TITLE: Expected value of number of events at time t
QUESTION [0 upvotes]: If an individual has never had a previous automobile accident, then the probability he or she has an accident in the next h time units is βh+o(h); on the other hand, if he or she has ever had a previous accident, then the probability is αh+o(h). Find the expected number of accidents an individual has by time t.
Let $X_1$ be the time of the first accident. $X_1$ ~ $Exp(\beta)$
$X_2, X_3$... are iid $Exp(\alpha)$
$N(t)$ be the number of accidents by time t   
$E[N(t)]$ = $E[E[N(t)|X_1]]\\
          = \int^\infty_0 E[N(t)|X_1=s]\beta e^{-\beta s}ds\\
          = \int^t_0E[N(t)-N(s)+1|X_1=s]\beta^{-\beta s}ds\\
          = \int^t_0 [1+\alpha(t-s)]\beta e^{- \beta s} \\$
I've been stuck on this for a while. I'm unsure as to why we are taking the expected value of the expected value of $N(t)$ given $X_1$, and how $E[N(t)|X_1=s]$ is equivalent to $E[N(t)-N(s)+1|X_1=s]$ I'm don't know where the $+1$ is coming from. Any help is appreciated. Thank you.

REPLY [1 votes]: There are two possibilities

there is no accident by time $t$  
there is at least one accident by time $t$, and we designate the time of the first accident as $t-s$, which needs to be between $0$ and $t$

In calculating the expectation, the first possibility contributes $0$ and can be ignored in the rest of the calculation
The second possibility requires you to find the expected number of accidents in the remaining time $s$ plus $1$ for the initial accident; multiply this by the probability density for the time of the initial accident and integrate over $s$
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