# Sum of truncated normal distribution

I need to estimate what's the probability that the sum of the height of 3 women + 7 men is greater than 17m considering that men follow a normal distribution [mu=1.8 sd=0.10] and women [mu=1.6 sd=0.08], with a catch: I have PRE-selected only men with height lower than 1.9m and women with height lower than 1.75m. Any ideas on how to proceed?

Thanks

-
I don't think closed form solution exists. You can try simulation procedure. – vinux Feb 27 '13 at 8:23
You could try numerical convolution, or simulation. – Glen_b Feb 27 '13 at 8:24
Please, add the Homework tag. – Zen Feb 28 '13 at 19:45

Let $Z\sim \mathrm{N}(\mu,\sigma^2)$. We say that $U$ has a truncated normal distribution $\mathrm{TN}(\mu,\sigma^2,(a,b))$ iff $$P(U\in B) = P(Z\in B\mid a<Z<b) \, ,$$ for every Borel set $B$. It is easy to prove that $$\mathrm{E}[U] = \mu + \frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)} \sigma \, ,$$ where $\alpha=(a-\mu)/\sigma$, $\beta=(b-\mu)/\sigma$, $\phi(t)=e^{-t^2/2}/\sqrt{2\pi}$, and $\Phi$ is the distribution function of a $\mathrm{N}(0,1)$ random variable.

In your problem, you have $$W_i=\mathrm{TN}(1.6,(0.08)^2,(0,1.75)) \, ,$$ for $i=1,2,3$, and $$M_i=\mathrm{TN}(1.8,(0.10)^2,(0,1.9)) \, ,$$ for $i=1,\dots,7$.

The following Theorem is due to the great Wassily Hoeffding.

Theorem. If $X_1,\dots,X_n$ are independent random variables such that $\mathrm{E}[X_i]=0$ and $a_i<X_i<b_i$, for $i=1,\dots,n$, then $$P\left(\sum_{i=1}^n X_i\geq \epsilon\right) \leq e^{-t\epsilon}\prod_{i=1}^n e^{t^2(b_i-a_i)^2/8}$$ for every $\epsilon>0$ and each $t>0$.

Now, define $X_i=W_i-\mathrm{E}[W_i]$, for $i=1,2,3$, and define $X_{j+3}=M_j-\mathrm{E}[M_j]$, for $j=1,\dots,7$. Note that $\mathrm{E}[X_i]=0$.

Now the bounds. For example, since $$0<W_i<1.75\, ,$$ you have $$-\mathrm{E}[W_i]<X_i<1.75-\mathrm{E}[W_i]\, ,$$ for $i=1,2,3$. Do the same for the men, and you will find all the $a_i$'s and $b_i$'s.

Using Hoeffding's inequality above, you can find, for each $t>0$, an upper bound (your estimate) for the probability $$P\left( \sum_{i=1}^3 W_i + \sum_{i=1}^7 M_i \geq 17 \right) = P\left( \sum_{i=1}^{10} X_i \geq 17 -3\mathrm{E}[W_1]-7\mathrm{E}[M_1]\right) \, .$$

Now, what is the value of $t$ that gives you the tightest upper bound?

Please, plug-in in the numbers and post the answer as a comment.

-
I wonder how well these bounds compare to the elementary Normal approximation based on computing the means and variances of the truncated distributions. – whuber Feb 28 '13 at 20:59

This is a quick simulation-based approach that vinux and Glen b allude to.

start <- Sys.time()
library("msm")
library("xts")
nsim <- 1000000
men <- rtnorm(7*nsim,mean=1.8,sd=0.1,upper=1.9)
women <- rtnorm(3*nsim,mean=1.6,sd=0.08,upper=1.75)
sums <- as.numeric(period.sum(men,seq(0,length(men),by=7)))+as.numeric(period.sum(women,seq(0,length(women),by=3)))
prob <- sum(sums>17)/nsim
prob
end <- Sys.time()
end-start


The probability is about 77.2% based on a simulation of 7M men and 3M women. This runs in about 6.4 seconds on my computer.

-
+1 This agrees with a normal approximation which gives the probability as 77.0% (the correct probability is 77.3%). – whuber Feb 28 '13 at 20:30