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Question 1: Let us first consider the univariate case:

Suppose we have $Y\in\{0,1\}$, suggesting that $Y$ is Bernoulli variable (and hence discrete) and $X\in \mathbb{R}$, then we know that \begin{equation} P(Y=y)=\int_{\mathbb{R}}P(Y=y\mid X=x)P(X=x)dx \end{equation}

What if now instead of $X\in \mathbb{R}$, we have $\mathbf{X}\in \mathbb{R}^n$. Would the notation
\begin{equation} P(Y=y)=\int_{\mathbb{R}^n}P(Y=y\mid \mathbf{X}=\mathbf{x})P(\mathbf{X}=\mathbf{x})d\mathbf{x} \end{equation}
be correct?

Question 2: Now consider the time-series $\{(Y_t,\mathbf{X}_t)\in\mathbb{R}\times\mathbb{R}^n, t=1,2,\cdots \}$, and suppose I wish to express the unconditional probability of $P(Y_t=y_t)$ by conditioning on all the past of $\mathbf{X}_t$, would the following expression be correct? \begin{equation} P(Y=y)=\int_{\mathbb{R}^n}\cdots\int_{\mathbb{R}^n}P(Y=y\mid\mathbf{X}_1=\mathbf{x}_1,\mathbf{X}_2=\mathbf{x}_2,\cdots,\mathbf{X}_n=\mathbf{x}_n)P(\mathbf{X}_1=\mathbf{x}_1,\mathbf{X}_2=\mathbf{x}_2,\cdots,\mathbf{X}_n=\mathbf{x}_n)d\mathbf{x}_1\cdots d\mathbf{x}_n \end{equation}

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  • $\begingroup$ Your notation is vague, which makes it risky to state whether the formulas are correct or not. E.g., in the first formula the left hand side and the conditional expression in the integrand must be probabilities. If, for consistency, we also interpret the "$P(X=x)$" factor in the integrand as a probability, then that is always zero, whence the integral is zero. It will help you to distinguish probabilities from probability densities. $\endgroup$
    – whuber
    Sep 23 at 18:31
  • $\begingroup$ @whuber That is actually what I was thinking. I had it in my initial draft of the question, but I removed it to make the question more concise. If I change the right hand side such that $P(X=x)$ is substituted with $f(x)$ would that make the notations correct? $\endgroup$
    – Carl
    Sep 23 at 18:40
  • $\begingroup$ It would help clarify it. "Correctness" is partly a matter of how much you explain. Especially in the second part of the question there is the possibility of adopting a clearer and more concise notation. There, too, it looks like all your uses of "$P$" might be densities. $\endgroup$
    – whuber
    Sep 23 at 20:46

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