# How do I convert complementary event rates into probabilities?

Suppose there are three complimentary events well, sick, dead. At anytime, if you are not sick nor dead, you are well.

From literature we know the rate of sick is 123 / 100,000 per year, and rate of dead is 456 / 100,000 per year.

1. What will be the probabilities of being sick and dead once in 5 years time respectively?
2. What if the two probabilities are given in different time frames? Say dead rate is 1 / 100,000 per month.
• Is it a homework? – Hossein Apr 4 '17 at 14:06
• No. I am running a Markov simulation, but to do that I need to convert 30-day mortality rates and 1-year readmission rates to probabilities at 6 month intervals. I thought this presentation is easier to understand than medical literature wordings though. – user2513881 Apr 4 '17 at 14:08
• To get complimentary events you just compute 1-P(A) where A represents the event. – Michael Chernick Apr 4 '17 at 14:09
• Literature pointed me to the equation $p_{t'} = 1-(1-p)^{t/t'}$. Using this equation to scale probabilities simply make the complement event blow up and become negative. – user2513881 Apr 4 '17 at 14:10
• In question one, do you mean probability of (sick OR dead)? – Hossein Apr 4 '17 at 14:11

Let be $r_1 = 0.00123$ and $r_2 = 0.00456$ the yearly rate of being sick and death respectively.

We have that the probability to survive one year is $$S(1) = 1-(r_1+r_2) = 0.99421.$$

In general the probability to survive until year $t$ is $$S(t) = S(1)^t,$$ so the probability to survive until year 5 is $S(5) = 0.9713833$.

Let's now calculate the probability of being sick. This probability is the sum of being sick during the first year plus being sick during the second year when surviving the first year plus being sick during the third year when surviving the first two years plus... In other words,

\begin{align}P(\text{"sick during 5 years"}) = P(\text{"sick during one year"}) + S(1) P(\text{"sick during one year"}) + \dots + S(4) P(\text{"sick during one year"})\end{align}

Therefore,

$$P(\text{"sick during 5 years"}) = r_1 + S(1) r_1 + S(1)^2 r_1 + S(1)^3 r_1 + S(1)^4 r_1 = 0.006079194.$$

In a similar way we can calculate the probability of dying during the five years

$$P(\text{"dying during 5 years"}) = r_2 + S(1) r_2 + S(1)^2 r_2 + S(1)^3 r_2 + S(1)^4 r_2 = 0.0225375$$

Notice that $0.9713833+0.006079194+0.0225375 = 1.$

If you have the rates in months you can apply the same reasoning but in months. To calculate "dying during 5 years" (5x12=60) instead of 5 terms you will have 60 terms. In months, you will get a more accurate result.

Here, the proposed solution is an adaptation of what is called the Aalen-Johansen estimator, which calculates the probability of event from observed data when the appearence of one event in one person provoques that the other event cannot appear or is not considered on that person (it is said that events are competing, competing risk survival analysis).

• This seems almost correct to me. Do note this assumes rates are stable and the state a person is in in one year does not influence his/her risk of specific other states during the next. Additionally and more importantly, aside from constant rates, for this answer you'd have to assume this is a sample which gets 'refilled' every year (those dying obviously cannot get sick or survive a year afterwards and would have to be replaced by new individuals to keep the probabilities you derived from the rates constant). – IWS Apr 4 '17 at 15:30
• yes, you are right, but without other restriction I think that for what is said in the question I should assume rates are stable. – marc1s Apr 4 '17 at 15:32
• Nice answer. I have only one problem with your method. @user2513881 wants the probability of being sick in EXACTLY one year, but in your method, you count the events that this person is sick in more than one year. Am I right? – Hossein Apr 4 '17 at 15:48
• I only check the first time is sick, afterwards the person is removed and not followed – marc1s Apr 4 '17 at 15:49
• What if, sick is in yearly rate but dead is in monthly rate? It is not unusual because 30-day mortality rates might be used for critical illness survivors. (Again, we assume sick is sick with minor ailment and is independent from dead.) – user2513881 Apr 5 '17 at 2:10

Let $s_i$, $d_i$, and $w_i$ be the events that this person is sick, dead, and well in the $i$-th year, respectively. We know that: $$\text{Pr}(s)=0.00123$$ $$\text{Pr}(d)=0.00456$$ $$\text{Pr}(\bar{d})=1-0.00456=0.99544$$ $$\text{Pr}(w)=0.99421$$ As you mentioned in the comments, you want the probability that a person is sick in exactly one year in 5 years. But to shorten the answer, assume that we are considering three years instead of five. The event of being sick in exactly one year is equal to occurring one of the following disjoint events:

$$s_1,w_2,w_3$$ $$s_1,w_2,d_3$$ $$s_1,d_2$$ $$w_1,s_2,w_3$$ $$w_1,s_2,d_3$$ $$w_1,w_2,s_3$$ To compute the probability of these events, you need more subtle information. For example, to compute the probability of the first event, you need $\text{Pr}(w_2|s_1)$. But with an independence assumption, we can simply compute the probability of each of the above events. For example, $$\text{Pr}(w_1,s_2,d_3)=0.99421\times 0.00123 \times 0.00456$$ Other probabilities can also be simply computed.

The event of dying in one year is equal to occurring one of the following events: $$d_1$$ $$\bar{d_1},d_2$$ $$\bar{d_1},\bar{d_2},d_3$$ These probabilities are also simply computed. For example, $$\text{Pr}(\bar{d_1},d_2)=0.99544\times 0.00456$$

Note that the made assumption is necessary to obtain the above results, although it is not a reasonable assumption. For example, when a person is not sick in 4 years, she/he is less likely to be sick in the fifth year.

• What does it mean $\text{Pr}(\bar{s_2} | s_1)$ when the event is death? Is the probability of being alive in year 2 when you have died in year 1? – marc1s Apr 4 '17 at 15:39
• It is the probability of being sick in the first year and not being sick in the second year. – Hossein Apr 4 '17 at 15:43
• The answer is now completely updated. – Hossein Apr 4 '17 at 16:20