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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 $Y$?

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    $\begingroup$ What you have described is not truncation. Do you mean that $Y=a$ when $X>a$ (i.e., that $Y=\min(X,a)$)? $\endgroup$
    – Ben
    Dec 7, 2021 at 8:02
  • $\begingroup$ For geometric distribution, I mean that we are sure that the event happens at most time $a$. $\endgroup$
    – Amin
    Dec 7, 2021 at 23:52
  • $\begingroup$ So don't you just have a point-mass in your distribution at time = a? Or do you mean that your "truncated" distribution is improper, with $F(\infty)<1$? $\endgroup$
    – EdM
    Dec 10, 2021 at 19:38
  • $\begingroup$ There is a difference between saying that you know $x \leq a$ and saying that all the extra probability mass gets added to $p(x=0)$. $\endgroup$
    – jbowman
    Dec 13, 2021 at 0:05
  • $\begingroup$ I have just answered this at or.stackexchange.com/questions/7386/… $\endgroup$ Dec 13, 2021 at 16:34

2 Answers 2

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Also posted at https://or.stackexchange.com/questions/7386/what-is-the-hazard-rate-of-a-truncated-probability-distribution/7440#7440 (It doesn't seem right to post this answer twice, but the question was posted twice (without cross-linking), and I believe the bounty points have already been spent by the OP, whether they are awarded or not.)

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 $p + (1-p)^a$.

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I'll start by showing you the answer to your question, but then I'll discuss a complication you need to be aware of when interpreting this answer. Suppose we have a non-negative continuous random variable $X$ and we truncate it to give the random variable:

$$Y=\min(X,a).$$

The hazard rate for $Y$ can be written in terms of the underlying hazard of $X$ as:

$$\begin{align} \lambda_Y(y) = \frac{f_Y(y)}{1-F_Y(y)} &= \begin{cases} \lambda_X(y) & & & \text{if } y < a, \\[12pt] 1 & & & \text{if } y = a, \\[12pt] \text{NA} & & & \text{if } y > a. \\[12pt] \end{cases} \end{align}$$

As you can see, the hazard function for the truncated random variable is the same as the hazard function for the underlying variable prior to the truncation point. At the truncation point the hazard jumps up to one (reflecting the fact that the instantaneous probability of failure is one at that point) and then it is non-defined (since failure has already occurred with probability one).

Now, while the hazard function itself is quite simple, the interpretation of the hazard function in this case is complicated. This is because you are looking at a density of a "mixed" random variable that is continuous over most of its range but with a discrete point-mass at one particular point. When you define the density of a random variable like this, the probability density over the continuous part is made with respect to Lebesgue measure and the density over the point-mass is made with respect to counting measure, so the values of the density in these different parts are comparing apples and oranges. So, it is perfectly simple to define the hazard function for a truncated random variable, but it is tricky to interpret this hazard rate.

In terms of interpretation, it is useful to go back to the notion of the hazard function as an "instantaneous" probability/density of failure given survival up to that time. In the present case, for any value $0 \leqslant x < a$ and an infinitesimal value $dx$ we have the conditional probabilities:

$$\begin{align} \mathbb{P}(x \leqslant Y \leqslant x + dx | Y \geqslant x) &= \lambda_Y(x) dx = \lambda_X(x) dx, \\[6pt] \mathbb{P}(Y=a | Y \geqslant a) &= \lambda_Y(a) = 1. \\[6pt] \end{align}$$

Observe here that the hazard function is giving a density (rate-of-change in probability) prior to the point $a$ but it is giving a probability at the point $a$. Consequently, you have to be careful when making hazard comparisons. If you have $\lambda_X(x)=1$ for some $0 \leqslant x<a$ that does not mean that the hazard is the same at points $x$ and $a$. (In technical terms, the hazard rates at these points are taken with respect to a different "dominating measure".)

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