Discretization of an exponential variable

Question

Given $X=Exp(\lambda)$, i have to define $Y=ceil(X)$ in order to prove the link between exponential and geometric variables.

By definition of ceiling $\forall x\in \mathbb{R},\exists n\in \mathbb{N}:x\leq n< x+1$, so:

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

Nevertheless, the result is $e^{-\lambda n}(1-e^{-\lambda})\Rightarrow Y\sim Geo(1-e^{-\lambda})$.

1) Where i wrong in the passages?

2) Geometric variable has $n-1$ at the exponent, not $n$. So, why that result?

Thanks for any help!

Answer 1

You have $e^{-\lambda n}(e^{\lambda}-1)$. This is equal to $e^{-\lambda(n-1)}-e^{-\lambda n}$. Now we can factor out $(e^{-\lambda})^{n-1}$

We have to add $\lambda (n-1)$ to both exponents.

• $-\lambda(n-1)+\lambda(n-1)=0$
• $-\lambda n+\lambda (n-1)=\lambda (-n+n-1)=-\lambda$

Therefore we get $(e^{-\lambda})^{n-1}\cdot (e^0-e^{-\lambda})=(e^{-\lambda})^{n-1}\cdot (1-e^{-\lambda})$

This is a geometric distribution with $p=1-e^{-\lambda}$

Answer 2

You did nothing wrong but can represent it differently:

$e^{-\lambda (n-1)}-e^{-\lambda n} = e^{-\lambda (n-1)}-e^{-\lambda (n-1) -\lambda} =e^{-\lambda (n-1)}(1-e^{-\lambda})\sim Geo(1-e^{-\lambda}) $

but

$e^{−λn}(1−e^{−λ})\neq e^{−λn}(e^{λ}-1)$.