Ceiling of exponential distribution
Let $X \sim \text{Exp}(\lambda)$ and $T = \lceil{X}\rceil$. What is the distribution of $T$?
My approach:$$ \begin{align*} P(T \leq t) &= P(\lceil{X}\rceil \leq t) \\ &= P(X \leq t-1) + P(t-1 < X \leq t) \\ &= \left(1 - e^{-\lambda(t-1)}\right) + \left(1-e^{-\lambda t}\right) - \left(1 - e^{-\lambda(t-1)}\right) \end{align*} $$But this obviously can't be true. How can I transform this inequality involving a ceil into an ordinary one I can apply my PDF on?
It should be a discrete random variable.
$\endgroup$ 11 Answer
$\begingroup$Your mistake lies in the second step, and becomes apparent if you choose a specific numeric value for $t$; e.g., $t = 2.5$: $$\Pr[T \le 2.5] = \Pr[\lceil X \rceil \le 2.5] \ne \Pr[X \le 1.5] + \Pr[1.5 \le X \le 2.5].$$ The last step above does not follow because for instance, if $X = 2.4 \le 2.5$, then $T = \lceil X \rceil = \lceil 2.4 \rceil = 3 > 2.5$. So the event $T \le 2.5$ is not satisfied, but the event $X \le 2.5$ is satisfied. What went wrong? Well, by working with the CDF, you're not restricting $t$ to be an integer.
Instead of doing it this way, it is much more preferable to work with the probability mass function of $T$. Specifically, let $t \in \{1, 2, 3, \ldots \}$ be a positive integer. Then $$\Pr[T = t] = \Pr[\lceil X \rceil = t] = \Pr[t-1 < X \le t] = (1 - e^{-\lambda t}) - (1 - e^{-\lambda(t-1)}) = e^{-\lambda (t-1)}(1- e^{-\lambda}).$$ Now if we let $p = e^{-\lambda}$, then we obtain $$\Pr[T = t] = p^{t-1} (1-p), \quad t \in \{1, 2, 3, \ldots\}.$$ What distribution does this look like?
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