ETC5521: Exploratory Data Analysis

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These slides are viewed best by Chrome or Firefox and occasionally need to be refreshed if elements did not load properly. See <a href=lecture-08A.pdf>here for the PDF </a>. 
]

---

# .monash-blue[ETC5521: Exploratory Data Analysis]

<h2 style="font-weight:900!important;">Sculpting data using models, checking assumptions, co-dependency and performing diagnostics</h2>

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Lecturer: *Emi Tanaka*

ETC5521.Clayton-x@monash.edu

Week 8 - Session 1

]

---

# Parametric regression

---

# Parametric regression

.flex[
.w-50[
* .monash-blue[**Parametric**] means that the researcher or analyst assumes in advance that the data fits some type of distribution (e.g. the normal distribution).

]
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]
]

---

# Parametric regression

.flex[
.w-50[
* .monash-blue[**Parametric**] means that the researcher or analyst assumes in advance that the data fits some type of distribution (e.g. the normal distribution).

]
.w-50.pl3[

**Example**

]
]

---

# Parametric regression

.flex[
.w-50[
* .monash-blue[**Parametric**] means that the researcher or analyst assumes in advance that the data fits some type of distribution (e.g. the normal distribution). 
* E.g. one may assume that
`$$\color{blue}{y_i} = \color{red}{\beta_0} + \color{red}{\beta_1} \color{blue}{x_i} + \color{red}{\beta_2} \color{blue}{x_i^2} + \epsilon_i,$$`
where `$\epsilon_i \sim NID(0, \color{red}{\sigma^2})$` for `$i = 1, ..., n$`,

* `$\color{red}{red} = \text{estimated}$`
  * `$\color{blue}{blue} = \text{observed}$`

]
.w-50.pl3[

**Example**

]
]

---

# Parametric regression

* `$\color{red}{red} = \text{estimated}$`
  * `$\color{blue}{blue} = \text{observed}$`

]
.w-50.pl3[

**Example**

]
]

---

# Parametric regression

* `$\color{red}{red} = \text{estimated}$`
  * `$\color{blue}{blue} = \text{observed}$`
* Because some type of distribution is assumed in advance, parametric fitting can lead to fitting a smooth curve that misrepresents the data.

]
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**Example**

]
]

---

# Parametric regression

]
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**Examples**

<img src="images/week8A/quad-good-fit-1.png" width="432" style="display: block; margin: auto;" />
Assuming a quadratic fit:

]
]

---

# Simulating data from parametric models

.flex[
.w-50.pr3[
* Say a model is 
`$$y = x^2 + e, \qquad e \sim N(0, 2^2).$$`
* Then we have
`$$y~|~x \sim N(x^2, 2^2).$$`

]
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]
]

---

# Simulating data from parametric models

.flex[
.w-50.pr3[
* Say a model is 
`$$y = x^2 + e, \qquad e \sim N(0, 2^2).$$`
* Then we have
`$$y~|~x \sim N(x^2, 2^2).$$`
* Let's draw `$200$` observations from this model.

]
.w-50.f4[

```r
set.seed(1)
df <- tibble(id = 1:200)

df
```

```
## # A tibble: 200 × 1
## id
## <int>
## 1 1
## 2 2
## 3 3
## 4 4
## 5 5
## 6 6
## 7 7
## 8 8
## 9 9
## 10 10
## # … with 190 more rows
```

]
]

---

# Simulating data from parametric models

.flex[
.w-50.pr3[
* Say a model is 
`$$y = x^2 + e, \qquad e \sim N(0, 2^2).$$`
* Then we have
`$$y~|~x \sim N(x^2, 2^2).$$`
* Let's draw `$200$` observations from this model.
* Suppose that `$x\in(-10,10)$` and that we have uniform coverage over the support.

