# CDF calculation

Help me please with the following question. I'm reading Gill's lectures on survival analysis http://www.math.leidenuniv.nl/~gill/stflour0.pdf On the page 27 he states that: by the Glivenko-Cantelli theorem CDF converge to its expectation. The problem is, I can't calculate expectation correctly. Let $T_i\sim F$ and $C_i\sim G$ are i.i.d. Consider the process $$F^1_n(t) = \frac 1n \sum _i \mathbf {1}(\min (T_i, C_i) < t, \mathbf 1(T_i\leq C_i) = 1)$$Its expectation should be $$\int (1-G)dF$$but I can't obtain it. My suggestions:\begin{align*}\mathbb EF^1_n(t) &= \mathbb P(\min (T_i, C_i) < t, \mathbf 1(T_i\leq C_i) = 1) \\&= \mathbb P(\min (T_i, C_i) < t\mid \mathbf 1(T_i\leq C_i) = 1)\mathbb P(1(T_i\leq C_i) = 1) \\ &= \mathbb P(T_i < t)\mathbb P(T_i < C_i) \\ &= F(t)\int (1-G(t))dF(t)\neq \int (1-G(t))dF(t)\end{align*}where am I wrong?

$$\mathbb{P}(T_i<t|T_i<C_i) \neq \mathbb{P}(T_i<t)$$ Knowing that $P_i$ was smaller than $C_i$ gives you information about the magnitude of $T_i$, indeed, it's more likely to be below $t$ than otherwise.
$$\mathbb{P}(P_i < t \wedge P_i < G_i) = \int_{-\infty}^{t} F'(x)(1-G(x))~\mathrm{d}x$$
The sum is over all possible values for $P_i$: $[-\infty, t)$, $F'(x)dx$ represent the probability for $T_i$ to be around $x$ and is multiplied by $1-G(x)$, the probability that $G_i$ was bigger.