# Can argument of a CDF be stochastic?

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?

• 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 for a salient example.
– whuber
Mar 3, 2020 at 21:24
• 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. Mar 3, 2020 at 21:34