Question

must be Rstudio code. can use other languages to do the problem but please convert to R code

The median is the middle of a dataset. For example, if your observations are c(1,2,3,8,10), the median is 3. Note ã = 1+2+3+8+10 4.8, so they're not necessarily equal. 5 The interquartile range is the difference between the 75% and 25% percentiles. The function IQR() in R can be used to compute it. Or, compute the 75% percentile with the function quantile(), and then the 25%, and subtract. This might be how you have to do it in MATLAB. Sometimes, IQR is used as an estimate for the standard deviatio o. Show plots and sumamrize your work for this problem. 1a) Using samples from the distribution N(0,1), create and implement a simulation to investigate whether the median is an unbiased and/or efficient estimator of the mean . 1b) Using samples from the distribution N(0,1), create and implement a simulation to investigate whether IQR is an unbiased and/or efficient estimator of the standard deviation o.

Answer 1

$1a.~The~pdf~of~N(\mu,\sigma^2)~is\\ f(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2} \right )\\\\ \log f(x)=-\frac{1}{2}\log 2\pi-\frac{1}{2}\log \sigma^2-\frac{(x-\mu)^2}{2\sigma^2}\\\\ \frac{\mathrm{d} \log f(x)}{\mathrm{d} \mu}=\frac{x-\mu}{\sigma^2}\Rightarrow \frac{\mathrm{d}^2 \log f(x)}{\mathrm{d} \mu^2}=-\frac{1}{\sigma^2};~\\\\ I(\mu)=Fisher's~information~for~\mu=-nE\left(\frac{\mathrm{d}^2 \log f(x)}{\mathrm{d} \mu^2} \right )=\frac{n}{\sigma^2}\\\\ Efficiency=\frac{\sigma^2/n}{Var(\tilde{x})}\times 100\%=\frac{1}{nVar(\tilde{x})}\times 100\%;~\\\\ where~\tilde{x}=sample~median,~n=sample~size,~\sigma=1.\\\\$

R code:

N=1000
n=10
me=1:1000*0
sigma=1:1000*0
for(i in 1:N)
{
x=rnorm(n)
me[i]=median(x)
}
round(mean(me),4)
var(me)
round((1/(n*var(me)))*100,2)

Estimate of $\mu$ =sample median=-0.0040 with efficiency=66.97%.

1b.

$1a.~The~pdf~of~N(\mu,\sigma^2)~is\\ f(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2} \right )\\\\ \log f(x)=-\frac{1}{2}\log 2\pi-\log \sigma-\frac{(x-\mu)^2}{2\sigma^2}\\\\ \frac{\mathrm{d} \log f(x)}{\mathrm{d} \sigma}=-\frac{1}{\sigma}+\frac{(x-\mu)^2}{\sigma^3}\Rightarrow \frac{\mathrm{d}^2 \log f(x)}{\mathrm{d} \mu^2}=\frac{1}{\sigma^2}-\frac{3(x-\mu)^2}{\sigma^4};~\\\\ I(\sigma)=Fisher's~information~for~\sigma=-nE\left(\frac{\mathrm{d}^2 \log f(x)}{\mathrm{d} \sigma^2} \right )=\frac{2n}{\sigma^2}\\\\ Efficiency=\frac{\sigma^2/2n}{Var(Q)}\times 100\%=\frac{1}{2nVar(Q)}\times 100\%;~\\\\ where~Q=sample~IQR,~n=sample~size,~\sigma=1.\\\\$

R code:

N=1000
n=10
Q=1:1000*0
sigma=1:1000*0
for(i in 1:N)
{
x=rnorm(n)
Q[i]=IQR(x)
}
round(mean(Q),4)
var(Q)
round((1/(2*n*var(Q)))*100,2)

Here Q=1.1564 with efficiency=28.61%.