# Exponential distribution from Poisson

In Poisson distribution, the probability of inter arrival time to be t or less is:

$$P(X\leq t)= 1 - P(X>t) = 1 - P(0 \mbox{ arrivals in } t) = 1 - e^{-\lambda t}$$

and probability of one arrival in t is:

$$P(k=1)= \lambda t e^{- \lambda t}$$

I wonder how the exponential distribution can be derived from Poisson to reach:

$$P(X\leq t)= \lambda e^{- \lambda t}$$

Thanks,

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By $P(X \leq t) = 1 - e^{-\lambda t}$, $t \geq 0$, $X$ is indeed exponential, and its density function is given by $\frac{d}{{dt}}(1 - e^{ - \lambda t} ) = \lambda e^{ - \lambda t}$, $t \geq 0$. –  Shai Covo Jan 25 '11 at 12:37
@Shai: you should move your comment to an aswer –  leonbloy Jan 25 '11 at 15:38
OK, I will..... –  Shai Covo Jan 25 '11 at 15:56

Suppose that $N = \{N_t: t \geq 0 \}$ is a Poisson process with rate $\lambda$. Let $X$ be the waiting time until the first occurrence in the process $N$. Then, for any $t > 0$, $${\rm P}(X \le t) = 1 - {\rm P}(X > t) = 1 - {\rm P}(N_t = 0) = 1 - e^{ - \lambda t}.$$ Hence, $X$ is an exponential random variable with density function $f_X$ given by $$f_X {(t)} = \frac{{\rm d}}{{{\rm d}t}}(1 - e^{ - \lambda t} ) = \lambda e^{-\lambda t},\;\; t > 0.$$