Discrete Choice

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# Discrete Choice
### AECN 396/896-002

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# Discrete Choice Analysis
+ Focus on understanding choices that are discrete (not continuous)
  * Whether you own a car or not (binary choice)
  * Whether you use an iPhone, Android, or other types of cell phones (Multinomial choice)
  * Which recreation sites you visit this winter (multinomial)
+ Linear models we have seen are often not appropriate

---
class: middle

# Binary Response Model

.content-box-green[**Binary response**]

`$y = 0 \;\; \mbox{(if you do not own a car)}$`

`$y = 1 \;\; \mbox{(if you own at least one car)}$`

.content-box-green[**Question we would like to answer**]

How do variables `$x_1,\dots,x_k$` affect the status of `$y$` (the choice of whether to own at least one car or not)?

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.content-box-green[**Binary response**]

We try to model the .red[probability] of `$y = 1$` (own at least one car)

`$Pr(y=1|x_1,\dots,x_k) = f(x_1,\dots,x_k)$`

as a function of independent variables.

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.content-box-green[**Linear Probability Model**]

`$Pr(y=1|x_1,\dots,x_k) = \beta_0+\beta_1 x_1+\dots+\beta_k x_k$`

.content-box-green[**Drawback**]

There is no guarantee that the predicted probability is bounded within [0, 1].

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.content-box-green[**How about this?**]

`$Pr(y=1|x_1,\dots,x_k) = G(\beta_0+\beta_1 x_1+\dots+\beta_k x_k)$`

where `$0<G(z)<1$` for all real numbers `$z$`

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Different choices of `$G()$` lead to different models.

.content-box-green[**Logit model**]

`$G(z) = exp(z)/[1+exp(z)] = \frac{e^z}{1+e^z}$`

where `$z = \beta_0+\beta_1 x_1+\dots+\beta_k x_k$`

.content-box-green[**Probit model**]

`$G(z) = \Phi(z)$`

where `$\Phi(z)$` is the standard normal cumulative distribution function

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This what `$G()$` looks like for logit and probit.

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`$Pr(y=1|x_1,\dots,x_k) = G(\beta_0+\beta_1 x_1+\dots+\beta_k x_k)$`

+ What do `$\beta$`s measure?
+ How do we interpret them?

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`$\mbox{Before:} x_1=0,\dots,x_k=0 \Rightarrow z=\beta_0$`

`$\mbox{After:}\;\; x_1=1 \;\;\mbox{and}\;\; x_2=0,\dots,x_k=0 \Rightarrow z=\beta_0+\beta_1$`

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+ `$\beta$`s measure how far you move along the x-axis

+ `$\beta$`s does not directly measure how independent variables influence the probability of `$y=1$`

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To understand the marginal impact of `$x_k$` on `$Prob(y=1)$` (how a change in `$x_k$` affects the likelihood of owning a car), you need to do a bit of math.

.content-box-green[**Model**]

`$Pr(y=1|x_1,\dots,x_k) = G(z)$`

`$z = \beta_0+\beta_1 x_1+\dots+\beta_k x_k$`

.content-box-green[**marginal impact**]

Differentiating both sides with respect to `$x_k$`,

`$\frac{\partial Pr(y=1|x_1,\dots,x_k)}{\partial x_k} = G'(z)\times \frac{\partial z}{\partial x_k}$`

`$\qquad\qquad\qquad\:\: = G'(z)\times \beta_k$`

---
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.content-box-green[**marginal impact**]

Differentiating both sides with respect to `$x_k$`,

`$\frac{\partial Pr(y=1|x_1,\dots,x_k)}{\partial x_k} = G'(z)\times \frac{\partial z}{\partial x_k}$`

`$\qquad\qquad\qquad\:\: = G'(z)\times \beta_k$`

.content-box-green[**Notes**]

+ The marginal impact of an independent variable depends on the values of all the independent variables: `$G(\beta_0+\beta_1 x_1+\dots+\beta_k x_k)$`

