The noncentrality parameter is $\delta^{2}$, the projection for the restricted model is $P_{r}$, $\beta$ is the vector of true parameters, $X$ is the design matrix for the unrestricted (true) model, $|| x ||$ is the norm:
$$
\delta^{2} = \frac{|| X \beta - P_{r} X \beta ||^{2}}{\sigma^{2}}
$$
You can read the formula like this: $E(y | X) = X \beta$ is the vector of expected values conditional on the design matrix $X$. If you treat $X \beta$ as an empirical data vector $y$, then its projection onto the restricted model subspace is $P_{r} X \beta$, which gives you the prediction $\hat{y}$ from the restricted model for that "data". Consequently, $X \beta - P_{r} X \beta$ is analogous to $y - \hat{y}$ and gives you the error of that prediction. Hence $|| X \beta - P_{r} X \beta ||^{2}$ gives the sum of squares of that error. If the restricted model is true, then $X \beta$ already is within the subspace defined by $X_{r}$, and $P_{r} X \beta = X \beta$, such that the noncentrality parameter is $0$.
You should find this in Mardia, Kent & Bibby. (1980). Multivariate Analysis.
Your "assume also" clause equates two quadratic forms in $\mathbb{R}^n$ (with $\mathrm{y}=(y_1,y_2,\ldots,y_n)$ the variable). Since any quadratic form is completely determined by its values at $1+n+\binom{n+1}{2}$ distinct points, their agreement at all points of $\mathbb{R}^n$ is far more than needed to conclude the two forms are identical, whence their coefficients must be the same.
The coefficients of $y_1^2$ are $1/\sigma^2$ and $1/\nu^2$, whence $\sigma=\pm \nu$. We always stipulate that $\sigma$ and $\nu$ are nonnegative, implying $\sigma=\nu$. (The "real" parameter should be considered to be $\sigma^2$ or $1/\sigma^2$ rather than $\sigma$ itself.)
The linear terms in $y_i$ are both proportional to $b_0+b_1 x_i = a_0 + a_1 x_i$. Letting $\mathrm{1} = (1,1,\ldots, 1)$ and $\mathrm{x} = (x_1, x_2, \ldots, x_n)$, we conclude
$$(a_0 - b_0)\mathrm{1} + (a_1 - b_1)\mathrm{x} = \mathrm{0}.$$
Thus either
$\mathrm{1}$ and $\mathrm{x}$ are linearly independent, which by definition implies both $a_0 = b_0$ and $a_1 = b_1$, or
$\mathrm{1}$ and $\mathrm{x}$ are linearly dependent, which means $x_1 = x_2 = \cdots = x_n = x$, say. In that case
- If $x \ne 0$, $a_0 - b_0 = (a_1 - b_1) x$ determines one of $(a_0, a_1, b_0, b_1)$ in terms of the other three, or
- Otherwise $a_0=b_0$ and $a_1$ and $b_1$ could have any values.
In case (1) all parameters are uniquely determined: this is the identifiable model. In case (2) $\sigma = \nu$ is identifiable no matter what and various linear combinations of $(a_0,a_1,b_0,b_1)$ can be identified.
Evidently, linear independence of $\mathrm{x}$ and $\mathrm{1}$ is both necessary and sufficient for identifiability.
This criterion easily generalizes to multiple regression, where the ordinary least squares model is identifiable if and only if the design matrix $X$ (whose columns are formed from $\mathrm{1}, \mathrm{x}$, and any other variables in any order) has full rank: that is, there is no linear dependence among its columns.
Best Answer
The reduced model is the restricted model.
In the first question, your restriction is that the slope coefficient equals $5$. You run the regression using this value and note the error sum of squares which you compare with the error sum of squares of the unrestricted model to see if the restriction is too costly, in which case $H_0:b_1=5$ is rejected. The degrees of freedom are clearly $n-1$ since only one parameter is estimated.
Following the same logic, the reduced model in your second question is
$$Y=2+5 x+\varepsilon$$
and you proceed as above to test the joint hypotheses. Now there are two restrictions instead of one since we have also forced the intercept to assume a certain value. Hence the degrees of freedom are $n$, as no parameter is estimated.