Truncated distribution

A truncated distribution is a conditional probability distribution obtained by restricting a random variable to a specified subset of its original support and renormalizing.

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 to lie in a specified subset of the real line. Its density is the original Gaussian density divided by the probability of the retained set.

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.