Cross Validated Asked by anto_zoolander on January 10, 2021
I’m searching for a book who treat GLM in a formal way, with a measure theoretic approach. Someone could help me?
i try to be more specific
Suppose $ (Omega,mathcal{F},mathbb{P})$ is a probability space, $vec{X} : Omega rightarrow mathbb{R}^n$ random vector and $Y:Omega rightarrow mathbb{R}$ random variable
i can write $Y(omega) = mathbb{E}[Y|vec{X}](omega) + mathcal{E}(omega)$ where $mathbb{E}[Y|vec{X}]$ is the only(a.s) r.variable that satisfy
$$int_{B}mathbb{E}[Y|vec{X}]dmathbb{P}=int_{B}Ydmathbb{P},forall B in sigma(vec{X})$$
by doob-dynkin lemma we can demonstrate that, for some $g: mathbb{R}^n rightarrow mathbb{R}$ measurable function,
$$mathbb{E}[Y|vec{X}](omega)=g(vec{X}(omega))=mathbb{E}[Y|vec{X}=vec{X}(omega)]$$
where $mathbb{E}[Y|vec{X}=vec{X}(omega)]=int_{Omega}Ydmathbb{P}(*|vec{X}=vec{X}(omega))$
ok now my opinion about GLM is to specify the following equations:
this means that $Y=g(vec{X}*vec{beta})+mathcal{E}$
$forall omega in vec{X}=vec{x}$ holds $Y(omega)=g(vec{x}*vec{beta})+mathcal{E}(omega)$
suppose $vec{x}_1,…,vec{x}_m$ are values of $vec{X}$ that for at least one $omega_i in vec{X}=vec{x}_i$, $y_i=Y(omega_i)$ is observable
Now i always see that $Y_i$ is the $i$– observation, maybe defined as $Y(omega_i)$, and specify the joint distribution saying that these are independent, but.. what? is there a passage to product measure? we can build a set of independent r.v defining on the product space the variable $$Y_i(vec{omega})=Y(omega_i)hspace{1em} mathrm{and} hspace{1em} vec{X}_i (vec{omega})=vec{X}(omega_i)$$
now ${(Y_i,vec{X}_i)}_{i=1,…,m}$ is an independent set of r.v and holds
$$mathbb{P}((Y_1,…,Y_m) in A_1 times…times A_m | vec{X}_1=vec{x}_1,…,vec{X}_m=vec{x}_m)=prod_{i=1}^{m}mathbb{P}(Y_i in A_i |vec{X}_i=vec{x}_i)$$
so the joint conditional density is $$f(vec{y}|vec{x}_1,…,vec{x}_m;vec{theta},varphi)=prod_{i=1}^{m}f_{Y|vec{X}}(y_i|vec{x}_i; theta_i,varphi)$$
this is some use of measure theory and this is what i need to find in a book.
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