[讨论]出两个小题给大家做做

是 否 +2 论坛币

k人 参与回答

经管之家送您一份

应届毕业生专属福利!

求职就业群

赵安豆老师微信：zhaoandou666

经管之家联合CDA

送您一个全额奖学金名额~ !

立即领取

感谢您参与论坛问题回答

经管之家送您两个论坛币！

+2 论坛币

1. Using R simulate n = 500 Cauchy random values and plot the values versus the cumulative mean.

2. Write an R program to simulate the expected number, E[N], of tests performed in group testing when:

• 1)  n = 100000, m = 10000, k = n/m, pi = 0.05, eta = 0.97, theta = 0.95 and
• 2)  n = 100000, m =   5000, k = n/m, pi = 0.05, eta = 0.97, theta = 0.95

扫码加我 拉你入群

请注明：姓名-公司-职位

以便审核进群资格，未注明则拒绝

分享0 收藏0 回帖

关键词：cumulative performed simulate Expected Perform 讨论

第一题可不可以这么做啊？

A = rcauchy(500)

B = cumsum(A)/(c(1:500))

plot(A~B)

恩，不错。不过我觉得用plot(B,A)可能符合题意

发第二题答案：

1)  n = 100000, m = 10000, k = n/m, pi = 0.05, eta = 0.97, theta = 0.95

> n = 100000

> m = 10000

> k = n/m

> Pi = 0.05

> Eta = 0.97

> Theta = 0.95

> p = 1 - ( Eta*Pi + (1-Theta)*(1-Pi) )

> N = 0

>

> for (i in 1:m){

+ if(rbinom(1,1,p^k) == 1){

+ N = N + 1

+ }

+ else{

+ N = N + k + 1

+ }

+ }

>

> ev.N = n*(1 + (1/k) - (p^k))

> ev.N

 

 

 

2)  n = 100000, m =   5000, k = n/m, pi = 0.05, eta = 0.97, theta = 0.95

 

> n = 100000

> m = 5000

> k = n/m

> Pi = 0.05

> Eta = 0.97

> Theta = 0.95

> p = 1 - ( Eta*Pi + (1-Theta)*(1-Pi) )

> N = 0

>

> for (i in 1:m){

+ if(rbinom(1,1,p^k) == 1){

+ N = N + 1

+ }

+ else{

+ N = N + k + 1

+ }

+ }

>

> ev.N = n*(1 + (1/k) - (p^k))

> ev.N