]
.w-50.f4[

```r
set.seed(1)
df <- tibble(id = 1:200) %>% 
 mutate(x = runif(n(), -10, 10))

df
```

```
## # A tibble: 200 × 2
## id x
## <int> <dbl>
## 1 1 -4.69
## 2 2 -2.56
## 3 3 1.46
## 4 4 8.16
## 5 5 -5.97
## 6 6 7.97
## 7 7 8.89
## 8 8 3.22
## 9 9 2.58
## 10 10 -8.76
## # … with 190 more rows
```

]
]

---

# Simulating data from parametric models

]
.w-50.f4[

```r
set.seed(1)
df <- tibble(id = 1:200) %>% 
 mutate(x = runif(n(), -10, 10),
 y = x^2 + rnorm(n(), 0, 2))
df
```

```
## # A tibble: 200 × 3
## id x y
## <int> <dbl> <dbl>
## 1 1 -4.69 20.8 
## 2 2 -2.56 6.63 
## 3 3 1.46 0.301
## 4 4 8.16 67.0 
## 5 5 -5.97 34.3 
## 6 6 7.97 67.0 
## 7 7 8.89 80.5 
## 8 8 3.22 12.2 
## 9 9 2.58 7.44 
## 10 10 -8.76 80.2 
## # … with 190 more rows
```
]
]

---

# Simulating data from parametric models

]
.w-50.f4[

```r
set.seed(1)
df <- tibble(id = 1:200) %>% 
 mutate(x = runif(n(), -10, 10),
 y = x^2 + rnorm(n(), 0, 2))
```
Plotting this:

```r
ggplot(df, aes(x, y)) +
  geom_point()
```

<img src="images/week8A/sim-plot-1.png" width="432" style="display: block; margin: auto;" />
]
]

---

# Logistic regression

---

# Logistic regression

{{content}}
]
--

* Logistic regression models the relationship between a set of explanatory variables `$(x_{i1}, ..., x_{ik})$` and a set of **binary outcomes** `$Y_i$` for `$i = 1, ..., n$`.
{{content}}
--

* We assume that `$Y_i \sim B(1, p_i)\equiv Bernoulli(p_i)$` and the model is given by

`$$\text{logit}(p_i) = \text{ln}\left(\dfrac{p_i}{1 - p_i}\right) = \beta_0 + \beta_1x_{i1} + ... + \beta_k x_{ik}.$$`
{{content}}
--

* Taking the exponential of both sides and rearranging we get
`$$p_i = \dfrac{1}{1 + e^{-(\beta_0 + \beta_1x_{i1} + ... + \beta_k x_{ik})}}.$$`
{{content}}
--

* The function `$f(p) = \text{ln}\left(\dfrac{p}{1 - p}\right)$` is called the **logit** function, continuous with range `$(-\infty, \infty)$`, and if `$p$` is the probablity of an event, `$f(p)$` is the log of the odds.

---

# Representation of data for binary outcomes

```r
mock_df
```

```
## # A tibble: 18 × 5
## Patient Smoker Sex Cancer CancerBinary
## <fct> <fct> <fct> <fct> <dbl>
## 1 1 Yes Female No 0
## 2 2 Yes Male No 0
## 3 3 No Female Yes 1
## 4 4 Yes Male No 0
## 5 5 Yes Female Yes 1
## 6 6 No Female No 0
## 7 7 Yes Female Yes 1
## 8 8 No Female No 0
## 9 9 No Female No 0
## 10 10 No Male No 0
## 11 11 Yes Male No 0
## 12 12 Yes Female Yes 1
## 13 13 Yes Male No 0
## 14 14 Yes Female No 0
## 15 15 No Male Yes 1
## 16 16 No Female Yes 1
## 17 17 No Male No 0
## 18 18 No Male Yes 1
```
]

]
.w-50[
Summarised data:

```r
mock_sumdf
```

```
## # A tibble: 4 × 4
## # Groups: Smoker [2]
## Smoker Sex Cancer Total
## <fct> <fct> <int> <int>
## 1 No Female 2 5
## 2 No Male 2 4
## 3 Yes Female 3 5
## 4 Yes Male 0 4
```
]
{{content}}
]]

* The summarised data here give the same information as the original data, except you lost the patient number
* Note the sample size, n,  is larger than the number of rows in the summarised data

