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$$\dfrac{1}{n}\sum_{i=1}^{n}1_{[y_{i}f(x_{i})<0]}$$ this gives me the number of how many times the expression in brackets is less then zero. But is it correct? I see it for the first time, so where(textbook) I can encounter it in an elucidating form? Or is it from a particular subject like statistics, and reading on calculus won't help me discover this notation?

Hrisip
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    Only the context will explain whether $y_{i}f(x_{i})$ is meaningful. Your whole expression calculates what proportion (not number) of these $y_{i}f(x_{i})$ are negative – Henry Apr 24 '18 at 11:14
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    Whether or not that expression says what you mean, I'd try to avoid it. It's hard to read because all the important information is in a subscript. Some words and perhaps an auxiliary definition would really help your readers. – Ethan Bolker Apr 24 '18 at 11:37

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I think there is (yet) no universally established notation for what you try to express. It is not uncommon to denote

$$1_{P}=\begin{cases} 1 & \text{if $P$ is true} \\ 0 & \text{otherwise}\end{cases}$$

for some statement $P$. In your case $P_i\equiv [y_i f(x_i)<0]$. This construct (indexed by $i$) is sometimes called "characteristic vector/indicator vector" or "characteristic function/indicator function" of some set or statement. I personally like the Iverson brackets (if no confusion can happen):

$$[P]:=\begin{cases} 1 & \text{if $P$ is true} \\ 0 & \text{otherwise}\end{cases}.$$

Anyways, in your case, the sum is counting the number of pairs $(x_i,y_i)$ for which $y_if(x_i)<0$. But this number is divided by $n$, so in the end it gives you the fraction of these pairs, and not the absolute number.

For counting or determining the size/cardinality of a set, there are other established notations like

$$|\{(x_i,y_i)\mid y_if(x_i)<0\}|\qquad\text{or}\qquad \#\{(x_i,y_i)\mid y_if(x_i)<0\}.$$

So e.g. $\#\{1,5,8\}=3$.

M. Winter
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