Derivation of pdf from cdf for integer part of exponential with rate $\lambda$

Question

For a variable X with integer part [X], I have found the CDF of [X] to be $1-e^{- \lambda (x+1)}$, where X is an exponentially distributed function with rate $\lambda$. How do I proceed from here to obtain the pmf of [X], $(e^{ \lambda } - 1)e^{- \lambda (x+1)}$? Thank you in advance.

Answer 1

$[X]$ is now a discrete random variable, so you should go back to the discrete formulas. Let $F(x) = P([X] \le x)$ be the cdf of $[X]$ and let $p(x) = P([X] = x)$ be the pmf. Then you have

$$p(x) = P([X] = x) = P([X] \le x) - P([X] \le x-1) = F(x) - F(x-1).$$

Note that you have to treat the case $p(0)$ separately because $F(-1) = P([X] \le -1)$ isn't given by the general formula you found, which only applies for nonnegative $x$.