---

# Logistic regression in R

* Fitting logistic regression models in R depend on the form of input data

```r
glm(Cancer ~ Smoker + Sex, 
    family = binomial(link = "logit"), 
    data = mock_df)
```

```
## 
## Call:  glm(formula = Cancer ~ Smoker + Sex, family = binomial(link = "logit"), 
##     data = mock_df)
## 
## Coefficients:
## (Intercept)    SmokerYes      SexMale  
##      0.2517      -0.5034      -1.1145  
## 
## Degrees of Freedom: 17 Total (i.e. Null);  15 Residual
## Null Deviance:	    24.06 
## Residual Deviance: 22.61 	AIC: 28.61
```
]
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```r
glm(cbind(Cancer, Total - Cancer) ~ Smoker + Sex, 
    family = binomial(link = "logit"),
    data = mock_sumdf)
```

```
## 
## Call:  glm(formula = cbind(Cancer, Total - Cancer) ~ Smoker + Sex, family = binomial(link = "logit"), 
##     data = mock_sumdf)
## 
## Coefficients:
## (Intercept)    SmokerYes      SexMale  
##      0.2517      -0.5034      -1.1145  
## 
## Degrees of Freedom: 3 Total (i.e. Null);  1 Residual
## Null Deviance:	    5.052 
## Residual Deviance: 3.604 	AIC: 15.82
```

]
]

---

# Simulating from a logistic regression model .f4[Part 1]

.flex[
.w-40.pr3[
* Let's suppose that the probability of having cancer are the following:
  * 0.075 for women smokers
  * 0.045 for men smokers
  * 0.005 for women non-smokers
  * 0.003 for men non-smokers
* We'll sample 500 people for each group
]
.w-60.f5[
{{content}}
]]

```r
df <- tibble(id = 1:2000) %>% 
 mutate(Smoker = rep(c("Yes", "No"), each = n() / 2),
 Sex = rep(c("Female", "Male"), times = n() / 2))

print(df, n = 4)
```

```
## # A tibble: 2,000 × 3
## id Smoker Sex 
## <int> <chr> <chr> 
## 1 1 Yes Female
## 2 2 Yes Male 
## 3 3 Yes Female
## 4 4 Yes Male 
## # … with 1,996 more rows
```

```r
df %>% 
  group_by(Smoker, Sex) %>% 
  count()
```

```
## # A tibble: 4 × 3
## # Groups: Smoker, Sex [4]
## Smoker Sex n
## <chr> <chr> <int>
## 1 No Female 500
## 2 No Male 500
## 3 Yes Female 500
## 4 Yes Male 500
```

---

# Simulating from a logistic regression model .f4[Part 1]

```r
df <- tibble(id = 1:2000) %>% 
 mutate(Smoker = rep(c("Yes", "No"), each = n() / 2),
 Sex = rep(c("Female", "Male"), times = n() / 2)) %>% 
 rowwise() %>% 
 mutate(CancerBinary = 
* case_when(Smoker=="Yes" & Sex=="Female" ~ rbinom(1, 1, 0.075),
* Smoker=="Yes" & Sex=="Male" ~ rbinom(1, 1, 0.045),
* Smoker=="No" & Sex=="Female" ~ rbinom(1, 1, 0.005),
* Smoker=="No" & Sex=="Male" ~ rbinom(1, 1, 0.003)),
 Cancer = ifelse(CancerBinary, "Yes", "No"))

