Suppose $X_1, \cdots ,X_n$ are independent and identically distributed random variables with probability density function $$f(x|θ) = θ\, \exp(−θx),\quad \text{ for } x ≥ 0\,.$$ Each variable is censored if it exceeds $C$, a known constant.

Then we have: then $$Pr(X > C) = \exp(−θC)\,.$$

Do anybody know why $$Pr(X > C) = \exp(−θC)?$$

  • $\begingroup$ What's the cdf that corresponds to the density you gave ($f$)? $\endgroup$ – Glen_b May 17 '17 at 11:03
  • $\begingroup$ What would you get if you integrate the density from C to Infinity? What does censoring mean in this context? Are you setting each value greater than C to C or are you discarding every X$_i$ that is greater than C in the sample? $\endgroup$ – Michael R. Chernick May 17 '17 at 11:21
  • $\begingroup$ @Michael Setting each value greater than C to C is winsorizing; discarding them is truncation -- and neither is strictly censoring, which would involve recording in some way that "$X>C$". However exactly what is done with the censored values doesn't relate to answering the question which is only about the probability that the value $C$ is exceeded in the first place. $\endgroup$ – Glen_b May 17 '17 at 11:31
  • $\begingroup$ @Glen_b The question as currently edited does appear to only address $Pr(X>C)$. But I still am not sure what the OP had in mind by censoring. Originally I thought the calculation was intended to determine what the distribution is after it is modified. $\endgroup$ – Michael R. Chernick May 17 '17 at 11:43
  • $\begingroup$ You can read the edit history as easily as I can and see every version that was posted including the first one. The question has not been changed in any substantive way since the first version; it asked what it now asks. $\endgroup$ – Glen_b May 17 '17 at 11:47

If $f(x|θ) = θ e^{−θx}$ then $F(x|θ) = 1- e^{−θx}$ (by integration).

Hence $P(X>C) = 1-P(X\leq C) = 1-F(C) = 1-(1-e^{-\theta C}) = e^{-\theta C}$.


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