This is continuation of my problem

Calculate variance of sum random variables

Suppose random variable $X$ takes 3 values $1, 2, 3$ with probability $\frac{1}{2}$, $\frac{1}{3}$ and $\frac{1}{6}$. Sample 1000 times from $X$ independently and do the following operations. Take a number $T=0$. If 1 comes, add 0.18 with $T$, if 2 comes add 0.05 with $T$ if 3 comes add -0.29 with $T$. Then expected value of $T$ will be 58.33 and variance will be 27.65. Now I want to find the probability $T<50$. For this I tried to use central limit theorem namely I approximated the distribution by Normal $N(58.33,27.65)$. Then integrate the density function of this normal distribution from -infinity to 50 to get the probability. Can we use better approximation than this? Since we know higher moments of $T$, can we somehow use these values?


1 Answer 1


You are adding many i.i.d. random variables with nice distribution, so the CLT will work fantastically. If you are after an approximately exact result, you could have a look at the asymptotic expansions mentioned in the comments or FFT.

An other (simple) option is simulation: Here, we draw 100'000 times from the distribution of interest and calculate mean, variance and the probability to be below 50.

R code

one_rv <- function(n = 1000) {
  sum(sample(c(0.18, 0.05, -0.29), size = n, 
             replace = TRUE, prob = c(1/2, 1/3, 1/6)))

many_rv <- replicate(1e5, one_rv())
mean(many_rv)                             # Mean: 58.3512
var(many_rv)                              # Variance: 27.69896
mean(many_rv < 50)                        # Prob < 50 by simulation: 0.056600
pnorm(50, mean = 58.33, sd = sqrt(27.65)) # By normal approx:        0.056579

The probabilities are very close (and both versions are approximations).


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