df %>% 
  filter(Cancer=="Yes")
```

```
## # A tibble: 53 × 5
## # Rowwise: 
## id Smoker Sex CancerBinary Cancer
## <int> <chr> <chr> <int> <chr> 
## 1 27 Yes Female 1 Yes 
## 2 32 Yes Male 1 Yes 
## 3 47 Yes Female 1 Yes 
## 4 83 Yes Female 1 Yes 
## 5 129 Yes Female 1 Yes 
## 6 136 Yes Male 1 Yes 
## 7 149 Yes Female 1 Yes 
## 8 218 Yes Male 1 Yes 
## 9 245 Yes Female 1 Yes 
## 10 251 Yes Female 1 Yes 
## # … with 43 more rows
```
]]

---

# Simulating from a logistic regression model .f4[Part 2]

* At times, you may want to **simulate the summary data directly** instead of the individual data
{{content}}

]
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]]

* Recall that if `$Y_i\sim B(1, p)$` for `$i = 1, ...k$` and `$Y_i$`s are independent,  `$$S = Y_1 + Y_2 + ... +  Y_k \sim B(k, p)$$`

---

# Simulating from a logistic regression model .f4[Part 2]

* At times, you may want to **simulate the summary data directly** instead of the individual data
* Recall that if `$Y_i\sim B(1, p)$` for `$i = 1, ...k$` and `$Y_i$`s are independent,  `$$S = Y_1 + Y_2 + ... +  Y_k \sim B(k, p)$$`

]
.w-60.f5[

```r
expand_grid(Smoker = c("Yes", "No"), Sex = c("Female", "Male")) %>% 
  rowwise() %>% 
  mutate(Cancer = 
*          case_when(Smoker=="Yes" & Sex=="Female" ~ rbinom(1, 500, 0.075),
*                    Smoker=="Yes" & Sex=="Male" ~ rbinom(1, 500, 0.045),
*                    Smoker=="No" & Sex=="Female" ~ rbinom(1, 500, 0.005),
*                    Smoker=="No" & Sex=="Male" ~ rbinom(1, 500, 0.003)),
*        Total = 500)
```

```
## # A tibble: 4 × 4
## # Rowwise: 
## Smoker Sex Cancer Total
## <chr> <chr> <int> <dbl>
## 1 Yes Female 35 500
## 2 Yes Male 23 500
## 3 No Female 0 500
## 4 No Male 3 500
```

]]

---

# .orange[Case study] .circle.bg-orange.white[1] Menarche

* In 1965, the average age of 25 homogeneous groups of girls was
recorded along with the number of girls who have reached
menarche out of the total in each group.

]
.panel[.panel-name[data]
.h200.scroll-sign.f4[

```r
data(menarche, package = "MASS")
skimr::skim(menarche)
```

```
## ── Data Summary ────────────────────────
##                            Values  
## Name                       menarche
## Number of rows             25      
## Number of columns          3       
## _______________________            
## Column type frequency:             
##   numeric                  3       
## ________________________           
## Group variables            None    
## 
## ── Variable type: numeric ──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
##   skim_variable n_missing complete_rate  mean     sd    p0   p25   p50   p75   p100 hist 
## 1 Age                   0             1  13.1   2.03  9.21  11.6  13.1  14.6   17.6 ▅▇▇▇▁
## 2 Total                 0             1 157.  195.   88     98   105   117   1049   ▇▁▁▁▁
## 3 Menarche              0             1  92.3 204.    0     10    51    92   1049   ▇▁▁▁▁
```

]]
.panel[.panel-name[R]
.scroll-sign[.s300.f4[

```r
# can't fit the model below using geom_smooth 
# so manually fit
menarche_model <- glm(cbind(Menarche, Total - Menarche) ~ Age, 
 data = menarche, 
 family = binomial(link = "logit"))
# length.out = 80 to match geom_smooth default
menarche_pred <- data.frame(Age = seq(min(menarche$Age),
 max(menarche$Age),
 length.out = 80))
menarche_pred <- menarche_pred %>% 
 mutate(fit = predict(menarche_model, newdata = menarche_pred, type = "response"))

ggplot(menarche, aes(Age, Menarche/Total)) + 
  geom_point() +
  geom_line(data = menarche_pred, aes(y = fit), color = "blue")
```
]]]

]

.footnote.f4[
Milicer, H. and Szczotka, F. (1966) Age at Menarche in Warsaw girls in 1965. Human Biology 38, 199–203.
]

---

# Simulating data from a fitted logistic regression model .f4[Part 1]

* Suppose we want to simulate from the fitted model

* We first fit the fitted model

