Truncated distribution

A truncated distribution is the conditional probability distribution obtained by restricting a random variable $X$ to a specified event $X\in A$ with positive probability. If its original law is $\mu$, the truncated law isThe denominator renormalizes the retained probability mass to one. If $X$ has a probability density function $f$, the truncated density is $f_A(x)=f(x)1_A(x)/\mu (A)$; the same restriction and normalization apply to a probability mass function.

For example, truncate a uniform distribution on $(0,10)$ to values at least six. The retained event has probability $2/5$, so the new density is $1/4$ on $(6,10)$, and its mean is eight. This is the law obtained by repeatedly sampling from the original law and rejecting values outside the retained set; rejection sampling becomes inefficient when the retained event is rare.

A zero-probability restriction cannot be normalized by the displayed formula. Conditioning a continuous random variable to equal a specified value requires an additional conditional-law construction or limiting convention, rather than division by zero.

Condition an exponential distribution of rate $d>0$ on falling below $t>0$. Normalizing its probability density function gives the displayed density. Integration gives expectation $1/d-t/(e^{dt}-1)$. As $d\downarrow 0$, the density tends to the uniform distribution on $(0,t)$ and the expectation tends to $t/2$.

// Target: probability-and-statistics.bigb

For a positive continuous density $f_{\theta }$ with strictly increasing cumulative distribution function $F_{\theta }$, observations from its upper-truncated law have likelihood ${\prod }_i{f}_{\theta }(t_i)/F_{\theta }(M{)}^n$ on $M>t_{(n)}$. Holding $\theta$ fixed, decreasing $M$ toward $t_{(n)}$ increases the likelihood; taking a supremum over $\theta$ preserves nonincreasing dependence on $M$. Provided the nuisance fit is finite, the usual closed-support endpoint estimate is $t_{(n)}$. With the literal open support $t<M$, only a limiting supremum need exist. A sample maximum is not a validated bound for future observations, and support depending on a parameter prevents automatic application of Wilks theorem.

For a nonnegative integer claim count $N$ with $p=\mathbb{P}(N=0)<1$, the conditional law given $N>0$ has masses $\mathbb{P}(N=n)/(1-p)$ for $n\ge 1$. Its probability generating function is $(G_N(z)-p)/(1-p)$. Truncation at zero is meaningful only when the positive-count event has positive probability.

Let $N\ge 1$ have probability mass function $p_n$ satisfying $p_n=(a+b/n)p_{n-1}$ for $n\ge 2$. Let independent positive integer claim sizes have probability mass function $f_j$, independently of $N$. The aggregate probability mass function $g_k$ satisfies the displayed recursion, with $g_0=0$. To prove it, write $P,F,G$ for the count, severity and aggregate probability generating functions. The count recurrence gives $(1-az)P^{\prime }(z)=(a+b)P(z)+p_1$. Since $G=P\circ F$, the chain rule gives $(1-aF)G^{\prime }=((a+b)G+p_1)F^{\prime }$. Comparing coefficients of $z^{k-1}$ gives the recursion. The forcing term $p_1{f}_k$ distinguishes this algorithm from the usual Panjer recursion. For a zero-truncated Poisson distribution with parameter $\lambda$, take $a=0$, $b=\lambda$ and $p_1=\lambda /(e^{\lambda }-1)$.

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.