+ Since `$G'()$` is always positive, the sign of the marginal impact of an independent variable on `$Prob(y=1)$` is always the same as the sign of its coefficient

---
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.content-box-green[**Estimation of Binary Choice Models**]

+ Linear models: OLS

+ Binary choice models: .blue[Maximum Likelihood Estimation (MLE)]

---
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.content-box-green[**OLS**]

Find parameters that makes the sum of residuals squared the smallest

.content-box-green[**MLE (very loosely put)**]

Find parameters `$(\beta$`s) that makes what we observed (collection of binary decisions made by different individuals) most likely (.blue[Maximum Likelihood])

---
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.content-box-green[**Observed decisions made by two individuals**]

+ Individual 1: `$y=1$` (own at least one car)

+ Individual 2: `$y=0$` (does not own a car)

.content-box-green[**Probability of individual decisions**]

+ `$\mbox{Individual 1}: Prob(y_1=1|\mathbf{x_1})= G(z_1)$`

+ `$\mbox{Individual 2}: Prob(y_2=0|\mathbf{x_2})= 1-G(z_2)$`

where

+ `$\mathbf{x_i}$` is a collection of independent variables for individual `$i$` `$(x_{1,i}, \dots, x_{k,i})$`.

+ `$z_i = \beta_0+\beta_1 x_{1,i}+\dots+\beta_k x_{k,i}$`

.content-box-green[**Probability of a collection of decisions**]

The probability that we observe a .blue[collection of choices] made by them (if their decisions are independent)

`$Prob(y_1=1|\mathbf{x_1})\times Prob(y_2=0|\mathbf{x_2}) = G(z_1)\times [1-G(z_2)]$`
  
which we call .blue[likelihood function].

---
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.content-box-green[**Probability of a collection of decisions**]

The probability that we observe a .blue[collection of choices] made by them (if their decisions are independent)

`$Prob(y_1=1|\mathbf{x_1})\times Prob(y_2=0|\mathbf{x_2}) = G(z_1)\times [1-G(z_2)]$`
  
which we call .blue[likelihood function].

.content-box-green[**MLE**]

`$Max_{\beta_1,\dots,\beta_k}\;\; G(z_1)\times [1-G(z_2)]$`

---
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.content-box-green[**MLE of Binary Choice Model in General**]

Maximize the likelihood function:

`$Max_{\beta_1,\dots,\beta_k}\;\; L$`

where `$L=\Pi_{i=1}^n \Big[y_i\times G(z_i)+(1-y_i)\times(1-G(z_i))\Big]$` is the likelihood function.

.content-box-green[**Log-likelihood function**]
    
`$LL = log\Big(\Pi_{i=1}^n \Big[y_i\times G(z_i)+(1-y_i)\times(1-G(z_i))\Big]\Big)$` 
`$\qquad = \sum_{i=1}^n log\Big(y_i\times G(z_i)+(1-y_i)\times(1-G(z_i))\Big)$`

.content-box-green[**MLE with `$LL$`**]

`$argmax_{\beta_1,\dots,\beta_k}\;\; L \equiv argmax_{\beta_1,\dots,\beta_k}\;\; LL$`

---
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# Implementation in R with an example

Participation of females in labor force:

`$Pr(inlf=1|\mathbf{x})= G(z)$`

where

`$z = \beta_0+\beta_1 nwifeinc+ \beta_2 educ+ \beta_3 exper$`

`$\quad\;\; + \beta_4 exper^2 + \beta_5 age + \beta_6 kidslt6 + \beta_7 kidsge6$`

+ `$inlf$`: 1 if in labor force in 1975, 0 otherwise
+ `$nwifeinc$`: earning as a family if she does not work
+ `$kidslt6$`: \# of kids less than 6 years old
+ `$kidsge6$`: \# of kids who are 6-18 year old

---
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```r
#--- import the data ---#
data <- read.dta13("MROZ.dta") %>%
 mutate(exper2 = exper^2)

