Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 27J / c / Solution

For one visitor, letand defineA visitor arriving at time $s\le t$ is in room $j$ at time $t$ exactly when its independent service-time mark satisfies $T_{j-1}\le t-s<T_j$. Thus it receives mark $j$ with probability $p_j(t-s)$; a further mark records that it has already left the gallery.

The Independent marking theorem for Poisson point processes makes the point processes carrying the different room marks independent. Consequently $V_1(t),\dots ,V_{10}(t)$ are independent, and Poisson thinning givesEquivalently,

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