0
$\begingroup$

[a] floors the value to a integer.

$$[4.3]=4; [4]=4; [-4.3]=-5$$

Prove that the discrete exponential distribution is geometric distribution.

Research: I have no idea how to represent the [j] values, where j is exponentialy distributed. And even I did how do I prove that this random variable is with geometric distribution?

1 Answers 1

5
  • Say $E\sim \exp(\lambda)$.
  • Then, $\mathbb{P}(E>x)=e^{-\lambda x}$ for $x\in[0,\infty)$
  • Set $G=\lfloor E \rfloor$, and let $m$ be a nonnegative integer.
  • Then, by definition of $G$, we have that $\mathbb{P}(G\ge m)=\mathbb{P}(\lfloor E\rfloor\ge m)$.
  • As $m$ is an integer, this is actually equal to $\mathbb{P}(E\ge m)$
  • By the second point, this is equal to $e^{-\lambda m}$

So, we have established that $\mathbb{P}(G\ge m)=e^{-\lambda m}$

  • Now, as an event, $\{G=m\}$ is the same as $\{G\ge m\}\cap\{G\ge m+1\}^C$
  • From this, $\mathbb{P}(G=m)=\mathbb{P}(G\ge m)-\mathbb{P}(G\ge m+1)$
  • By our earlier calculations, this is equal to $e^{-\lambda m}-e^{-\lambda (m+1)}=e^{-\lambda m}(1-e^{-\lambda})$

Thus, we obtain that $\mathbb{P}(G=m)=e^{-\lambda m}(1-e^{-\lambda})$.

From this, we can say that $G$ is a geometric random variable with parameter $e^{-\lambda}$.

  • 0
    What does P(E>x) represent – 2017-01-08
  • 0
    The probability that the random variable $E$ is greater than the value $x$ – 2017-01-08
  • 0
    and why P(G≥m)=P(⌊E⌋≥m)=P(E≥m)=e−λmP(G≥m)=P(⌊E⌋≥m)=P(E≥m)=e−λm => P(G=m)=P(G≥m)−P(G≥m+1)=e−λm−e−λ(m+1) – 2017-01-08
  • 0
    @Harton i've updated my answer to add greater detail ... had you understood any of the answer previously, though? you basically copied out half of the answer in your comment :S – 2017-01-08