Show that $Y=\lfloor X\rfloor+1$ where $X$ is exponential is a geometric distribution?

Question

Given the exponential distribution, $X$ with rate $\lambda$ define $Y=\lfloor X\rfloor+1$. Show that $Y$ is geometric with $p=1-e^{-\lambda}$.

Work

$$f(y) = P(Y=y)$$ $$=P(\lfloor X \rfloor + 1=y) $$ $$= P(\lfloor X \rfloor=y-1) $$ $$= P(y-1\leq X <y) $$ $$= \int_{y-1}^y\lambda e^{-\lambda x}dx $$ $$= [-e^{-\lambda x}]_{y-1}^y $$ $$= (e^{-\lambda})^y(e^\lambda-1)$$

Which is not correct. Could someone point me in the right direction?

Thanks

Answer 1

$ e^{-y\lambda}(e^{\lambda}-1) = e^{-(y-1)\lambda}(1-e^{-\lambda})$ $ = (1-p)^{y-1}p$