Truncated distribution

A truncated distribution is a conditional distribution that conditions on the random variable in question. It is derived from some other probability distribution


 * $$f(X|a < X \leq b) = \frac{f(x)}{F(b)-F(a)}$$

where $$f(x)$$ is a probability density function where we have redefined the support from $$(-\infty, \infty)$$ to $$(a, b)$$ and $$F(x)$$ is the cumulative distribution function with standard support, both of which are assumed to be known.

There are two commonly used cases. First, a truncated distribution where the bottom of the distribution has been removed is as follows:


 * $$f(X|X>y) = \frac{f(x)}{1-F(y)}$$

where $$f(x)$$ is a continuous distribution function where we have redefined the support from $$(-\infty, \infty)$$ to $$(y, \infty)$$ and $$F(x)$$ is the cumulative distribution function with standard support, both of which are assumed to be known.

Second, a truncated distribution where the top of the distribution has been removed is as follows:


 * $$f(X|X \leq y) = \frac{f(x)}{F(y)}$$

where $$f(x)$$ is a continuous distribution function where we have redefined the support from $$(-\infty, \infty)$$ to $$(\infty, y)$$ and $$F(x)$$ is the cumulative distribution function with standard support, both of which are assumed to be known.

Notice that in fact $$f(X|a < X \leq b)$$ is a distribution:
 * $$\int_{a}^{b} f(X|a < X \leq b))dx = \frac{1}{F(b)-F(a)} \int_{a}^{b} f(x) dx = 1 $$.

Expectation of Truncated Random Variable
The expectation of a truncated random variable is thus:

$$ E(X|X>y) = \frac{\int_y^\infty x f(x) dx}{1 - F(y)} $$

Letting $$ a $$ and $$ b $$ be the lower and upper limits respectively of support for $$f(x)$$ properties of $$ E(u(X)|X>y) $$ where $$u(X)$$ is some continuous function of $$ X $$ with a continuous derivative and where $$ f(x) $$ is assumed continuous include:

(i) $$ \lim_{y \to a} E(u(X)|X>y) = E(u(X)) $$

(ii) $$ \lim_{y \to b} E(u(X)|X>y) = u(b) $$

(iii) $$ \frac{\partial}{\partial y}[E(u(X)|X>y)] = \frac{f(y)}{1-F(y)}[E(u(X)|X>y) - u(y)] $$

(iv) $$ \lim_{y \to a}\frac{\partial}{\partial y}[E(u(X)|X>y)] = f(a)[E(u(X)) - u(a)] $$

(v) $$ \lim_{y \to b}\frac{\partial}{\partial y}[E(u(X)|X>y)] = \frac{1}{2}u'(b) $$

Provided that $$ \lim_{y \to c} u'(y) = u'(c) $$, $$ \lim_{y \to c} u(y) = u(c) $$ and $$\lim_{y \to c} f(y) = f(c) $$ where $$ c $$ represents either $$a$$ or $$ b$$.

The Tobit model employs truncated distributions.

Random Truncation
Suppose we have the following set up: a cut-off value, $$t$$, is selected at random from a density, $$g(t)$$. Then a value, $$x$$, is selected at random from the density, $$f(x|t)$$. Suppose we wish to know the density of the cut-off given an observed value of $$x$$.

First, by definition:

$$f(x)=\int_{x}^{\infty} f(x|t)g(t)dt $$, and $$F(a)=\int_{-\infty}^a[\int_{x}^{\infty} f(x|t)g(t)dt]dx $$

Notice that $$t$$ must be greater than $$x$$, hence when we integrate over $$t$$, we set a lower bound of $$x$$. $$f(x)$$ and $$F(x)$$ are the unconditional density and unconditional cumulative distribution function, respectively.

By Bayes Rule:

$$g(t|x)= \frac{f(x|t)g(t)}{f(x)}$$

which expands to:

$$g(t|x) = \frac{f(x|t)g(t)}{\int_{x}^{\infty} f(x|t)g(t)dt}$$

Example, Two Uniform Distributions
Suppose we know that t is uniformly distributed from [0,T] and x|t is distributed uniformly from [0,t]. Let g(t) and f(x|t) be the densities that describe t and x respectively. Suppose we observe a value of x and wish to know the distribution of t given that value of x.

$$g(t|x) =\frac{f(x|t)g(t)}{f(x)} = \frac{1}{t(ln(T) - ln(x))}$$ $$ \forall t > x$$