Rademacher errors show orthogonality alone gives neither independence nor exact t and F laws

For two coordinates, apply $U=2^{-1\mathrm{/}2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$ to $(R_1,R_2{)}^T$. The events $A=\{\mathrm{\mid }(UR{)}_1\mathrm{\mid }=\sqrt{2}\}$ and $B=\{\mathrm{\mid }(UR{)}_2\mathrm{\mid }=\sqrt{2}\}$ each have probability $1\mathrm{/}2$, while $A\cap B=\mathrm{\varnothing }$: $A$ occurs exactly when $R_1=R_2$ and $B$ exactly when $R_1=-R_2$. If the transformed coordinates were independent, $P(A\cap B)$ would be $1\mathrm{/}4$, contradicting its actual value $0$.

For three independent Rademacher observations, the event $E=\{R_1=R_2=R_3\}$ has probability $2\mathrm{/}8=1\mathrm{/}4$. On $E$, $\bar{R}=\pm 1$ and $S^2=\frac{1}{2}{\sum }_{i=1}^3(R_i-\bar{R}{)}^2=0$, so the usual studentized mean $\bar{R}\mathrm{/}(S\mathrm{/}\sqrt{3})$ is not a real-valued statistic on an event of positive probability. It therefore cannot have the $t_2$ law, which is a probability law on $\mathbb{R}$ with a positive chi-square denominator almost surely by [A1, F1].

For $n=2$, take the full fixed design with one intercept column $(1,1{)}^T$ and the reduced zero-dimensional design whose mean space is $\{0\}$. Under the null intercept $0$, the errors are $(R_1,R_2{)}^T$. On $E_{+}=\{R_1=R_2\}$, which has probability $1\mathrm{/}2$, the full-model residual sum is $RS{S}_U=0$ while the reduced residual sum is $RS{S}_R=2$. Thus the nested F ratio has positive numerator and zero denominator (or is $+\mathrm{\infty }$ under extended-value convention), so it is not an almost surely finite $F_{1,1}$ variable.