Q&A

# Acceptable, usual to write $\ge 2$ pipes simultaneously?

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I'm NOT asking for the solution to this exercise that's publicly accessible. Rather, pls see the green and red underlines. If I apply the author's green definition to the red underline, then $\tilde P({\color{red}{L \mid M_2}}) \equiv P(\color{red}{L \mid M_2} \quad \color{limegreen}{\mid M_1})$.

Is it natural or wont to write $\ge 2$ Conditional Probability pipes simultaneously? Blitzstein, Introduction to Probability (2019 2 edn), Ch 2, Exercise 26, p 87.

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Say No To Multiple Pipes (1 comment)
Accessibility (1 comment)

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As mentioned in the answers you referenced in earlier versions of your question, $(-\mid-)$ is not standalone notation in usual probability theory notation. Instead, $P(A\mid B)$ is just an unusual notation for a binary function. A common definition of this binary function in terms of a separate unary, unconditional probabilty function also called $P$ and the one I believe Blitzstein is using is $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$

While I don't know how clear Blitzstein makes this elsewhere, the way to read what's being written is we're defining a new unary, unconditional probability function $\tilde P$ via $\tilde P(A) = P(A \mid M_1)$. The binary conditional probability function $\tilde P(A \mid B)$ is still defined as usual, i.e. $$\tilde P(A \mid B) = \frac{\tilde P(A \cap B)}{\tilde P(B)}$$ NOT as the meaningless $P((A \mid B) \mid M_1)$. If we want to expand $\tilde P(A \mid B)$ in terms of $P$, we get $$\tilde P(A \mid B) = \frac{P(A \cap B \mid M_1)}{P(B \mid M_1)} = \frac{P(A \cap B \cap M_1)}{P(B \cap M_1)}$$ which doesn't present any issues. The answer to (c) follows readily.

1. I have seen it used, e.g. in Jaynes' "Probability Theory: The Logic of Science", as standalone notation, but not in a way such that $P(A\mid B)$ meant $P((A\mid B))$ where $P$ was some unconditional probability measure. ↩︎

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