7
$\begingroup$

I have not seen the following notation before (the 1 with a subscript in the density):

Consider the problem of sampling from the truncated normal distribution $\mathcal{N}_t (\mu, 1, a)$, given by the random variable $X ∼ \mathcal{N} (\mu, 1)$ conditional on the event $\{X \geq a\}$. Its density is proportional to

$$f(x) \propto \mathrm{exp}\{-\frac{(x-\mu)^2}{2 \sigma^2}\}1_{x \geq a}$$

What exactly does that mean?

Improve this question
$\endgroup$
1

2 Answers 2

Reset to default
14
$\begingroup$

$\mathbb{1}_{x\ge a}$ is an indicator function, that is equal to $1$ when $x\ge a$ and zero otherwise. Multiplying by it is a fancy, math way of saying that everything else is equal to zero. In this case, it says that only cases greater or equal to $a$ can be observed.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you - that makes perfect sense with how i'd understood the truncated distribution. $\endgroup$
    – jm22b
    Commented Apr 5, 2018 at 16:09
  • $\begingroup$ @jm22b see stats.stackexchange.com/a/198481/35989 for more details on truncation and censoring $\endgroup$
    – Tim
    Commented Apr 5, 2018 at 16:11
7
$\begingroup$

The I in the formula is the indicator function. In this case it equals 1 when x>=a and zero otherwise.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Note that this is an example of a normal distribution truncated on the left. The normal distribution can also be truncated on the right or on both side. These can also be specified using the indicator function. $\endgroup$
    – Michael R. Chernick
    Commented Apr 5, 2018 at 21:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.