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Suppose that, for all $n$, $B_{j,n}$ for $j=1,...,n$ is a triangular array of Bernoulli variates. Define

$$ p_{j,n} = \mathbb{P}\left[B_{j,n}=1\mid B_{j-1,n}\right] $$

and assume that

$$ p_{j,n} = p_{j-1,n}+\alpha\,(\psi-p_{j-1,n})+\nu\,B_{j-1,n}=p_{j-1,n}\,(1-\alpha)+\alpha\,\psi_n+\nu\,B_{j-1,n} $$

with $p_{0,n}=\psi>0$, $\alpha>0$ and $\nu>0$. In essence, the Bernoulli process is self-exciting in the sense that a $B_{j-1,n}=1$ increases the probability of having $B_{j,n}=1$.

My purpose is to compute $\mathbb{E}\left[B_{j,n}\right]$ recursively. I can find a recursive formula for $p_{j,n}$

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whence, by induction,

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Nevertheless I cannot find a similar formula for $\mathbb{E}\left[B_{j,n}\right]$, any idea?

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    Yes, but I need a recursive formula. I have such a formula only for the conditional expected value, what I need is the unconditional one. – 2017-01-26
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    By definition, $p_{j,n}$ is $\sigma(B_{j-1,n})$-measurable. However, the rhs of the recursive definition, namely the term $p_{j-1,n}$, is not. – 2017-01-26
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    Maybe I am not understanding your question, but don't you already have a recursive formula for $\mathbb E[B_{j, n}]$ since $$ \mathbb E[B_{j, n}] = \mathbb E[ \mathbb E[ B_{j, n} \mid B_{j-1, n}]] = \phi + \frac{\nu}{1-\alpha} \sum^{j -1}_{l = 0} \mathbb E[B_{l, n}](1-\alpha)^{j-l} $$? – 2017-01-26
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    You'r right, could you write as an answer for the bounty? It expires in just three hours. – 2017-01-27

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By law of total expectation, you readily have a recursive formula for the expectation:

$$ \mathbb E[B_{j, n}] = \mathbb E[\mathbb E[B_{j, n} \mid B_{j-1, n}]] = \phi + \frac{\nu}{1 - \alpha} \sum^{j - 1}_{l = 0} \mathbb E[B_{l, n}](1-\alpha)^{j-l} $$