# waiting for two events with given average and stddev

Imagine that I am working in a fire station. For the truck to leave, I need all my firefighters. They receive their alarm at the same time. Their time to come to the station can be modelled as a random normal variable having the average $\mu_i$ and standard deviation $\sigma_i$.

Is it possible from those values to determine the average waiting time for the truck to leave and its standard deviation?

Edit on 15/05:

I tried to compute it the following (wrong) way: I can compute the probability that each variable is greater than another one $P(A-B>0) = 1-CDF(0;\mu_A - \mu_b; \sqrt{\sigma_A ^2 + \sigma_B ^2})$. Next step I do is to compute this for every possible pair. I then compute the probability that a given variable is the maximum value $P(max = A) = P(A-B>0) \times P(A-C>0) \times ...$.

From there, I thought that I could determine the average time to wait as $P(max = A) \times \mu_A + P(max = B) \times B + ...$.

This is wrong because if all random variable are equals, the sum of all $P(max = .)$ is not 1.

I don't know what is wrong exactly, but I get the feeling that I may have to take into account the fact that two variable can be equal to each other. But in such case, I don't have a real intuition on how this should be taken into account.

• Note that a random normal variable might get negative values. A rectified normal variable might be better to model the problem. But anyway, are you asking your question for a specific distribution, or for any kind of distribution ? May 14 '18 at 9:14
• Indeed, a rectified normal variable models the problem better May 14 '18 at 10:12

One generic solution would consist in running a significant amount of simulations and calculate the resulting averages and standard deviations of your experiments.

Example with R code and three rectified normal distributions.

sim <- data.frame(apply(data.frame(a = rnorm(100000,3,1), # simulate normal distributions
b = rnorm(100000,5,2),
c = rnorm(100000,6,4)), 2,
function(x) ifelse(x > 0, x, 0))) # rectify
sim$final <- apply(sim, 1, max) # final results corresponds to last arrival mean(sim$final)
# 7.395576
sd(sim\$final)
# 2.75843


Obviously, you will need to run this simulation a couple of times to get a reliable estimation of average/standard deviation.

• This is indeed a nice pragmatic solution. However, it is too computationally expensive for my case. May 14 '18 at 10:15
• Really ? This script did not even take a second on a 8GB RAM machine. May 14 '18 at 10:23
• This problem will rise extremely often and therefore, the faster the solution, the better. May 14 '18 at 11:10
• This feels like a job for the weibull. Since the truck can’t leave until full, it’s an extreme value distribution. May 14 '18 at 22:06