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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?

amoeba
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jm22b
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    See https://stats.stackexchange.com/questions/145945/indicator-function and https://stats.stackexchange.com/questions/239055. For the use of the indicator function in describing truncated distributions see https://stats.stackexchange.com/questions/330771, https://stats.stackexchange.com/a/308228/919, and https://stats.stackexchange.com/a/77961/919. – whuber Apr 05 '18 at 20:18

2 Answers2

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$\mathbb{1}_{x\ge a}$ 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.

Tim
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  • Thank you - that makes perfect sense with how i'd understood the truncated distribution. – jm22b Apr 05 '18 at 16:09
  • @jm22b see https://stats.stackexchange.com/a/198481/35989 for more details on truncation and censoring – Tim Apr 05 '18 at 16:11
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The I in the formula is the indicator function. In this case it equals 1 when x>=a and zero otherwise.

Michael R. Chernick
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  • 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. – Michael R. Chernick Apr 05 '18 at 21:45