Truncated normal distribution ($T{N}_A(\mu ,{\sigma }^2)$)

A truncated normal distribution is a normal distribution conditioned on belonging to a retained measurable set $A$ of positive probability. If $X\sim N(\mu ,{\sigma }^2)$, $\sigma >0$, and $Z=\mathbb{P}(X\in A)>0$, its density iswhere $\varphi$ is the standard normal density. The indicator is essential: the density is zero outside $A$. For an interval $A=(a,b)$, possibly with infinite endpoints,The same expression holds for any endpoint inclusion convention because the original normal distribution has no atoms.

For a standard normal variable restricted to positive values, $Z=1/2$ and the density is $2\varphi (x)$ for $x>0$. Its mean is $\sqrt{2/\pi }$: integrate $2x\varphi (x)$ using ${\varphi }^{\prime }(x)=-x\varphi (x)$. Symmetry gives second moment one, hence variance $1-2/\pi$. Truncation therefore changes the mean and variance; they are not generally the original parameters $\mu ,{\sigma }^2$.

sign-truncated normal sampling uses inverse transform sampling to generate this conditional law directly. Rejection from the original normal is also exact but may be slow when the retained set has very small probability.

For $Z$ with standard normal distribution, $m\in \mathbb{R}$, and $a>0$, integrate $(m+az)\varphi (z)$ over $z>-m/a$. The identity ${\varphi }^{\prime }(z)=-z\varphi (z)$ gives the displayed expectation of the positive part. This is useful for option pricing when a financial payoff is the positive part of a normal random variable, including a log-call option. At $a=0$ the formula is interpreted by its limit, $m^{+}$.

To draw $N(\mu ,1)$ restricted to a positive value, let $a=\Phi (-\mu )$ and return $\mu +{\Phi }^{-1}(a+(1-a)U)$ for uniform $U\in (0,1)$. For restriction to a negative value use $\mu +{\Phi }^{-1}(aU)$. These are inverse transform sampling constructions. Repeatedly drawing an untruncated normal until its sign agrees is also exact, but inefficient for rare signs.