Within a Poisson arrival process, the arrival rate is given as 4 per year. Let's say there is a bet that rewards the holder a function dependent on the arrival time $t$ each time an arrival occurs such as $p(t) = \frac{1}{2t}$. In that case what is the expected reward gained in a year?
I was thinking that within a year the expected arrival times are $0.25, 0.5, 0.75, 1.0$ and the expected rewards from these are $2 + 1 + 0.66 + 0.5$. But I do not know if this is a solid approach. Can you help me?
Assume $n$ arrivals occur in $[0,1)$ where $n\geq 1$ is a fixed positive integer. Take $T_k$ as the arrival time of the $k^{\text{th}}$ arrival in $[0,1)$. It is known that $T_k\sim \text{Beta}(k,n-k+1)$.
If $X$ represents your reward in a year, then $X|N=n$ has the same distribution as $p(T_1)+\dots+p(T_n)$ where $N\sim \text{Poisson}(4)$. Moreover, $$\begin{eqnarray*}\mathbb{E}(X)&=&\sum_{n=0}^\infty\mathbb{E}(X|N=n)\mathbb{P}(N=n) \\ &\geq& \mathbb{E}(X|N=1)\mathbb{P}(N=1) \\ &=&\mathbb{E}\left(\frac{1}{2T_1}\right)\mathbb{P}(N=1) \\ &=& \int_0^1 \frac{\mathrm{d}x}{2x}\cdot \frac{e^{-4}\cdot 4^1}{1!} \\ &=&+\infty\end{eqnarray*}$$