#--- take a look ---#
dplyr::select(data, inlf, nwifeinc, kidslt6, kidsge6, educ) %>%
  head()
```

```
##   inlf  nwifeinc kidslt6 kidsge6 educ
## 1    1 10.910060       1       0   12
## 2    1 19.499981       0       2   12
## 3    1 12.039910       1       3   12
## 4    1  6.799996       0       3   12
## 5    1 20.100058       1       2   14
## 6    1  9.859054       0       0   12
```

---
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For individual 1 (row 1 of the data),

`$z_1 = \beta_0+\beta_1 10.91+ \beta_2 12+ \beta_3 14 + \beta_4 196 + \beta_5 32 + \beta_6 1 + \beta_7 0$`

The probability that individual 1 would make the decision he/she made given `$\beta$`s is:

`$G(z_1)$` (a function of `$\beta$`s)

---
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```
##     inlf nwifeinc kidslt6 kidsge6 educ
## 748    0    5.330       0       2   12
## 749    0   28.200       0       2   13
## 750    0   10.000       2       3   12
## 751    0    9.952       0       0   12
## 752    0   24.984       0       0   12
## 753    0   28.363       0       3    9
```

For individual 753 (row 753 of the data),

`$z_{753} = \beta_0+\beta_1 28.36+ \beta_2 9+ \beta_3 12 + \beta_4 144 + \beta_5 39 + \beta_6 0 + \beta_7 3$`

The probability that individual 753 would make the decision he/she made given `$\beta$`s is:

`$1 - G(z_{753})$` (a function of `$\beta$`s)

---
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Multiply all the probabilities of observed choices given `$\beta$`s,

`$L = G(z_1) \times G(z_2) \times \dots [1-G(z_753)]$`

`$LL = log\Big(G(z_1) \times G(z_2) \times \dots [1-G(z_753)]\Big)$`

Solve the following problems to estimate `$\beta$`s:

`$Max_{\beta_1, \dots, \beta_7} \quad LL$`

---
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.content-box-green[**Estimating binary choice model using `$R$`**]

You can use the `glm()` function (no new packages installation necessary) when using cross-sectional data

+ `glm` refers to Generalized Linear Model, which encompass linear models we have been using

+ you specify the `family` option to tell what kind of model you are estimating

---
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.content-box-green[**Probit model estimation**]

```r
probit_lf <- glm(
 #--- formula ---#
 inlf ~ nwifeinc + educ + exper + exper2 + age + kidslt6 + kidsge6,
 #--- data ---#
 data = data,
 #--- models ---#
 family = binomial(link = "probit")
)
```

.content-box-green[**family option**]

+ `binomial()`: tells R that your dependent variable is binary

+ `link = "probit"`: tells R that you want to use the cumulative distribution function of the standard normal distribution as `$G()$` in `$Prob(y=1|\mathbf{x})=G(z)$`

---
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.left5[

```r
msummary(
  probit_lf,
  stars = TRUE,
  gof_omit = "IC|F",
  output = "flextable"
) %>%
  fontsize(
    size = 9,
    part = "all"
  ) %>%
  autofit()
```
]

