= Solution
{c}
<Independent increments> give independent Poisson variables $N[0,s]$ and $N(s,t]$ with means $\lambda s$ and $\lambda(t-s)$. Dividing their joint probability by $\mathbb P(N[0,t]=n)$ yields
$$
\mathbb P(N[0,s]=k\mid N[0,t]=n)
=\binom nk(s/t)^k(1-s/t)^{n-k},\quad0\leq k\leq n.
$$
Thus the <conditional distribution> is <binomial distribution> with parameters $n,s/t$.
The event $T_i>s$ means that at most $i-1$ events have occurred by $s$. Consequently
$$
\boxed{\mathbb P(T_i>s\mid N[0,t]=n)
=\sum_{k=0}^{\min(i-1,n)}\binom nk(s/t)^k(1-s/t)^{n-k}.}
$$
For $1\leq i\leq n$, the conditional event time lies in $[0,t]$. Integrate its <survival function> and substitute $u=s/t$:
$$
\mathbb E(T_i\mid N[0,t]=n)
=t\sum_{k=0}^{i-1}\binom nk\int_0^1u^k(1-u)^{n-k}du
=t\sum_{k=0}^{i-1}\frac1{n+1}
=\boxed{\frac{it}{n+1}}.
$$
The index range $i\leq n$ is implicit in the source's last assertion. For $i>n$, the $i$th event is after $t$, and independent future increments instead give expectation $t+(i-n)/\lambda$; the displayed $it/(n+1)$ formula does not apply.
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