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This question may sound very trivial and obvious, however I need to clarify this. Say we have a model: \begin{equation} y_t=\beta x_{t-1}+u_t \end{equation} where $x_{t-1}$ is stochastic and generated by the process \begin{equation} x_t=\rho x_{t-1}+\varepsilon_t \end{equation} Now assume we want to find $P[y_t\geq 0]$. Can we express (and considering a normal CDF) \begin{eqnarray} P[y_t\geq 0]&=&P[\beta x_{t-1}+u_t\ge0]\\ &=&P[u_t\geq-\beta x_{t-1}]\\ &=&1-P[u_t<-\beta x_{t-1}]\\ &=&1-\Phi(-\beta x_{t-1})\\ &=&\Phi(\beta x_{t-1})? \end{eqnarray} As $\beta x_{t-1}$ is stochastic as opposed to a constant, would such expression be correct?

Carl
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    By definition, the probability of an event like "$y_t\ge 0$" is a (definite) *number.* Thus, what you seek when you say you "want to find" its probability is an answer that is a *number,* not a random variable. The general question in your title, though, is answered affirmatively: see [the Probability Integral Transform](https://stats.stackexchange.com/questions/424169) for a salient example. – whuber Mar 03 '20 at 21:24
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    You can do the calculation conditioning on $X_{t-1} = x_{t-1}$. If you are not in reality working with conditional probabilities and really are intersted in the marginal distribution $Pr(y_t \geq 0)$ rather than $Pr(y_t \geq 0\lvert x_{t-1})$ then as Whuber says $Pr(y_t \geq 0)$ is a number but you end up with $\Phi(\beta x_{t-1})$ which is a random variable if you really are treating $x_{t-1}$ as random. My guess is you are really interested in conditional probabilities. – Jesper for President Mar 03 '20 at 21:34

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