Question

Let Xi be iid with E(Xi) = 0 and Var(Xi) = 1 and let Sn = X1 + … + Xn. Consider the limiting behaviors of Sn/n and of Sn /n. Does either of these correspond to the LLN? to the CLT? Demonstrate using UNIF(–3, 3).

Answer 1

# Weak law of large number Basically, weak law of large number is nothing but sample mean equate to population mean And Sn=x1+x2+......+ xn And Sn/n=x1+x2+......... + xn /n =xbar=sample mean & given that El Xi ] =0 therefore n=1000 xbar=c() for(i in 1:n) xbar[i]=mean(rnorm(i.mean=0,sd=1)) xbar plot(xbar) 1.0 0.5 xbar 0.0 0.5 0 200 400 600 800 1000 Index Scanned with CamScanner

## central limit theorem n=1000 mu=0 S=1 xbar=c() Z=C() for(i in 1:n) xbar[i]=mean(rnorm(n,mu,s)) z(i)=(xbar(i)-mu)/(s/sqrt(n)) xbar hist(z.prob=TRUE) Histogram of z 0.3 Density 0.2 0.1 0.0 -2 2 NO- Scanned with CamScanner

## Weak law of large number using uniform (-3,3) n=1000 xbar=c() for(i in 1:n) xbar[i]=mean(runif(i,-3,3)) xbar plot(xbar) 1.0 So 0.0 xbar 1.0 -2.0 oo oo of O 0 200 400 600 800 3/ 4000 Index Scanned with CamScanner

## central limit theorem n=1000 a=-3 b=3 mu=(a+b)/2 var=((b-a)^2)/12 s=sqrt(var) xbar=c() Z=c() for(i in 1:n) xbar[i]=mean(runif(n,a,b)) z[i]=(xbar[i]-mu)/(s/sqrt(n)) xbar hist(z,prob=TRUE) Histogram of z Density 0.0 0.1 0.2 0.3 0.4 4 -3 -2 -1 0 1 2 3 Scanned with CamScanner
here, I write code in R because better visualization please understand all codes and see corresponding graphs

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