SOLUTION TO hw5
===============

(a) Exact answer = 
> integrate(function(x) sqrt(1+x+x^4), 0,1, abs.tol=1e-10)
1.288417 with absolute error < 1.4e-14

(b) Exact discrete-probability answer to probability distribution
of # Z of distinct elements of (X1,..,X6) for iid discrete-uniform
variates X in 0:9 is 

   k        1        2        3        4       5        6
Prob(Z=k)  1e-5   .00279    .0648    .3276   .4536    .1512

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In both parts: it is OK to overestimate variances, i.e. bound 
them conservatively, in providiing confidence intervals widths.
It is also OK to make a preliminary estimate of variance based 
on a moderate-sized sample (say n0=400), then choose the 
actual n toattain the desired precision, and then to confirm by
re-estimating variance using the full sample that your CI has the
desired precision.

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Simulations for (a): 
> fa = function(x) sqrt(1+x+x^4)
  sqrt(var(fa(runif(1000))))  ### = .20321
## Using this + 0.2 as the estimated variance, use sample size 
##  n for simulation found by solving 
##     qnorm(.995)*.22321/sqrt(n) < .01 , or  n >= 3306
> ftmp = fa(runif(3306))
## now CI estimate = 
> mean(ftmp) + c(-1,1)*qnorm(.995)*sqrt(var(ftmp)/3306)
[1] 1.277210 1.295301     #### OK, the mean 1.2863 is close !

Simulation for (b): find sample size by using conservative 
bound 1/4 for variance of each of the indicators, and solving
##  n >= (qnorm(1-.01/6)/.01)^2*(.25) = 21538.5

## Here are two ways to do the simulation, both with n=22000:
> tmpmat <- array(0, dim=c(22000,10))
    for(i in 1:6) {
     tmpvec <- sample(1:10,22000, replace=T)
     tmpmat[cbind(1:22000,tmpvec)] = 1 }
  dvec <- c(tmpmat %*% rep(1,10))
> round(table(dvec)/22000, 4)
     2      3      4      5      6
0.0029 0.0647 0.3223 0.4565 0.1536
### Compare with true analytically calculated values:
 0.0028 0.0648 0.3276 0.4536 0.1512
### NB. The probability that only 1 distinct value occurs = 1e-5.

> max(abs(c(.00279,.0648,.3276,.4536,.1512)-table(dvec)/22000))
[1] 0.005327273

### The largest absolute error turns out to have been: .0053
### which indicates that our Bonferroni-inequality calculation
### was not terribly conservative.