.right5[

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style="overflow-wrap:break-word;"><td class="cl-42e7ccf5"></td><td class="cl-42e7ccfe">(0.508)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7cd58">nwifeinc</td><td class="cl-42e7cd59">-0.012*</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7ccf5"></td><td class="cl-42e7ccfe">(0.005)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7cdd0">educ</td><td class="cl-42e7cdc6">0.131***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7ccf5"></td><td class="cl-42e7ccfe">(0.025)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7cddb">exper</td><td class="cl-42e7cdda">0.123***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7ccf5"></td><td class="cl-42e7ccfe">(0.019)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7cdee">exper2</td><td class="cl-42e7cde4">-0.002**</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7ccf5"></td><td class="cl-42e7ccfe">(0.001)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7cd08">age</td><td class="cl-42e7cd09">-0.053***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7ccf5"></td><td class="cl-42e7ccfe">(0.008)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7cd1c">kidslt6</td><td class="cl-42e7cd12">-0.868***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7ccf5"></td><td class="cl-42e7ccfe">(0.118)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7cd1d">kidsge6</td><td class="cl-42e7cd26">0.036</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7ccf5"></td><td class="cl-42e7ccfe">(0.044)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7cd30">Num.Obs.</td><td class="cl-42e7cd31">753</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7cd44">Log.Lik.</td><td class="cl-42e7cd3a">-401.302</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-42e7cd4e">RMSE</td><td class="cl-42e7cd45">1.04</td></tr></tbody><tfoot><tr style="overflow-wrap:break-word;"><td colspan="2"class="cl-42e7cdf8">+ p &lt; 0.1, * p &lt; 0.05, ** p &lt; 0.01, *** p &lt; 0.001</td></tr></tfoot></table></div></template>
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---
class: middle

.content-box-green[**Logit model estimation**]

```r
logit_lf <- glm(
 #--- formula ---#
 inlf ~ nwifeinc + educ + exper + exper2 + age + kidslt6 + kidsge6,
 #--- data ---#
 data = data,
 #--- models ---#
 family = binomial(link = "logit")
)
```

.content-box-green[**family option**]

+ `binomial()`: tells R that your dependent variable is binary

+ `link = "logit"`: tells `$R$` that you want to use `$G(z) = \frac{e^z}{1+e^z}$` in `$Prob(y=1|\mathbf{x})=G(z)$`

---
class: middle

.left5[

```r
msummary(
  logit_lf,
  stars = TRUE,
  gof_omit = "IC|F",
  output = "flextable"
) %>%
  fontsize(
    size = 9,
    part = "all"
  ) %>%
  autofit()
```
]

.right5[

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0.00);margin-bottom:0;margin-top:0;margin-left:0;margin-right:0;}.cl-6d62d794{width:61.5pt;background-color:transparent;vertical-align: middle;border-bottom: 2pt solid rgba(102, 102, 102, 1.00);border-top: 2pt solid rgba(102, 102, 102, 1.00);border-left: 0 solid rgba(0, 0, 0, 1.00);border-right: 0 solid rgba(0, 0, 0, 1.00);margin-bottom:0;margin-top:0;margin-left:0;margin-right:0;}.cl-6d62d795{width:56.6pt;background-color:transparent;vertical-align: middle;border-bottom: 2pt solid rgba(102, 102, 102, 1.00);border-top: 2pt solid rgba(102, 102, 102, 1.00);border-left: 0 solid rgba(0, 0, 0, 1.00);border-right: 0 solid rgba(0, 0, 0, 1.00);margin-bottom:0;margin-top:0;margin-left:0;margin-right:0;}</style><table class='cl-6d6a361a'><thead><tr style="overflow-wrap:break-word;"><td class="cl-6d62d794"> </td><td class="cl-6d62d795">Model 1</td></tr></thead><tbody><tr style="overflow-wrap:break-word;"><td class="cl-6d62d654">(Intercept)</td><td class="cl-6d62d668">0.425</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d669"></td><td class="cl-6d62d672">(0.860)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d727">nwifeinc</td><td class="cl-6d62d728">-0.021*</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d669"></td><td class="cl-6d62d672">(0.008)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d74e">educ</td><td class="cl-6d62d744">0.221***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d669"></td><td class="cl-6d62d672">(0.043)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d759">exper</td><td class="cl-6d62d758">0.206***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d669"></td><td class="cl-6d62d672">(0.032)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d76c">exper2</td><td class="cl-6d62d762">-0.003**</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d669"></td><td class="cl-6d62d672">(0.001)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d67c">age</td><td class="cl-6d62d6ae">-0.088***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d669"></td><td class="cl-6d62d672">(0.015)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d6c2">kidslt6</td><td class="cl-6d62d6b8">-1.443***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d669"></td><td class="cl-6d62d672">(0.204)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d6c3">kidsge6</td><td class="cl-6d62d6c4">0.060</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d669"></td><td class="cl-6d62d672">(0.075)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d6cc">Num.Obs.</td><td class="cl-6d62d6d6">753</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d6fe">Log.Lik.</td><td class="cl-6d62d6f4">-401.765</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-6d62d726">RMSE</td><td class="cl-6d62d71c">1.04</td></tr></tbody><tfoot><tr style="overflow-wrap:break-word;"><td colspan="2"class="cl-6d62d776">+ p &lt; 0.1, * p &lt; 0.05, ** p &lt; 0.01, *** p &lt; 0.001</td></tr></tfoot></table></div></template>
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---
class: middle

