How to translate an individual confidence interval to a confidence interval describing a large population? I have a population and way to stratify risk in that population. For each risk group, I have the probability of an event happening at 5 years and a 95% CI generated from R's survfit (i.e. Kaplan-Meier estimate) function with conf.type =  "log":
Probability of an individual patient relapsing by 5 years (95% CI):
Risk Category      Probability
     1             4%  (2.1-7.5)
     2             12% (8.2-15)
     3             23% (20-30)
     4             45% (33-45)

Assuming the probability table above is true, I can calculate the expected number of events to happen within 5 years by mapping each patient to their risk category and then mapping the risk category to it's population and summing it all up. Can I do something similar with the 95% Confidence interval. 
So I can say: "226 patients in our population are expected to relapse by 5 years, 95% CI: 199-256".
 A: Let $y_{cat_i}$ be the probability of a patient in category $i$ relapsing in $5$ years.
Let $y_{total}$ be the probability of a patient relapsing in $5$ years. 
If we knew all $y_{cat_i}$, then we could compute $y_{total}$ using:
$$y_{total} = \sum{P(relapse|cat_i)*P(cat_i)} = \sum{y_{cat_i}*P(cat_i)},$$
assuming all risk categories are mutually exclusive.
Let $X_{cat_i}$ be the estimator of $y_{cat_i}$. Let the confidence intervals associated with the estimators $X_{cat_i}$ be $[CI_{LB_i}, CI_{UB_i}]$.
Using the definition of Confidence Intervals, we have
$$P(CI_{LB_i} <= y_{cat_i} < CI_{UB_i}) = 0.95$$
This says the event that the estimates of $X_{cat_i}$ between $CI_{LB_i}$ and $CI_{UB_i}$ containing the true value $y_{cat_i}$ has the probability 0.95
which implies
$$P(CI_{LB_i}*P(cat_i) <= y_{cat_i}*P(cat_i) < CI_{UB_i}*P(cat_i)) = 0.95$$
Let $E_i$ represent this event. So $P(E_i = 0.95)$
Assuming these events for different risk categories are independent, the probability of all of the events $E_1$, $E_2$, $E_3$ and $E_4$ occurring is $0.95^4$.
$$P(\cap_i{E_i}) = 0.95^4$$
This event can be represented by the following:
$$CI_{LB_1}*P(cat_1) <= y_{cat_1}*P(cat_1) < CI_{UB_1}*P(cat_1)$$
$$CI_{LB_2}*P(cat_2) <= y_{cat_2}*P(cat_2) < CI_{UB_2}*P(cat_2)$$
$$CI_{LB_3}*P(cat_3) <= y_{cat_3}*P(cat_3) < CI_{UB_3}*P(cat_3)$$
$$CI_{LB_4}*P(cat_4) <= y_{cat_4}*P(cat_4) < CI_{UB_4}*P(cat_4)$$
If we sum these inequalities we get,
$$\sum_i(P(cat_i)*CI_{LB_i}) <= y_{total} < \sum_i(P(cat_i)*CI_{UB_i})$$
which occurs with probability 0.95^4 (~ 0.81)
So if you want $0.95$ CI, you might have to find out $0.95^{0.25} = ~0.98$ CI for the individual risk categories and use them to get $0.95$ CI for the full population.
