FS2.Rmd

---
title: 'FS2'
date: "`Sys.Date()`"
output:
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    df_print: paged
    highlight: kate
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---
# Pre-requisites
```{r}
library(matlib)
library(data.table)
library(tidyverse)
```


# Linear Regression
```{r}
# Example 1
# Create the data.table
dt <- data.table(a=c(20,40,60,80,100), 
                 b=c(48, 55.1, 56.3, 61.2, 68))
# Create the intercept
dt <- dt %>% mutate(one=1)

# Get into matrices X and y respectively
X <- dt %>% select(one,a) %>% as.matrix()
y <- dt %>% select(b) %>% as.matrix()
# Solve for the regression coefficients
library(matlib)
inv(t(X)%*%X) %*% t(X)%*% y

18238 - 57.72*300
922/4000
57.72 - (0.2305)*60

c(1, 58) %>% t() %*% inv(t(X)%*%X) %*% t(X)%*% y
```

```{r}
# Example 2
dt <- data.table(x1=seq(2,14,2), 
                 y=c(3.2, 5.2, 5.5, 6.4, 7.2, 8.5, 9.8))
dt <- dt %>% mutate(one=1)


X <- dt %>% select(one, x1) %>% as.matrix()
Y <- dt %>% select(y) %>% as.matrix()
inv(t(X)%*%X) %*% t(X) %*% Y

422.6 - 8*45.8
56.2/112
6.542 - (56.2/112)*8 # 2.527714

```

```{r}
# Example 3
dt <- data.table(x1=seq(1, 3.8, 0.4),
                 y=c(0.96, 1.33, 1.75, 2.14, 2.58, 2.97, 3.38, 3.75))

dt <- dt %>% mutate(one = 1)
X <- dt %>% select(one, x1) %>% as.matrix()
Y <- dt %>% select(y) %>% as.matrix()
b <- inv(t(X)%*% X) %*% t(X) %*% Y


dt[,sum(x1*y)]
(52.04 - (1/8)*19.2*18.86) /(52.8 - (1/8) * 19.2^2)


dt %>% 
    ggplot(aes(x=x1, y=y)) + 
    geom_point(size=4, shape=3) + 
    geom_smooth(method="lm")

t(c(1, 2)) %*% b # 1.954166
```

```{r}
# 1A.10
b1 <- (41444 - (1/12)*540 * 780)/(30786 - (1/12)*540^2)

780/12 - b1*540/12 # 20.9852

b <- c(20.9852, b1)
t(c(1,40))%*%b # 60.10947
```

```{r}
# 1A.11
1844-0.1*112^2 # 589.6
6850 - 0.1 * 112 * 480 # 1474
b1 <- 1474/589.6 # 2.5
48-2.5*11.2 # 20
# p_i = 20 + 2.5 n_i

# B1 is the increase in cost associated with another 100 leaflets
# 
```
```{r}
dt <- data.table(one=1, x1=1:5, y=c(1.4, 2.9, 4.1, 5.8, 7.2))
X <- dt %>% select(one, x1) %>% as.matrix()
Y <- dt %>% select(y) %>% as.matrix()
b <- inv(t(X)%*%X)%*%t(X)%*%Y
b # 1.45 is the additional years of protection given by an additional coat of paint. 
# Not suitable when extrapolating, because you are unsure if a linear specification remains correct
t(c(1,7)) %*% b # 10.08 years

## 
dt <- rbind(dt, data.table(one=1, x1=c(6,8), y=c(8.2, 9.9)))
dt
X <- dt %>% select(one, x1) %>% as.matrix()
Y <- dt %>% select(y) %>% as.matrix()
b <- inv(t(X)%*%X)%*%t(X)%*%Y
t(c(1,7))%*%b # 9.204911

# Interpolating, have more data points. 
```
```{r}
# Example 4
204-(1/8)*36^2 # 42
35+28+24+16+15+12+8+6
478-(1/8)*36*144 # -170

b1=-170/42 # -4.047619
b1
a=144/8 - (-170/42)*(36/8)
a

700+5*(a+b1*(2.5))

