Question

# 3. The following code draws a sample of size $n=30$ from a chi-square distribution with 15 degrees of freedom, and then puts $B=200$ bootstrap samples into a matrix M. After that, the 'apply' function is used to calculate the median of each bootstrap sample, giving a bootstrap sample of 200 medians. "{r} set.seed (117888) data=rchisq(30,15) M=matrix(rep(0,30*200), byrow=T, ncol=30) for (i in 1:200 M[i,]=sample(data, 30, replace=T) bootstrapmedians=apply(M,1,median) (3a) Use the 'var' command to calculate the variance of the bootstrapped medians. (This provides valid estimate of the variance of the median of a random variable.) ANSWER: "{r} # add your own R code below this linel (1 point for variance) (36) Use the apply command to calculate bootstrapped means rather than bootstrapped medians, and then use 'var' to calculate an estimate of the variance of the mean (hint: use 'apply' on the 'mean'). ANSWER: " '{r} # add your own R code below this line
$Modify the function slightly so that it returns the value: $\sum_{i=1}^n e^{x_i}$ In other words, we want to write a function$

Answer 1

3a) R code

***{r} #add you own code below this line var (bootstrapmedians)

get this

[1] 1.276356

ans: the variance of the bootstrapped estimate medians is 1.2764

3b) R code

***{r} #add you own code below this line set. seed (117888) data=rchisq(30,15) M=matrix(rep (9,30*200), byrow=T, ncol=30) for (i in 1:200) M[I,]=sample (data, 30,replace=T) bootstrapmeans=apply(M,1,mean) var (bootstrapmeans) 1

get this

[1] 1.258912

ans: The variance of bootstrapped mean is 1.2589

3c) The theoretical variance of the sample mean is 1. The variance of bootstrapped mean is 1.2589.

ans: Yes, the bootstrap has done a reasonable job of estimating the variance of the mean as the estimated variance is 1.2589 and it is close to the theoretical variance, which is 1.

4) As given below,

- **{r} myfun=function(){ if(length(x)=-1){return(x) return x, when there is only one element in x } else return(x[1]+myfun(x[-1])) recursively call mayfunction(), by passing, without the first element of x and add the result to the first element of x myfun(c(1:1000)) #here's what happens when it is applied to 1:1000

The line x[1]+myfun(x[-1]), adds the first element of x to the output from myfun(x[-1]), where x[-1] indicates the vector x without the first element. That means for each recursive call of myfun, we keep adding the first element of the remaining x to the sum of first elements. We do this till there is only one element left in x. Hence myfun(x) gives the sum of the elements of the vector x.

For example, if we run the code, we should get the sum of 1+2+...+1000=(100+1)*1000/2=500500

R code

myfun=function() { if(length(x)==1){return(x) }else return(x[1]+myfun (x[-1])) myfun(C(1:1000)) sum(c(1:1000)) }

get this

[1] 500500 [1] 500 500

indicating both myfun(x) and sum(x) give the sum of the elements of vector x.

ans: myfun(x) gives the sum of the elements of the vector x. In fact it simulates the function sum(x)

R code to get

$\sum_{i=1}^ne^{x_i}$

$1596246058983_image.png$

get this

[1] 31.6847

All the code in text format

code given
```{r}
set.seed(117888)
data=rchisq(30,15)
M=matrix(rep(9,30*200),byrow=T,ncol=30)
for (I in 1:200) M[I,]=sample(data,30,replace=T)
bootstrapmedians=apply(M,1,median)
```

part 3a)

```{r}
#add you own code below this line
var(bootstrapmedians)
```

part 3b)

```{r}
#add you own code below this line
set.seed(117888)
data=rchisq(30,15)
M=matrix(rep(9,30*200),byrow=T,ncol=30)
for (I in 1:200) M[I,]=sample(data,30,replace=T)
bootstrapmeans=apply(M,1,mean)
var(bootstrapmeans)
```

part 4)

```{r}
myfun=function(x){
if(length(x)==1){return(exp(x))
}else
return(exp(x[1])+myfun(x[-1]))
}
myfun(c(2,1,3,0.4))
```