Hey guys, I was just wondering why in my textbook(A First Course in Probability, 8th edition) and basically everywhere I've looked at when we have some random variable(assume for the sake of the following example that it is discrete) denoted by $X$ and we want to write down the expression for the probability of $X$ assuming the value of $k$ we do it as follows: $$P\{X=k\} = ...$$
and not as
$$P(X=K)=...$$ Why the curly brackets? Does it have anything to do with sets or is it just convention? The same question applies to expected values - why $E[X]$ and not $E(X)$ although variance is usually denoted as $Var(X)$ ? $\endgroup$ 13 Answers
$\begingroup$It is just a matter of preference. For the expected value some authors even use $\mathbb E X$ without any brackets at all. Note that $P(X=k)$ is actually short-hand for $P(\{X=k\})$, i.e. it is the probability of the event $\{X=k\}$ happening. So the notation $P\{X=k\}$ is probably emphasizing this without any additional brackets.
$\endgroup$ 2 $\begingroup$Note that an event (e.g., $\{X = k\}$) is, by definition, a (measurable) set, hence the use of the curly brackets. The notation for the measure (probability) of a measurable set (event) may come with or without the parentheses by convention when there is no confusion (i.e., $\mu E$ is the same as $\mu(E)$). So people write both $P\{X=k\}$ and $P(\{X=k\})$.
The use of square brackets with the expectation operator $\mathbb{E}$ is a more specific convention I think. Most operators are not often seen used with square brackets.
$\endgroup$ $\begingroup$For the expectation $\mathbb{E}[\cdot]$, as I understand it this gets square braces because it is an operator (... so totally arbitrary, but...). Expectation is more basic than the variance, since the variance is actually a function involving the expectation of random variables $Var(X)=\mathbb{E}[g(X)] = \mathbb{E}[(X - \mathbb{E}X)^2]$. Seeing variance as a function, and thus having round braces made me happier at least.
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