```r
fit1 <- 
 glm(cbind(Menarche, Total - Menarche) ~ Age, 
 family = "binomial", 
 data = menarche)
(beta <- coef(fit1))
```

```
## (Intercept)         Age 
##  -21.226395    1.631968
```

* The fitted regression model is given as:
`$$\text{logit}(\hat{p}_i) = \hat{\beta}_0  + \hat{\beta}_1 x_{i1}.$$`
* Rearranging we get
`$$\hat{p}_i = \dfrac{1}{1 + e^{-(\hat{\beta}_0  + \hat{\beta}_1 x_{i1})}}.$$`
]
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* Simulating from first principles:

```r
menarche %>% 
  rowwise() %>% 
  mutate(
    phat = 1/(1 + exp(-(beta[1] + beta[2] * Age))),
*   simMenarche = rbinom(1, Total, phat))
```

```
## # A tibble: 25 × 5
## # Rowwise: 
## Age Total Menarche phat simMenarche
## <dbl> <dbl> <dbl> <dbl> <int>
## 1 9.21 376 0 0.00203 1
## 2 10.2 200 0 0.0103 3
## 3 10.6 93 0 0.0187 2
## 4 10.8 120 2 0.0279 3
## 5 11.1 90 2 0.0413 1
## 6 11.3 88 5 0.0609 6
## 7 11.6 105 10 0.0888 9
## 8 11.8 111 17 0.128 12
## 9 12.1 100 16 0.181 17
## 10 12.3 93 29 0.249 23
## # … with 15 more rows
```

]

---

# Simulating data from a fitted logistic regression model .f4[Part 2]

* An easier way to do this is to use the `simulate` function which works for many model objects in R
* Below it's simulating 3 sets of responses (i.e. counts of "success" and "failure" events) from `fit1` logistic model object

```r
simulate(fit1, nsim = 3) 
```

```
##    sim_1.Menarche sim_1.V2 sim_2.Menarche sim_2.V2 sim_3.Menarche sim_3.V2
## 1               0      376              0      376              0      376
## 2               2      198              1      199              0      200
## 3               4       89              0       93              3       90
## 4               4      116              2      118              6      114
## 5               8       82              5       85              3       87
## 6               6       82              3       85              6       82
## 7              14       91              7       98              5      100
## 8              13       98             14       97             16       95
## 9              21       79             18       82             20       80
## 10             25       68             21       72             27       66
## 11             47       53             32       68             33       67
## 12             37       71             43       65             41       67
## 13             47       52             54       45             59       40
## 14             57       49             63       43             68       38
## 15             76       29             82       23             84       21
## 16             87       30             96       21             91       26
## 17             83       15             82       16             84       14
## 18             90        7             88        9             91        6
## 19            107       13            108       12            113        7
## 20             97        5             98        4             97        5
## 21            115        7            119        3            119        3
## 22            108        3            108        3            108        3
## 23             93        1             93        1             92        2
## 24            112        2            112        2            114        0
## 25           1048        1           1049        0           1048        1
```
]

---

# Diagnostics for logistic regression models

* One diagnostic is to compare the observed and expected proportions under the logistic regression fit.

```r
df1 <- menarche %>% 
 mutate(
 pexp = 1/(1 + exp(-(beta[1] + beta[2] * Age))),
 pobs = Menarche / Total)
```
<img src="images/week8A/plot-logistic-1.png" width="288" style="display: block; margin: auto;" />

]
.w-50.pl3[

]

* Goodness-of-fit type test is used commonly to assess the fit as well.

* E.g. Hosmer–Lemeshow test, where test statistic is given as

`$$H = \sum_{i = 1}^r \left(\dfrac{(O_{1i} - E_{1g})^2}{E_{1i}} + \dfrac{(O_{0i} - E_{0g})^2}{E_{0i}}\right)$$`
where `$O_{1i}$` `$(E_{1i})$` and `$O_{0i}$` `$(E_{0i})$` are observed (expected) frequencies for successful and non-successful events for group `$i$`, respectively.

