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Let's say we have a Poisson process starting at time $0$ with an average arrival rate of $\lambda$ arrivals per hour. The arrival rate doubles immediately after each arrival. Let $X_1$ and $X_2$ be the first and second inter-arrival times. We have to find the expectation and variance of $X_1+X_2$.

Now, I was thinking of modelling this situation as a nonhomogeneous Poisson process with a variable rate $\lambda(t) = \lambda$ for $0 \le t < X_1$ and $\lambda (t) = 2\lambda$ for $X_1 \le t < X_1+X_2$. So far , I know how to deal with variable rates that are deterministic functions of time, which is not the case here as $\lambda (t)$ depends on $X_1$ and that is a random variable.

Maybe, I'm making this unnecessarily complicated. Any ideas on how to proceed would be really appreciated.

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You can find expectation by:$$\mathbb E(X_1+X_2)=\mathbb EX_1+\mathbb EX_2$$ Since $X_1$ and $X_2$ are independent you can find variance by:$$\text{Var}(X_1+X_2)=\text{Var}X_1+\text{Var}X_2$$

Here $X_1$ has exponential distribution with parameter $\lambda$ and $X_2$ has exponential distribution with parameter $2\lambda$.

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  • $\begingroup$ Thanks. I'm more concerned about why $X_1$~ $Exp(\lambda)$ and $X_2$~$Exp(2\lambda)$. $\endgroup$
    – Andre.J
    Aug 12 '17 at 11:22
  • $\begingroup$ If $N_t$ denotes the Poissonproces then $\{X_1>x\}=\{N_x=0\}$ so that $P(X_1>x)=P(N_x=0)=e^{-\lambda x}$. Also take a look here. $\endgroup$
    – drhab
    Aug 12 '17 at 11:27
  • $\begingroup$ Quote: "The Poisson counting process can also be defined by stating that the time differences between events of the counting process are exponential variables with mean $\frac1{\lambda}$." $\endgroup$
    – drhab
    Aug 12 '17 at 11:33
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    $\begingroup$ That is a correct description. I could imagine that you have some trouble with a formal verification of that, but not so much that you have trouble with "seeing" that. Essential is that the interarrival times have exponential distribution. This with parameters $\lambda$, $2\lambda$, $4\lambda$..This is the only thing we need, and we it deduce from the datum that we are dealing with a Poisson proces. The words "Poisson process" can in fact be left out if they are replaced sensibly by "exponential distribution". $\endgroup$
    – drhab
    Aug 12 '17 at 11:58
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    $\begingroup$ A slight modification on the quote I gave you: "This Poisson counting process can also be defined by stating that the time differences between events of the counting process are exponential variables with mean $\frac1{\lambda}$, $\frac1{2\lambda}$, $\frac1{4\lambda}$ etc." $\endgroup$
    – drhab
    Aug 12 '17 at 12:07
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You have to specify what you want to achieve. If your just want numerically calculate the pdf, just convolve two exponential distributions.

Trying to find the MGF of the sum of one exponential distribution with $\lambda$ and one with $2\lambda$, $$ \frac{\lambda}{\lambda-t} \cdot \frac{2\lambda}{2\lambda-t} $$ I don't think this matches any common distribution. If you multiply their characteristic functions instead, you can do inverse Fourier to get the explicit expression of pdf, but it seems at least to involve special functions. Also, beware of the convention of the calculation of CF.

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