I came across this problem with the Rayleigh distribution where this notation was used:
$f(x|\theta) = \displaystyle\frac{x}{\theta^2}\exp(-\displaystyle\frac{x^2}{2\theta^2})1_{[0,\infty)}(x)$
What does the notation $1_{[0,\infty)}(x)$ mean?
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2 Answers
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It is likely the indicator function, which in this case is defined by
$$ 1_{[0,\infty)}(x) = \begin{cases} 1 & \text{if}\ x\in[0,\infty), \\ 0 & \text{otherwise}. \end{cases} $$
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$\begingroup$ +1 for first correct answer. Why not just $x \ge 0$ for the first condition? $\endgroup$ Sep 27, 2017 at 0:07
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1$\begingroup$ @EthanBolker Thank you! I wanted to mimic the way the indicator function of an arbitrary set works. $\endgroup$– JagriffSep 27, 2017 at 0:11
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$1_A$ is an indicator function. $1_A(x) = 1$ if $x\in A$ and $1_A(x) = 0$ otherwise.
answered Sep 27, 2017 at 0:05