n-r)$. I now have a $U$, $V$, and $r$ that cancel out the unknown $\sigma^2$, but I still have to show that $U$ and $V$ are independent...

This is easy if $b'Y$ is the BLUE because then $b'Y = c'\hat{\theta} =c'P_xY$. All that's required is to show $Cov(P_xY,(I-P_x)Y)=P_xCov(Y,Y)(I-P_x)'=\sigma^2P_x(I-P_x)=0$ due to the idempotence of $P_x$, and that suffices to show independence.

However, in this problem, I need to show that this holds for any $b'Y$ regardless of if it's the BLUE or not...that's where I'm stumped. I can't just show independence with $Cov(b'Y,(I-P_x)Y)$ anymore, and I haven't found any other way to do so...

I know this is a long winded one, but if anyone can give me any insight as to how to finish this that'd be amazing!

UPDATE

I may have figured it out, but I'm not sure if all of my assumptions are correct.

I think I successfully proved that $c'\theta$ is estimable if and only if $c$ is contained in the column space of $X'X$. Thus, we can write $c = X'Xv$ for some vector $v$.

Since $c=X'b$ that means $X'Xv = X'b$ so $b = Xv$.

Now, $Cov(b'Y,(I-P_x)Y) = \sigma^2(b'(I-P_x)') = \sigma^2(v'X'(I-P_X)) = \sigma^2(v'X'-v'X'P_x)$ and since $P_xX=X$ doesn't $X'P_x = X'$?

• This sound like homework? Perhaps teh self-study tag is in order?
– user30490
Oct 14 '15 at 0:54
• Please add the self-study tag indeed. Oct 14 '15 at 9:56
• My bad... I'm kinda new to this place. Oct 22 '15 at 18:46
• Are you sure that your projection matrix is defined correctly. Did you mean $P_{x} = X(X'X)^{-1}X'$ or $P_{x} = (X'X)^{-1}X'$? Oct 22 '15 at 19:23
• While I haven't worked this out, my suggestion would be to would be to play with $Cov[b'Y,(I-P_{x})Y]= E[(b'Y)((I-P_{x})Y)] - E[b'Y] E[(I-P_{x})Y]$ a little, keeping in mind that $b'Y$ in an unbiased estimator of $c'\theta$. See where that takes you. Oct 22 '15 at 19:50

You're correct in that you don't know that $b'Y$ is the BLUE, but you do know that it is a linear unbiased estimator. Do the properties of covariance you wish to use for $b'Y$ rely on being the best linear unbiased estimator, or simply the linear and unbiased properties of the estimator?

You are, however, incorrect in your specification of your projection matrix. The projection matrix $P$ is generally $X(X'X)^{-1}X'$, unless your $X$ is not full rank, in which case it should be $X(X'X)^-X'$, where $(X'X)^-$ indicates the generalized inverse of $X'X$. Per the problem, since $X$ is $m\times n$ where $m<n$, $X$ cannot be invertible and you must use the g-inverse of $(X'X)$.

(I would have put this in a comment instead of an answer, but I apparently do not yet have enough reputation to comment.)

• It's given that $X$ is not necessarily of full rank, but I'm starting to get a couple ideas... I'll update as I go along Oct 22 '15 at 20:08
• I'll edit my answer to update this, but X is necessarily not full rank. For a matrix to be of full rank, it must be invertible. For it to be invertible, it must be a square matrix. Since $X$ is $m\times n$ and $m<n$, $X$ cannot be full rank. Thus $(X'X)$ has infinite g-inverses but no unique inverse. Oct 22 '15 at 20:17