```

```{r}
# 1B.6 - you got this wrong. Be careful. Take it slowly. 
dt <- data.table(one=1, x1=c(1,5,10,16,17), y=c(9,12,16,21,23))
X <- as.matrix(dt %>% select(one, x1))
Y <- as.matrix(dt %>% select(y))
b <- inv(t(X)%*%X)%*%t(X)%*%Y
a1 <- 5*b[1] - 4*b[2] # 35.94446
b1 <- b[2]/2 # 0.4250525
a1+b1*32 # 49.54614
```

```{r}
# 1B.7
b1 <- 15.26/10.21
a1 <- 9.88 - b1*4.58
hold <- function(x){
    return(a1+b1*x)
}
hold(x=(42-4)/8) # 10.13408
```

## 1.2
```{r}
# Example 7
# S_yy
371.75 - 0.2*42.9^2 # 3.668
# S_xx
589.09 - 0.2*53.7^2 # 12.352
# S_yy
467.45 - 0.2*53.7*42.9 # 6.704
# 

```


```{r}
# Question 2
dt <- data.table(x=c(2,5,8,9,11,15), 
                 m=c(110, 105, 103, 101, 96, 88))
dt <- dt %>% mutate(mhat=114.3 - 1.655*x,
                    res=m-mhat)
dt %>% select(res) %>% sum()

dt %>% 
    ggplot() + 
    geom_point(aes(x=x, y=res)) + 
    geom_hline(yintercept=0, linetype="dashed") 

# No clear pattern, think we're all good? 

```

```{r}
dt <- data.table(t=c(5.1, 5.7, 6.3, 6.4, 7.1, 7.2, 8.0, 8.3, 8.7, 9.1),
                 p=c(79, 81, 85, 86, 89, 84, 95, 96, 98, 99))
dt <- dt %>% mutate(phat=51.04 + 5.308*t, 
                    res=p-phat)
dt[abs(res)==max(abs(res)),] # t=7.2, p=84 is outlier. 
# Doesn't appear to follow the same pattern, so can remove?

dt2 <- dt[res!=-5.2576]

b <- dt2 %>% 
    lm(formula=p~t, data=.) %>%  # 51.036, t=5.308
    coefficients()
t(c(1, 7.8)) %*% b # 92.43787
```
```{r}
# Question 4 
dt <- data.table(x=c(1.2, 1.7, 2.4, 3.1, 3.8, 4.2, 5.1),
                 y=c(13.1, 12.5, 10.9, 9.4, 7.9, NA, 5.8))
dt %>% mutate(yhat=15.7 - 2.02*x, 
              res=y-yhat) %>% 
    select(res) %>% 
    sum(na.rm=T) # 0.346

dt <- dt %>% mutate(yhat=15.7 - 2.02*x, 
              res=y-yhat) # Therefore res = - 0.346 implies y = yhat + res = 7.216 - 0.346 

dt[x==4.2, y:=7.216-0.346]
dt %>% mutate(res=y-yhat) %>% select(res) %>% sum() # Almost 0
# a = 6.87
dt # There are three on one side, and four on the other so could be a good model. 

