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I'd like to ask for clarification of the following example in my textbook.

Example:

Suppose events are occurring at random with average rate $\lambda$ per unit of time. What is the probability density function of the random variable $X$, the waiting time to the second event?
We first find $F(x)$.

$F(x) = Pr[X$ $\leq$ $x] = Pr[$waiting time less than $x$ units of time for the 2nd event$]$

$=$ $1-Pr[$waiting more than $x$ units of time for the 2nd event$]$
$=$ $1-Pr[$less than 2 events take place in $x$ units of time$]$
$=$ $1-Pr[$0 events in $x$ units$] - Pr[$1 event in $x$ units$]$
$=$ $1-e^{-\lambda x}-\lambda x e^{-\lambda x}$,
using the Poisson distribution. We now differentiate $F(x)$ to find the density function $f(x)$.

$f(x) =\dfrac {dF(x)}{dx} = \lambda e^{-\lambda x} - \lambda e^{-\lambda x} + \lambda ^2 x e ^{-\lambda x } = \lambda ^2 x e ^{-\lambda x }$

Thus the density function of the random variable $X$, the waiting time to the second event, is given by

$f(x) = \lambda ^2 x e ^{-\lambda x } \qquad x\geq 0$
$\:\qquad = 0 \qquad \qquad \quad otherwise$

This is an extension of the exponential distribution which is the density function of the waiting time to the first event.

(End of example)

How did they transition from the first part to the second part here "using the Poisson distribution"? It looks like the exponential dist. to me.
Edit: To clarify, I wish to know how: $Pr[$0 events in $x$ units$]$ becomes $-e^{-\lambda x}$ and $Pr[$1 event in $x$ units$]$ becomes $-\lambda x e^{-\lambda x}$.

$1-Pr[$0 events in $x$ units$] - Pr[$1 event in $x$ units$]$
$=$ $1-e^{-\lambda x}-\lambda x e^{-\lambda x}$,

If you can explain this process, I'd appreciate your help.

Thanks in advance.

Soo Kyung Ahn
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  • **You cannot find $F$ from this information alone.** Although it is likely the author intended you to assume the events are independent and have zero probability of coinciding, they did not state either of these critical assumptions. The need for them is discussed at https://stats.stackexchange.com/questions/214421, which might answer the questions you have. BTW, to differentiate $F$ you need to use the product rule: the derivative of $1$ was not even written down because it is obvious it is zero. – whuber Jan 15 '19 at 21:19
  • Thanks for your reply! I now understood the differentiation part. But assuming independence and zero probability of coinciding, how do they end up with the other part of my question? I have shortly read the answer you linked to, but I feel that I need to review some math before I carry on. If need be, I am willing to invest the time to understand that post, but I'd appreciate it if you can tell me whether it does answer my question :) – Soo Kyung Ahn Jan 16 '19 at 11:48
  • I'm not sure what your question is. Is it why the Poisson distribution is computed with these formulas or could it be why the Poisson distribution is even applicable? Regardless, the intent of the answer I linked to was to explain both of those. – whuber Jan 16 '19 at 12:40
  • My question is why the $Pr[$0 events in $x$ units$]$ would yield $-e^{-\lambda x}$ given the probability mass function for a Poisson process is ${e^-\lambda * \lambda ^x}/x!$. To be specific, I am thinking that plugging in 0 for x in the pmf would result in just $e^-\lambda$ without the $x$. I have the same question for the $Pr[$1 events in $x$ units$]$, shouldn't it result in $e^-\lambda * \lambda$? instead of $-\lambda x e^{-\lambda x}$ – Soo Kyung Ahn Jan 16 '19 at 14:28

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