I am studying consecutive independent events that last more or less 11 hours. The duration of the event follows a Normal distribution ($\mu_0 = 11$, $\sigma_0 = 1.1$). When an event finishes, a new one starts immediately, forming a chain of events. At $t=0min$, all chains of events (thousands of them) are perfectly synchronized, but of course they de-synchronize slowly.
Playing with the data, I realized that the end time of an event follows a Normal distribution too ($\mu = \mu_0n$, $\sigma = \sigma_0\sqrt{n}$ with $n$ the generation of the event in its chain).
I am able to reproduce this observation numerically using R:
s0 <- 1.1 # Standard deviation
l0 <- 11 # Average duration of events (in hours)
num <- 5 # Number of generations to compute
# Initial length (t) of all chain of events is zero
t <- rep(0, 10000)
for (i in seq(num)) {
# Add the result of random draws to the time length of each chain of events
t <- rnorm(length(t), mean = l0 + t, sd = s0)
# Display an histogram of the end time of each chain of events
hist(t, main = paste0("Histogram of generation #", i), freq = F)
# Draw the theoretical end time
xs <- seq(min(t), max(t), 0.05)
lines(xs, dnorm(xs, i*l0, sqrt(i)*s0), col = 'red')
}
But I cannot figure out a proof for that. I am not even sure what is happening mathematically when the R script is running.
Any help would be very much appreciated to point me in the right direction.