```

```{r}
# Question 5
4821 - (1/8)*193^2 # 164.875 = SYY
7720-(1/8)*236^2 # 758 = SXX
6046 - (1/8)*193*236 # 352.5 = SXY

164.875 - (352.5)^2/758 # 0.9485488

# Question 6
1.949 - (23.13)^2/289.4 # 0.1003583
# October has lower RSS with same units for outcome implies more likely for October.
```
```{r}
dt <- data.table(x=c(0.7, 1.3, 1.8, 2.3, 2.9, 3.8), 
                 y=c(15.4, 13.5, 12.1, 10.1, 8.5, 5.8), 
                 one=1)
b <- dt %>% 
    lm(data=., formula=y~x) %>% coefficients()

b # y = 17.549041 -3.116738 x
# a is the estimated value of the car when it is new
t(c(1,2))%*%b # 11.31557, so 11,316 pounds


model <- glm(data=dt, formula=y~x)
predict(model, newdata=as.data.frame(dt$x), type="response")
?predict
dt <- dt %>% mutate(yhat=17.549041-3.116738*x, 
                    res=y-yhat)
dt[,sum(res)] # Looks pretty good! Sum to zero
dt %>% 
    ggplot() + 
    geom_point(aes(x=x, y=res)) # No obvious trend

773.72 - (1/6)*65.4^2 - model$coefficients[2] * (120.03 - (1/6)*12.8*65.4) # 0.1147814 


```
Mixed exercises

```{r}
# Mixed exercises 1
b <- (31185 - (1/12)*553*549)/6193
a <- 45.73-b*46.083

a
b
t(c(1,50))%*%c(a,b)
```
```{r}
# Question 3
dt <- data.table(x=c(3.4, 7.7, 12, 75, 58, 67, 113, 131),
                 y=c(55, 240, 390, 1100, 1390, 1330, 1400, 1900), 
                 one=1)
dt
dt[,sum((x-mean(x))^2)] # S_xx = 16350.05
dt[,sum((x-mean(x))*(y-mean(y)))] # 210330.6
dt[,sum((x-mean(x))*(y-mean(y)))] /dt[,sum((x-mean(x))^2)] # 12.86422
dt[,mean(y)-12.86422*mean(x)] # 224.5154

B <- glm(data=dt, formula=y~x) %>% coefficients()
t(c(1,100)) %*% B # 1510.937

B
(3500-B[1])/B[2] # 254.6198 consumption
# No good! Can't invert a prediction by OLS

84.25 - (1/6)*13.5 * 25.5 # 26.875
26.875/59.88 # 0.4488143
(1/6)*13.5 - (26.875/59.88)*(1/6)*25.5 # 0.3425392
B <- c((1/6)*13.5 - (26.875/59.88)*(1/6)*25.5 + 2,0.4488143/2)
t(c(1,10)) %*% B

```
```{r}
# Question 6c
(0.16 + 0.79*50)*1.2
```
```{r}
# Question 7
dt <- data.table(x=c(140, 150, 170, 180, 180, 200, 220, 220), 
                 y=c(150, 180, 190, 220, 240, 290, 300, 310))
dt %>% 
    ggplot(aes(x=x, y=y)) + 
    geom_point(shape=3, colour="black", size=4) +
    geom_smooth(method="lm", se=F, linetype="dashed", size=0.5, )

dt %>% mutate(x2=x/10, y2=y/10) %>% glm(data=., formula=y2~x2) %>% coefficients()
B <- dt %>% mutate(x2=x/10, y2=y/10) %>% glm(data=., formula=y2~x2) %>% coefficients()
# recoding
B[1] <- B[1]*10
t(c(1,210)) %*% B # 289.5528
```

```{r}
dt <- data.table(n=c(10, 13, 22, 15, 24, 16, 19), 
                 w=c(2800, 3600, 6000, 3600, 5200, 4400, 5200))
dt <- dt %>% mutate(x=n-10, y=w/400)
dt[,sum((x-mean(x))^2)] # 148
dt[,sum((x-mean(x))*(y-mean(y)))] # 78

dt %>% 
    glm(data=., formula=y~x) %>% coefficients()

400*(7.310811 - 0.527027*10) # 816.2164
400*.527027 # 210.8108

dt %>% 
    glm(data=., formula=w~n)

preds1 <- function(x){
    return(7.310811 + 0.527027 * x)}
preds2 <- function(x){
    return(816.2164 + 210.8108 * x)
}

dt %>% mutate(yhat=preds1(x), res1=y-yhat,
              what=preds2(n), res2=w-what)