```r
vcdExtra::HLtest(fit1)
```

```
## Hosmer and Lemeshow Goodness-of-Fit Test 
## 
## Call:
## glm(formula = cbind(Menarche, Total - Menarche) ~ Age, family = "binomial", 
##     data = menarche)
##  ChiSquare df   P_value
##  0.1088745  8 0.9999996
```

---

# Diagnostics for linear models

---

# Assumptions for linear models

`$$Y_i = \beta_0 + \beta_1x_{i1} + ... + \beta_{k}x_{ik} + \epsilon_i,$$`
where `$\epsilon_i \sim NID(0, \sigma^2)$` or in matrix format,

`$$\boldsymbol{Y} = \mathbf{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}, \quad \boldsymbol{\epsilon} \sim N(\boldsymbol{0}, \sigma^2 \mathbf{I}_n)$$`

<div style="font-size:12pt;padding-left:20px;">
where

<ul>
<li> `$\boldsymbol{Y} = (Y_1, ..., Y_n)^\top$`,</li>
<li> `$\boldsymbol{\beta} = (\beta_0, ..., \beta_k)^\top$`, </li>
<li> `$\boldsymbol{\epsilon} = (\epsilon_1, ..., \epsilon_n)^\top$`, and </li>
<li> `$\mathbf{X} = \begin{bmatrix}\boldsymbol{1}_n & \boldsymbol{x}_1 & ... & \boldsymbol{x}_k \end{bmatrix}$`, where </li>
<li> `$\boldsymbol{x}_j =(x_{1j}, ..., x_{nj})^\top$` for `$j \in \{1, ..., k\}$`</li>
</ul>
</div>

]
.w-50.pl3[

]

---

# Assumptions for linear models

`$$Y_i = \beta_0 + \beta_1x_{i1} + ... + \beta_{k}x_{ik} + \epsilon_i,$$`
where `$\color{red}{\epsilon_i \sim NID(0, \sigma^2)}$` or in matrix format,

`$$\boldsymbol{Y} = \mathbf{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}, \quad \color{red}{\boldsymbol{\epsilon} \sim N(\boldsymbol{0}, \sigma^2 \mathbf{I}_n)}$$`

<div style="font-size:12pt;padding-left:20px;">
where

]
.w-50.pl3[
This means that we assume

1. `$E(\epsilon_i) = 0$` for `$i \in \{1, ..., n\}.$`
2. `$\epsilon_1, ..., \epsilon_n$` are independent.
3. `$Var(\epsilon_i) = \sigma^2$` for `$i \in \{1, ..., n\}$` (i.e. homogeneity).
4. `$\epsilon_1, ..., \epsilon_n$` are normally distributed.

]

So how do we check it?

---

# Model diagnostics for linear models

.flex.h-45[

.w-50.pa3[
Plot `$Y_i$` vs `$x_i$` to see if there is `$\approx$` a linear relationship between `$Y$` and `$x$`.

]
.w-50.bg-gray80.pa3[

A boxplot of the residuals `$R_i$` to check for symmetry.
<img src="images/week8A/unnamed-chunk-11-1.svg" width="180" style="display: block; margin: auto;" />

]]
.flex[
.w-50.bg-gray80.pa3[
To check the homoscedasticity assumption, plot `$R_i$` vs `$x_i$`. There should be no obvious patterns.

<img src="images/week8A/unnamed-chunk-12-1.svg" width="158.4" style="display: block; margin: auto;" />
]
.w-50.pa3[
A normal Q-Q plot, i.e. a plot of the ordered residuals vs `$\Phi^{-1}(\frac{i}{n+1})$`.

]
]

---

# Assessing (A1) `$E(\epsilon_i) = 0$` for `$i=1,\ldots,n$`

* It is a property of the least squares method that $$\sum_{i=1}^n R_i = 0,\quad
\text{so}\quad \bar R_i  = 0$$ for `$R_i = Y_i - \hat{Y}_i$`, hence (A1)  will always appear valid "overall". 
--

* Trend in residual versus fitted values or covariate can indicate "local"
failure of (A1). 
--

* What do you conclude from the following plots?

---

# Assessing (A2)-(A3)

### (A2) `$\epsilon_1, \ldots ,\epsilon_n$` are independent

* If (A2) is correct, then residuals should appear randomly scattered
about zero if plotted against fitted values or covariate.
* Long sequences of positive residuals followed by sequences of negative residuals in `$R_i$` vs `$x_i$` plot suggests that the error terms are not independent.
]
.w-50.pa3[

### (A3) `$Var(\epsilon_i) = \sigma^2$` for `$i=1,\ldots,n$`

* If (A3) holds then the spread of the residuals should be roughly the same across the fitted values or covariate.

]
]

---

# Assessing (A4) `$\epsilon_1, \ldots ,\epsilon_n$` are normally distributed

* The function `qqnorm(x)` produces a Q-Q plot of the ordered vector `x` against the quantiles of the normal distribution.
* The `$n$` chosen normal quantiles `$\Phi^{-1}(\frac{i}{n+1})$` are easy to calculate but more sophisticated ways exist:
 * `$\frac{i}{n+1} \mapsto \frac{i-3/8}{n+1/4}$`, default in `qqnorm`. 
 * `$\frac{i}{n+1} \mapsto \frac{i-1/3}{n+1/3}$`, recommended by Hyndman and Fan (1996).

]
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### In R