.content-box-green[**Important**]

+ You cannot directly compare the coefficient on the same variable from probit and logit! The fact that the coefficient on `educ` is higher from the logit model does not mean the logit model is suggesting `educ` is more influential than the probit model suggests. They are on different scales.

---
class: middle

# Post-estimation operations and diagnostics

---
class: middle

.content-box-green[**Log-likelihood (fitted)**]

`$LL =\sum_{i=1}^n log\Big(y_i\times G(\hat{z}_i)+(1-y_i)\times(1-G(\hat{z}_i))\Big)$`

+ `$\hat{z}_i=\hat{\beta}_0+\hat{\beta}_1 x_1+\dots+\hat{\beta}_k x_k$`

+ `$G(\hat{z}_i)$` is the fitted value of `$Prob(y=1|\mathbf{x})$`

.content-box-green[**Example**]

+ `$G(\hat{z}_i) = 0.9$`: predicted that individual `$i$` is very likely to own a car 
+ `$y_i = 0$`: in reality, individual `$i$` does not own a car

`$\Rightarrow$`

`$log\Big(0\times 0.9 +(1-0)\times(1-0.9)\Big) = log(0.1) = -2.3$`

---
class: middle

.content-box-green[**Log-likelihood (fitted)**]

`$LL =\sum_{i=1}^n log\Big(y_i\times G(\hat{z}_i)+(1-y_i)\times(1-G(\hat{z}_i))\Big)$`

+ `$\hat{z}_i=\hat{\beta}_0+\hat{\beta}_1 x_1+\dots+\hat{\beta}_k x_k$`

+ `$G(\hat{z}_i)$` is the fitted value of `$Prob(y=1|\mathbf{x})$`

.content-box-green[**Example**]

+ `$G(\hat{z}_i) = 0.9$`: predicted that individual `$i$` is very likely to own a car 
+ `$y_i = 1$`: in reality, individual `$i$` indeed owns a car

`$\Rightarrow$`

`$log\Big(1\times 0.9 +(1-1)\times(1-0.9)\Big) = log(0.9) = -0.11$`

---
class: middle

.content-box-green[**Log-likelihood (fitted)**]

So, the better your prediction (model fit) is, the .blue[the greater (less negative) LL is.]

---
class: middle

.content-box-green[**McFadden's**] pseudo- `$R^2$`

A measure of how much better your model is compared to the model with only the intercept.

`$pseudo-R^2=1-LL/LL_0$`

where `$LL_0$` is the log-likelihood when you include only the intercept.

---
class: middle

.content-box-green[**R code**]

```r
logit_lf_0 <- glm(
 inlf ~ 1,
 data = data,
 family = binomial(link = "logit")
)

#--- extract LL using the logLik() function ---#
(LL0 <- logLik(logit_lf_0))
```

```
## 'log Lik.' -514.8732 (df=1)
```

```r
#--- extract LL using the logLik() function from your preferred model ---#
(LL <- logLik(logit_lf))
```

```
## 'log Lik.' -401.7652 (df=8)
```

```r
#--- pseudo R2 ---#
1 - LL / LL0
```

```
## 'log Lik.' 0.2196814 (df=8)
```
---
class: middle

