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Write the linear mean model asThe errors are independent but heteroscedastic. Minimizing the unweighted residual sum of squaresgives the normal equationThus, provided ${\sum }_i{x}_i^2>0$,Since $\mathbb{E}{Y}_i=\theta x_i$,Hence it is an unbiased estimator. This is the least-squares part of the estimators for a Poisson exposure model.

With the normalization used in the question, ridge regression solvesDifferentiating with respect to $\mu$ and using the centered columns $X^{T}1=0$ gives $\hat{\mu }=\overline{Y}$. The normal equation for $\beta$ issoThe push-through identity then gives the equivalent dual form

Let ${\Xi }_{ik}=⟨X_i,B_k⟩$ and define the roughness penalty matrixFor $c=(c_1,\dots ,c_K{)}^{\mathsf{T}}$, the loss is the quadratic functionIts normal equations areWhenever ${\Xi }^{\mathsf{T}}\Xi +\rho \Omega$ is positive definite, the unique minimizer isFor the usual choice $\rho \ge 0$, the penalty matrix is positive semidefinite, so full column rank of $\Xi$ suffices. Since the question permits arbitrary real $\rho$, a sufficiently negative value can make the quadratic form indefinite; then the loss is unbounded below and no minimizer exists. In the singular positive-semidefinite case, the Moore-Penrose inverse describes the minimum-norm solution whenever the normal equations are consistent.

Because the columns of the design matrix are centred, ridge regression with an unpenalized intercept solvesThe normal equations give, for $\lambda >0$,The fitted values are consequentlyIf the objective is normalized by $n$, the same formulas hold after replacing $\lambda$ by $n\lambda$.