How Does One Construct a Statistic that Follows a t Distribution?

Suppose we have a normal linear model $Y \sim N_n(X\theta, \sigma^2I_n),$ where $X_{n \times m}$ has rank $r \leq m < n.$ Suppose $\sigma^2$ is unknown and $b^\prime Y$ is an unbiased estimator of $\theta,$ though not the best linear unbiased estimator. Construct a statistic $(b^\prime Y - c^\prime\theta)/H$ that follows a $t$ distribution.

I am completely lost. A hint would be greatly appreciated; a full solution is not necessary.

The question's source: a course's assignment.

• what is the "BLUE" supposed to be? Oct 17, 2016 at 0:10
• Best Linear Unbiased Estimator
– uipo
Oct 17, 2016 at 0:11
• Where you say "best linear unbiased estimator" do you mean of $\theta$? I'd have said that. Also, do you mean that $b'Y$ is in fact an unbiased estimator of $\theta$? I'd have said that too. $\qquad$ Oct 17, 2016 at 0:23
• @MichaelHardy You're right. I did mean to say that. Thanks for the correction.
– uipo
Oct 17, 2016 at 0:26
• ok, It seems you want $H$ to be a statistic that depends on $Y$ and follows a scaled chi-square distribution. $\qquad$ Oct 17, 2016 at 0:35

You have $Y$, a random variable taking values in $\mathbb R^{n\times 1}$ and $\theta$, a constant (i.e. non-random) vector in $\mathbb R^{m\times 1}$, and $X$, a constant matrix in $\mathbb R^{n\times m}$ with $n\ge m$.

For the problem to make sense, it must have been intended that $b'Y$ is a linear unbiased estimator of $c'\theta$. Thus $b\in\mathbb R^{n\times1}$ and $c\in\mathbb R^{m\times 1}$.

The only linear scalar-valued functions of $\theta$ that are estimable are linear combinations of the elements of $\theta$ for which the coefficients form a vector in the row space of $X$. To see that consider that you need $$c'\theta = \operatorname{E}(b'Y) = b'X\theta,$$ and that must hold for every value of $\theta$ in $\mathbb R^{m\times1}$. That means in particular it must hold if $\theta$ is any column of the $m\times m$ identity matrix $I_m$, so we have $$c' I_m = b'X I_m$$ or $$c' = b'X,$$ i.e. $c'$ is in the row space of $X$.

So $b'Y-c'\theta$ has expected value $0$ (and since this is true regardless of the value of $\theta$, it is an unbiased estimator of $0$.

Lemma: Every unbiased estimator of $0$ is uncorrelated with the least-squares estimator of every linear estimable function of $\theta$.

Proof: Exercise.

(This lemma is central to the proof of the Gauss‒Markov theorem, which has far weaker assumptions than what we have here. In that theorem, normality is not assumed. Finite variance is assumed. Identical distribution is not assumed, but identical variance is. Independence is not assumed by uncorrelatedness is. These weaker assumptions are all you need in proof of the lemma.)

Observe that $$\operatorname{var}(b'Y-c'\theta) = b' \Big(\sigma^2 I_n\Big) b = \sigma^2 b'b \in\mathbb R,$$ so $$b'Y-c'\theta \sim N(0, \sigma^2 b'b).$$ As the chi-square statistic in the denominator of your t-distributed statistic, you need the square of the norm of a vector that is uncorrelated with $b'Y-c'\theta$, and it must have expected value $\sigma^2$ times some known constant, so that the unobservable $\sigma^2$ appears in both the numerator and the denominator and cancels out. Uncorrelatedness with $b'Y-c'\theta$ is uncorrelatedness with $b'Y$, since $c'\theta$ is constant. So when is a linear function of $Y$ uncorrelated with $b'Y$? Suppose $DY$ is such a linear function, and $D\in\mathbb R^{\ell \times n}$ $$\operatorname{cov}(b'Y, DY) = b'\Big(\operatorname{cov}(Y,Y)\Big) D' = b'\Big(\sigma^2 I_n \Big) D' = \sigma^2 b'D',$$ so we need $b'D'=0$.

For now I'll leave this here. Except to observe that $\ell$ can't be too big: we only have so many degrees of freedom. I'm not sure we need the "lemma" I stated.