Exact equivalence of LR and Wald in linear regression under known error variance Is it true that the LR statistic and the Wald statistic are numerically equivalent when testing a nested hypothesis in a linear regression when the error variance is known? Hence, is a squared t-statistic equal to the corresponding LR statistic?
 A: Yes.
[Remark: I took the liberty to ask and answer this question in order to be able to supply the missing parts in this answer. As I believe the present question could be of independent interest I thought it might be helpful to state it separately.]
Consider a partitioned linear regression model
$$y=X_1\beta_{01}+X_2\beta_{02}+\epsilon,$$ where $y|X\sim\mathcal{N}(X\beta_0,\sigma^2I)$.
We test
\begin{equation}
H_0:\beta_{02}=0
\end{equation}
The Wald test statistic is given by (see, e.g., here for the general formula)
\begin{eqnarray*}
\mathcal{W}&=&n\widehat{\beta}_2'\left[n\sigma^2\left[X_2'M_{X_1}X_2\right]^{-1}\right]^{-1}\widehat{\beta}_2\\
&=&\frac{\widehat{\beta}_2'X_2'M_{X_1}X_2\widehat{\beta}_2}{\sigma^2}\\
&=&\frac{y'M_{X_1}X_2(X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}X_2(X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}y}{\sigma^2}\\
&=&\frac{y'M_{X_1}X_2(X_2'M_{X_1}X_2)^{-1}X_2'M_{X_1}y}{\sigma^2}\\
&=&\frac{y'P_{M_{X_1}X_2}y}{\sigma^2}\\
&=:&\frac{y'P_{X_{2\bot1}}y}{\sigma^2},
\end{eqnarray*}
where the third equality follows from the Frisch-Waugh-Lovell theorem. Here, $M_A$ and $P_A$ denote the usual residual maker and projection matrices on $A$.
We now give an expression for the Likelihood ratio test statistic under known error variances.
Inserting the restricted and unrestricted estimators, denoted $\widehat{\beta}$ and $\widehat{\beta}_R$, into the sample log-likelihood yields, using
\begin{eqnarray*}
L(\widehat{\beta})&=&-\frac{n}{2}\log\left(2\pi\sigma^2\right)-\frac{(y-X\widehat{\beta})'(y-X\widehat{\beta})}{2\sigma^2}
\end{eqnarray*}
and analogously for $L(\widehat{\beta}_R)$, the following expression for the $\mathcal{L}\mathcal{R}$-test statistic:
\begin{eqnarray*}\mathcal{L}\mathcal{R}&=&2[L(\widehat{\theta})-L(\widehat{\theta}_R)]\\
&=&\frac{(y-X\widehat{\beta}_R)'(y-X\widehat{\beta}_R)-(y-X\widehat{\beta})'(y-X\widehat{\beta})}{\sigma^2}\\&=&\frac{y'(I-P_{X_1})y-y'(I-P_{X})y}{\sigma^2}
\end{eqnarray*}
We now show that $\mathcal{W}$ may also be written in this format,
\begin{eqnarray*}
\mathcal{W}&=&\frac{y'(I-P_{X_1})y-y'(I-P_{X})y}{\sigma^2}
\end{eqnarray*}
We first show that
$$
P_{X}=P_{X_1}+P_{X_{2\bot1}},
$$
as a partition of $X$, $$X=(X_{A}\vdots X_{B}),$$ in orthogonal matrices $X_{A}$, $X_{B}$ ($X_{A}'X_{B}=0$) satisfies that
$$
P_{X}=P_{A}+P_{B}
$$
This is so because
\begin{eqnarray*}
P_{X}&=&\left(\begin{array}{cc}
                    X_{A} & X_{B} \\
                 \end{array}
               \right)
               \left(
                 \begin{array}{cc}
                   X_{A}'X_{A} & X_{A}'X_{B} \\
                   X_{B}'X_{A} & X_{B}'X_{B} \\
                 \end{array}
               \right)^{-1}
               \left(
                 \begin{array}{c}
                   X_{A}' \\
                   X_{B}' \\
                 \end{array}
               \right)\\
               &=&\left(\begin{array}{cc}
                    X_{A} & X_{B} \\
                 \end{array}
               \right)
               \left(
                 \begin{array}{cc}
                   X_{A}'X_{A} & 0 \\
                   0 & X_{B}'X_{B} \\
                 \end{array}
               \right)^{-1}
               \left(
                 \begin{array}{c}
                   X_{A}' \\
                   X_{B}' \\
                 \end{array}
               \right)\\
               &=&\left(\begin{array}{cc}
                    X_{A} & X_{B} \\
                 \end{array}
               \right)
               \left(
                 \begin{array}{cc}
                   (X_{A}'X_{A})^{-1} & 0 \\
                   0 & (X_{B}'X_{B})^{-1} \\
                 \end{array}
               \right)
               \left(
                 \begin{array}{c}
                   X_{A}' \\
                   X_{B}' \\
                 \end{array}
               \right)\\
               &=& X_{A}(X_{A}'X_{A})^{-1}X_{A}'+X_{B}(X_{B}'X_{B})^{-1}X_{B}'\\
               &=&P_{A}+P_{B}
\end{eqnarray*}
We can apply this intermediate result to $X_1$ and $X_{2\bot1}$, as $X_{2\bot1}'X_1=0$. Hence, $$P_{X_{2\bot1}}=P_{X}-P_{X_1}.$$ Adding and subtracting $y'Iy$ in the numerator of the Wald statistic completes the proof.
It may be seen from, e.g., this answer that $\mathcal{W}=t^2$, with $t$ the t-ratio for some single coefficient, when we replace $s^2$ by $\sigma^2$, assumed known.
When we need to estimate $\sigma^2$, it is however no longer true that $\mathcal{W}=\mathcal{L}\mathcal{R}$.
It may be established that the score test statistic is also numerically equivalent, which is skipped here for brevity. 
