Expected Poisson Increments with Stochastic Intensity
Summary
For a Poisson counting process whose intensity varies randomly over time, the note asks how to find the expected increment. It contrasts a fixed intensity, where the conditional expected increment over a small time interval is the intensity times the interval, with a stochastic intensity.
The proposed method conditions on the current intensity and then takes the unconditional expectation. This gives an expected increment equal to the mean intensity times the time interval. The reasoning uses iterated expectation and does not require the intensity process to be Gaussian. The note offers no derivation of the counting process assumptions or treatment of dependence, predictability, or interval-level variation in intensity, so those details must be checked in applications.
Key ideas
- Condition on the stochastic intensity to evaluate the expected counting increment.
- Iterated expectation reduces the unconditional increment to the expected intensity multiplied by the time interval.
- The result depends on the mean intensity, rather than requiring a particular distribution such as a Gaussian process.
- The note does not discuss the process assumptions needed for the conditional increment formula.
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# How to calculate the expectation of Poisson process when its intensity is also stochastic # How to calculate the expectation of Poisson process when its intensity is also stochastic How to calculate the expectation of Poisson process $N_t$ when its intensity is also stochastic? Since when intensity $\lambda_t$ is non-random, then we have $$E[dN_t] = \lambda_tdt.$$ But how about the stochastic $\lambda_t?$ I have no idea to calculate it. You can simply assume $\lambda_t$ is a Gaussian process. ## Answer by Ezy (score 7, accepted) https://quant.stackexchange.com/a/42235 You can condition on the value of $\lambda_t$. So $E[dN_t] = E[E[dN_t|\lambda_t]] = E[\lambda_t dt] = E[\lambda_t] dt$
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