```r
fit <- lm(y ~ x)
```

By "hand"

```r
plot(qnorm((1:n) / (n + 1)), sort(resid(fit)))
```

By `base`

```r
qqnorm(resid(fit))
qqline(resid(fit))
```

By `ggplot2`

```r
data.frame(residual = resid(fit)) %>% 
  ggplot(aes(sample = residual)) + 
  stat_qq() + stat_qq_line(color="blue")
```

]
]

.footnote.f4[
Reference: Hyndman and Fan (1996). *Sample quantiles in statistical packages*, American Statistician, 50, 361--365.
]

---

# Examining simulated data

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Simulation scheme

```r
n <- 100
x <- seq(0, 1, length.out = n)
y1 <- x + rnorm(n) / 3 # Linear
y2 <- 3 * (x - 0.5) ^ 2 + 
 c(rnorm(n / 2)/3, rnorm(n / 2)/6) # Quadratic
y3 <- -0.25 * sin(20 * x - 0.2) + 
 x + rnorm(n) / 3 # Non-linear

M1 <- lm(y1 ~ x); M2 <- lm(y2 ~ x); M3 <- lm(y3 ~ x)
```

---

# Take away messages

<ul class="fa-ul">
{{content}}
</ul>

]
]

<li>Parametric models assume some distribution in advance 
</li>

{{content}}
--

<li>Logistic models can be used to model explanatory variables with binary outcomes
</li>

{{content}}
--

<li>You should be able to simulate from parametric models
</li>

{{content}}
--

<li>You can perform basic model diagnostics
</li>

{{content}}
--

<li>You can use simulation to analyse model properties
</li>

---

# Resources and Acknowledgement

- Some of these slides were inspired by STAT3012 Applied Linear Models at The University of Sydney by Prof Samuel Muller
- Cook & Weisberg (1994) 
"An Introduction to Regression Graphics"
- Data coding using [`tidyverse` suite of R packages](https://www.tidyverse.org) 
- Slides constructed with [`xaringan`](https://github.com/yihui/xaringan), [remark.js](https://remarkjs.com), [`knitr`](http://yihui.name/knitr), and [R Markdown](https://rmarkdown.rstudio.com).

---

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<a rel="license" href="http://creativecommons.org/licenses/by-sa/4.0/"><img alt="Creative Commons License" style="border-width:0" src="https://i.creativecommons.org/l/by-sa/4.0/88x31.png" /></a> This work is licensed under a <a rel="license" href="http://creativecommons.org/licenses/by-sa/4.0/">Creative Commons Attribution-ShareAlike 4.0 International License</a>.

.bottom_abs.width100[

Lecturer: *Emi Tanaka*

ETC5521.Clayton-x@monash.edu

Week 8 - Session 1

]