Poisson distribution and exponential distribution Here is some notes taken for probablity:
A random variable X has a exponential distribution with mean $\theta=\frac{1}{\lambda}$ if it has PDF: $f(x)=\frac{1}{\theta} e^{\frac{-x}{\theta}}$ where $x \geq 0$
Wondering how is it different from Poisson distribution.
Is it, for example, that, the difference is "the number of customers that arrive in a specific time period has Poisson distribution. The waiting time between successive customer arrivals has exponential distribution"?
 A: Let $X_n$, $n=1,2,\ldots$ be i.i.d. random variables with $\mathbb P(X_1>0)=1$ and $\mathbb E[X_1]<\infty$. The sequence $\{S_n:n=1,2,\ldots\}$ defined by $S_0=0$ and $S_n = \sum_{k=1}^n X_k$ is called a renewal sequence, with the associated counting process $N(t) = \sup\{n\geqslant 0: S_n\leqslant t\}$. In the case where $X_1$ has exponential distribution with mean $\frac1\lambda$, $N(t)$ is known as a Poisson process, and has the property that $\mathbb P(N(t) = n) = e^{-\lambda t}\frac{(\lambda t)^n}{n!}$, that is, $N(t)$ has Poisson distribution with mean $\lambda t$. This is a fundamental relationship between the Poisson and exponential distributions which you should know.
A: The exponential distribution  indicates the probability of the distance between two successive, poisson distributed events. The random variable $X$ thus designates the interval between two successive events and is a continuous random variable.
The probability that $X$ takes at most the value $x$ is calculated as
$P (X \leq x) = 1 - P (\textrm{no event in the interval of length x})$
Now $P (\textrm{no event in the interval of length x})$ is equal to the probability that the poisson distributed random variable $Y$ takes the value zero in the interval of length $x$. We have
$P(Y=y)=e^{-\lambda \cdot x}\cdot \frac{(\lambda \cdot x)^y}{y!} \Rightarrow P(Y=0)=e^{-\lambda \cdot x}\cdot \frac{(\lambda \cdot x)^0}{0!}=e^{-\lambda  \cdot x}$ 
Next we get the link to the expoenential distribution.
$P(X\leq x)=1-e^{-\lambda  \cdot x}$
Differentiating
$P(X= x)=\lambda \cdot e^{-\lambda  \cdot x}$
