2 added 125 characters in body edited Nov 16 '15 at 9:22 Xi'an 62.7k88 gold badges100100 silver badges385385 bronze badges The major problem with theyour question is that taking limits does not straightforwardly applyextend to measures and probability distributions. There are many different types of convergence associated with measures. Hence, considering the conjugate $$\beta|h \sim \mathcal{N}(0,cI), h\sim \mathcal{G}(s^{-2},\nu)$$ and letting $$\nu$$ and $$c$$ go to $$0$$ and $$\infty$$, respectively, does not have a proper or unique mathematical meaning. If Now, if you consider the improper prior $$\pi(\beta,h)\propto\frac{1}{h}$$ there is notno posterior distribution associated with the likelihood $$L(\beta,h|X,y)=\exp\{-h(y-X\beta)^\text{T}(y-X\beta)/2\}h^{T/2}$$ because itthe potential posterior does not integrate in $$\beta$$ conditional on $$h$$. There is no $$\hat{\Sigma}=(X^\text{T}X)^{-1}$$$$\hat{\Sigma}=(X^\text{T}X)^{-1}$$ either because the inverse does not exist and no well-defined distribution on $$A\beta$$. The major problem with the question is that taking limits does not straightforwardly apply to measures and probability distributions. Hence, considering the conjugate $$\beta|h \sim \mathcal{N}(0,cI), h\sim \mathcal{G}(s^{-2},\nu)$$ and letting $$\nu$$ and $$c$$ go to $$0$$ and $$\infty$$, respectively, does not have a proper mathematical meaning. If you consider the improper prior $$\pi(\beta,h)\propto\frac{1}{h}$$ there is not posterior distribution associated with the likelihood $$L(\beta,h|X,y)=\exp\{-h(y-X\beta)^\text{T}(y-X\beta)/2\}h^{T/2}$$ because it does not integrate in $$\beta$$ conditional on $$h$$. There is no $$\hat{\Sigma}=(X^\text{T}X)^{-1}$$ because the inverse does not exist and no well-defined distribution on $$A\beta$$. The major problem with your question is that taking limits does not straightforwardly extend to measures and probability distributions. There are many different types of convergence associated with measures. Hence, considering the conjugate $$\beta|h \sim \mathcal{N}(0,cI), h\sim \mathcal{G}(s^{-2},\nu)$$ and letting $$\nu$$ and $$c$$ go to $$0$$ and $$\infty$$, respectively, does not have a proper or unique mathematical meaning. Now, if you consider the improper prior $$\pi(\beta,h)\propto\frac{1}{h}$$ there is no posterior distribution associated with the likelihood $$L(\beta,h|X,y)=\exp\{-h(y-X\beta)^\text{T}(y-X\beta)/2\}h^{T/2}$$ because the potential posterior does not integrate in $$\beta$$ conditional on $$h$$. There is no $$\hat{\Sigma}=(X^\text{T}X)^{-1}$$ either because the inverse does not exist and no well-defined distribution on $$A\beta$$. 1 answered Nov 14 '15 at 10:26 Xi'an 62.7k88 gold badges100100 silver badges385385 bronze badges The major problem with the question is that taking limits does not straightforwardly apply to measures and probability distributions. Hence, considering the conjugate $$\beta|h \sim \mathcal{N}(0,cI), h\sim \mathcal{G}(s^{-2},\nu)$$ and letting $$\nu$$ and $$c$$ go to $$0$$ and $$\infty$$, respectively, does not have a proper mathematical meaning. If you consider the improper prior $$\pi(\beta,h)\propto\frac{1}{h}$$ there is not posterior distribution associated with the likelihood $$L(\beta,h|X,y)=\exp\{-h(y-X\beta)^\text{T}(y-X\beta)/2\}h^{T/2}$$ because it does not integrate in $$\beta$$ conditional on $$h$$. There is no $$\hat{\Sigma}=(X^\text{T}X)^{-1}$$ because the inverse does not exist and no well-defined distribution on $$A\beta$$.