```
```{r}
dt <- data.table(t=c(2,3,5,6,7,9,10), 
                 TT=c(72, 68, 59, 54, 50, 42, 38))
dt <- dt %>% mutate(TThat=80.445-4.289*t, 
              res=TT-TThat)
dt %>% select(res) %>% sum() # 0.023 # Rounding error


957.43 - (-223)^2/52 # 1.103077

-sum(c(-0.177, -0.196, 0.526, 0.124, -0.129, -0.216, -0.032)) # 0.1

```

# Chapter 2 
```{r}
# Question 10
74458.75 - (1/8)*354.5^2 # 58749.97

5667.5/sqrt(11148*58749.97) # 0.2214569

# Question 11
Sxy=180.37 - (1/7)*12*97.7
Sxx = 22.02 - (1/7)*12^2
Syy=1491.69 - (1/7)*97.7^2

Sxy/sqrt(Sxx*Syy) # 0.9459204

dt <- data.table(x=c(1.1, 1.3, 1.4, 1.7, 1.9, 2.1, 2.5),
                 y=c(6.2, 10.5, 12, 15, 17, 18, 19))
cor(dt$x, dt$y, method="pearson") # 0.9459204
cor(dt$x, dt$y, method="spearman") 
dt %>% mutate(yhat=-1.2905 + 8.8945*x, 
              res=y-yhat)
# Appears to be a concave polynomial, so arguable trend in residuals and not good linear fit. 
```
```{r}
1-6*22/990 # 0.8666667

c(1,2,3,4,5,2, 2) %>% rank(.,ties.method = "average")

1-6/(6*7*5)
```
```{r}
dt <- data.table(a=1+2*runif(n=100, min=-1, max=1), 
           b=-1 + 1.5* runif(n=100, min=-1, max=1))

dt[,cor(a,b)] # -0.2912456
dt[,cor.test(a,b, method="kendall")]
```
```{r}
# 2C.1
sxx=131-(1/7)*29^2
syy=140-(1/7)*28^2
sxy=99-(1/7)*29*28

sxy/sqrt(sxx*syy) # -0.9750169

# Question 2
sxx = 2586 - (1/11)*168^2
syy = 320019 - (1/11)*1275^2
sxy = 20704 - (1/11)*168*1275
sxy/sqrt(sxx*syy) # 0.6604098

# Question 4
1-60/(7*8*9) # 0.8809524

# Question 5
1-276/(8*7*9)

# Question 6 
1-6*58/(11*12*13) # 0.7972028
# Question 9 
1-324/990 # 0.6727273

# Question 10
1-6*64/(5*6*7) # -0.8285714
```

## Mixed exercises
```{r}
# Question 1
sxx = 465 - 0.1*67*65
syy=429 - 0.1*65^2
sxy=434 - 0.1*67*65
sxy/sqrt(sxx*syy) # -0.1083237
# It's the same as correlation is unaffected by affine transformations
# Probably not, but wouldn't stand up to statistical test. 

# Question 2
# Not worth doing

# Question 3
spp = 85.5
spq=-11.625
sqq=77.0375 - (1/8)*(491/20)^2
spq/sqrt(sqq*spp) # -0.9643293
# There's a clear negative, almost linear, correlation between them. Rank correlation of one. 

# Question 4
# Discrete data, already using ranks. Not from bivariate normal distribution.
1-6*58/990 # 0.6484848
# Use their tables, two tailed therefore alpha = 0.025. 

# Question 5
# Non-parametric alternative to PMCC, so with discrete data sets.
1-6*28/(8*9*10) # 0.7666667


#  Question 7 
1-16/56 # 0.7142857

# Question 8 
# You assume the data generating process is bivariate normal

# Question 10
dt <- data.table(x=c(49, 76, 69, 71, 50, 64, 78, 74),
                 y=c(25, 88, 80, 62, 37, 86, 89, 67))
cor.test(dt$x, dt$y, method="spearman")

# Question 17
1-6*26/(8*9*10) # 0.7833333

```