# Can we express the following unconditional probability as follows?

Some of you may be aware that I have been asking a nagging question for quite a while on this forum, in different shapes and forms. Although I may have been a nuisance, may I thank you as this has been a great learning curve for me and led me to conduct more research. I now have a different solution and have approached the question in a different manner:

Let us say we have a predictive regression of the form $$$$y_t=\beta_1 X_{t-1}+\varepsilon_t$$$$ where $$\varepsilon_t$$ for $$t=1,\cdots, T$$ are independent error terms, such that $$\varepsilon_t\sim N(0,1)$$ and $$X_t$$ is (for now) a strictly stationary stochastic variable. Say, we wish to evaluate

$$$$P[y_t<0]$$$$ for $$t=1,\cdots, T$$. In other words, we wish to evaluate $$$$P[\beta_1X_{t-1}+\varepsilon_t<0]$$$$ Can we using the Bayes Theorem to express the above probability as

$$$$P[\beta_1X_{t-1}+\varepsilon_t<0]=\frac{P[\beta_1X_{t-1}+\varepsilon_t<0\mid X_{t-1}=x_{t-1}]P[X_{t-1}=x_{t-1}]}{P[X_{t-1}=x_{t-1}\mid \beta_1 X_{t-1}+\varepsilon_t<0]}$$$$ and if so then since $$\varepsilon_t\sim N(0,1)$$, we may express the first probability expression in the numerator as $$\phi(\beta_1x_{t-1})$$ and as the process is strictly stationary, the second probability in the numerator and the one in the denominator are invariant to time $$t$$. Am i correct? So can it be said that

$$$$P[\beta_1X_{t-1}+\varepsilon_t<0]=\phi(\beta_1x_{t-1})\times w$$$$ where

$$\begin{eqnarray} w= P[X_{t-1}=x_{t-1}]/P[X_{t-1}=x_{t-1}\mid \beta_1X_{t-1}+\varepsilon_t<0],\forall t \end{eqnarray}$$

?

The reasoning is flawed: defining the random variable $$Z_t=\mathbb I_{\beta X_{t-1}+\epsilon_t>0}$$, the pair $$(X_{t-1},Z_t)$$ admits a density wrt the product measure made of the Lebesgue measure on $$\mathbb R$$ and the counting measure on $$\{0,1\}$$, namely $$p(x,z)=\varphi(x)\Phi(-\beta x)^z\Phi(\beta x)^{1-z}$$ The conditional density of $$X_{t-1}$$ given $$Z_t$$ is thus given by $$p(x|z) = \dfrac{\varphi(x)\Phi(-\beta x)^z\Phi(\beta x)^{1-z}}{\underbrace{\int \varphi(x)\Phi(-\beta y)^z\Phi(\beta y)^{1-z}\,\text{d}y}_{m(z)}}$$ where the denominator is the marginal density of $$Z_T$$. Bayes' formula does not simplify in the denominator: \begin{align*}\text{Prob}(\overbrace{\beta X_{t-1}+\epsilon_t>0}^{Z_t=1})&=\dfrac{\varphi(x)\Phi(-\beta x)^1\Phi(\beta x)^{1-1}}{\left\{\dfrac{\varphi(x)\Phi(-\beta x)^1\Phi(\beta x)^{1-1}}{m(1)}\right\}}\\&=m(1)\\&=\int \varphi(x)\Phi(-\beta y)^1\Phi(\beta y)^0\,\text{d}y\end{align*}