Poisson process with variable intensity

1.1k Views Asked by At

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.

2

There are 2 best solutions below

0
On

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.

7
On

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 $2\lambda$.