What is the hazard-rate of a truncated probability distribution?

Question

Suppose $X$ is a random variable with pdf and cdf in forms of $f(X)$ and $F(X)$, with hazard-rate $h(X)$. Now, we define a new variable $Y$ which is a truncated random variable, $Y=X, \quad \text{if } X \le a$, and $Y=0, \quad \text{if } X > a$. I am wondering what is the hazard-rate of the random variable $Y$? For example, if $X$ is a geometric random, what is the hazard-rate of $Y$?

Answer 1

The blog entry Defining Hazard Rate at a Point Mass. "Applied Probability and Statistics in Actuarial Science and Financial Economics", contains what is needed to answer the question.

Define the Hazard Rate Function $h_T(t) = \frac {P(T = t)}{P(T \ge t)}$, at discrete mass points t.

First I will assume $X$ is a nonnegative continuous variable, as implied by mention of pdf of $X$. I will also assume $a$ is positive.

The question seems to make more sense if $Y = a$ when $X \ge a$, i.e., $Y = \text{min}(X,a)$. In that case, there is a discrete mass point of $Y$ at $a$, and the hazard rate function of $Y$ is $1$ at $a$, and the usual hazard rate applies for $X < a$.

If instead, $Y = 0$ when $X \ge a$, then there is a discrete mass point of $Y$ at $0$, and the hazard rate function of $Y$ is $P(X \ge a)$ at $0$ (because the denominator of the hazard rate function is $1$); and the usual hazard rate applies for $X < a$.

Now, in order to handle your question about $X$ being Geometric, assume that $X$ is a discrete nonnegative variable, and $a$ is a positive integer.

If $X$ is Geometric with parameter $p$, the hazard rate function of $Y$ is $p$ for all nonnegative integers, except at $a$. If $Y = \text{min}(X,a)$, the hazard rate function of $Y$ is $1$ at $Y = a$. If, instead, $Y = 0$ when $X \ge a$, the hazard rate function of $Y$ at $0$ is $p + (1-p)^a$.