# Geometric distribution obtained from exponential distribution

I was trying to solve the following:

Let $\lambda>0$ and $X$ a random variable with $X \sim exp(\lambda)$. Show that $Y=[X]+1$ has geometric distribution of parameter $p=1-e^{-\lambda}$, where $[x]$ denotes the integer part of a number.

I did the following:

$$P(Y=n)=P([X]+1=n)$$$$=P(n-1 \leq X <n)$$$$=F_X(n)-F_X(n-1)$$$$=1-e^{-\lambda n}-(1-e^{-\lambda (n-1)})$$$$=e^{-\lambda (n-1)}-e^{-\lambda n}$$$$=e^{-\lambda (n-1)}(1-e^{-\lambda})$$

If I call $p=1-e^{-\lambda}$, then we have $$P(Y=n)=(1-p)^{n-1}p$$

which is exactly the density function of a geometric random variable with parameter $p=1-e^{-\lambda}$.

I was having doubts with my solution, could anyone tell me if my answer is correct?

• Your answer is correct. – A.S. Oct 16 '15 at 5:09
• Clearly explained, well done. – André Nicolas Oct 16 '15 at 6:08