Question

Recall that log 2 = 1 0 1 x+1 dx. Hence, by using a uniform(0,1) generator, approximate log 2. Obtain an error of estimation in terms of a large sample 95% confidence interval.

Answer 1

∫101/(x+1)dx=(1−0)∫101/(x+1)dx/(1−0)=∫101/(x+1)f(x)dx=E(1/(x+1))

Where f(x)=1, 0

And then I calculated log 2 from the calculator and got 0.6931471806

From R, I got 0.6920717

So, from the weak law of large numbers, we can see that the sample mean is approaching the actual mean as n gets larger.