Question about a point process Q) Let the number of calls we receive on a telephone have a mean density of $1$ per hour. Suppose that we don't get calls at the same time and the number of calls in disjoint intervals of time are independent of each other.
1) We survey $10$ days of calls. How many pairs of calls do we expect to see within $5$ minutes of each other. (If there are $k$ calls in a given $5$ minute interval, there are ${k \choose 2}$ pairs).
2) Suppose that all the calls get delayed by a random amount which is an exponential distribution with mean $10$ minutes. Show that the resulting call times have the same distribution as before.
I am not sure if the question is missing information about the point process being a Poisson point process. I am going to assume that for now.
Let $X$ be the number of calls in a given $5$ minute interval, then $X\sim Pois(1*\frac{5}{60})$ and thus the expected number of calls in a given $5$ minute interval $= \frac{1}{12}$. In $10$ days, we have $\frac{10\times 24\times 60}{5}$ disjoint intervals of $5$ minutes. So would the answer to $(1)$ be $\frac{10\times 24\times 60}{5}\times {1/12 \choose 2}$?
Not sure how to address $(2)?$

For $(1)$ let $n$ be the number of days surveyed, $T$ the number of minutes in the interval, and $\lambda$ the rate of the arrival process. Then the expected pairs of calls within $T$ minutes of each other is given by
\begin{align}
\mathbb E[N(n,T,\lambda)] &= \int_{\frac T{60}}^{8n}\left( \sum_{k=0}^\infty \binom k2\cdot\mathbb P\left(N(t) - N\left(t-\frac T{60}\right) = k \right)\right)\ \mathsf dt\\
&= \int_{\frac T{60}}^{8n}\left( \sum_{k=0}^\infty \binom k2 e^{-\lambda T/60}\frac{(\lambda T/60)^k}{k!}\right)\ \mathsf dt\\
&= \int_{\frac T{60}}^{8n}\left(\left(\frac{\lambda T}{60}\right)^2\sum_{k=0}^\infty e^{-\lambda T/60}\frac{(\lambda T/60)^k}{k!}\right)\ \mathsf dt \\
&= \int_{\frac T{60}}^{8n} \left(\frac{\lambda T}{60}\right)^2\ \mathsf dt\\
&= \int_{\frac T{60}}^{8n}\frac{\lambda  T}{60}\ \mathsf dt\\
&= \frac{\lambda ^2 T^2 \left(8 n-\frac{T}{60}\right)}{7200}
\end{align}
(Note that $T$ has unit $\mathrm{min}$ and $\lambda$ has unit $\mathrm{hour^{-1}}$, hence the division by $60$.) Substituting in your values of $n=8$, $T=5$, and $\lambda = 1$, we have $N= \frac{767}{3456}$.
$(2)$ simply follows from the memoryless property of the exponential distribution. That is, if $X\sim\mathrm{Expo}(\lambda)$ then for $s,t>0$:
\begin{align}
\mathbb P(X>s+t\mid X>s) &= \frac{\mathbb P(X>s+t,X>s)}{\mathbb P(X>s)}\\
&= \frac{\mathbb P(X>s+t)}{\mathbb P(X>s)}\\
&= \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}}\\
&= e^{-\lambda t}\\
&= \mathbb P(X>t).
\end{align}
The times between arrivals in a Poisson process are exponentially distributed, so delaying the arrivals by an exponentially distributed amount of time does not affect their distribution.
