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 \begin{equation} y_t=\beta_1 X_{t-1}+\varepsilon_t \end{equation} 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

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

\begin{equation} 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]} \end{equation} 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

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



1 Answer 1


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*}


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