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I have a truncated normal distribution with mode $0$ and variance $\sigma^2$ that only consists of non negative values. What is the density of this distribution at some non negative $x$? I have just been doubling the density of a normal distribution of mean $0$, variance $\sigma^2$, but I'm unsure if that is correct.

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Hint: what is the distribution (density) of |X| when $X \sim \mathcal{N}(0,\sigma^2)$? –  dtldarek Mar 15 '12 at 4:23
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It's not entirely clear from the question whether you mean that the distribution is a truncated version of a distribution with variance $\sigma^2$ or that the truncated distribution itself has variance $\sigma^2$. In the first case, you're right to simply double the non-negative part of the distribution. In the second case, in $\sigma^2=\langle x^2\rangle-\langle x\rangle^2$ the first term is unchanged but the second term is now non-zero, so the variance is reduced by the square of the mean and you'd have to adjust for this.

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It was the first, thanks. –  Andrew Mar 18 '12 at 2:20
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