.content-box-green[**Alternatively**]

```r
#--- or more easily ---#
1 - logit_lf$deviance / logit_lf$null.deviance
```

```
## [1] 0.2196814
```

```r
#--- what are deviances? ---#
logit_lf$null.deviance # = -2*LL0
```

```
## [1] 1029.746
```

```r
logit_lf$deviance # = -2*LL
```

```
## [1] 803.5303
```

+ `null.deviance` `$= -2\times LL_0$`
+ `deviance` `$= -2\times LL$`

---
class: middle

.content-box-green[**Testing joint significance**]

You can do Likelihood Ratio (LR) test:

`$LR = 2(LL_{unrestricted}-LL_{restricted}) \sim \chi^2_{df\_restrictions}$`

where `$df\_restrictions$` is the number of restrictions.

.content-box-green[**Note**]

LR test is very similar conceptually to F-test.

.content-box-green[**Example**]

+ `$H_0:$` the coefficients on `$exper$`, `$exper2$`, and `$age$` are `$0$`

+ `$H_1:$` `$H_0$` is false

---
class: middle

```r
#--- unrestricted ---#
logit_ur <- glm(
 inlf ~ nwifeinc + educ + exper + exper2 + age + kidslt6 + kidsge6,
 data = data, family = binomial(link = "logit")
)

#--- restricted ---#
logit_r <- glm(
 inlf ~ nwifeinc + educ + kidslt6 + kidsge6,
 data = data, family = binomial(link = "logit")
)

#--- LR test using lrtest() from the lmtest package ---#
library(lmtest)
lrtest(logit_r, logit_ur)
```

```
## Likelihood ratio test
## 
## Model 1: inlf ~ nwifeinc + educ + kidslt6 + kidsge6
## Model 2: inlf ~ nwifeinc + educ + exper + exper2 + age + kidslt6 + kidsge6
## #Df LogLik Df Chisq Pr(>Chisq) 
## 1 5 -464.92 
## 2 8 -401.77 3 126.32 < 2.2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
```
---
class: middle

# Prediction

After estimating a binary choice model, you can easily predict the following two

+ `$\hat{z}=\hat{\beta}_0+\hat{\beta}_1 x_1+\dots+\hat{\beta}_k x_k$`

+ `$\widehat{Prob(y=1|\mathbf{x})}=G(\hat{z})=G(\hat{\beta}_0+\hat{\beta}_1 x_1+\dots+\hat{\beta}_k x_k)$`

---
class: middle

.content-box-green[**R code**]

```r
#--- z hat ---#
z <- predict(probit_lf, type = "link")
head(z)
```

```
##         1         2         3         4         5         6 
## 0.5071349 0.6624576 0.5116317 0.7423429 0.1972781 0.8837878
```

```r
#--- G(z) hat ---#
Gz <- predict(probit_lf, type = "response")
head(Gz)
```

```
##         1         2         3         4         5         6 
## 0.6939699 0.7461610 0.6955456 0.7710602 0.5781950 0.8115946
```

---
class: middle

# Marginal effect of an independent variable

+ Coefficient estimates across different models (probit and logit) are not meaningful because the same value of a coefficient estimate means different things

+ They are the estimates of `$\beta$`s, not the direct impact of the independent variables on the `$Prob(y=1)$`

---
class: middle

Marginal effect of an independent variable

`$\frac{\partial Pr(y=1|x_1,\dots,x_k)}{\partial x_k} = G'(z)\times \beta_k$`

+ the marginal impact depends on the current levels of all the independent variables
+ we typically report one of the two types of marginal impacts
  * (becoming obsolete) the marginal impact .blue[at the mean] (average person): when all the independent variables take on their respective means
  * the average of the marginal impacts calculated for each of all the individuals observed

---
class: middle

.content-box-green[**Marginal impact at the mean**]

`$\frac{\partial Pr(y=1|x_1,\dots,\bar{x_k})}{\partial x_k} = G'(\beta_0+\beta_1 \bar{x_1}+\dots+\beta_k\bar{x_k})\times \beta_k$`

.content-box-green[**Mean marginal impact (MME)**]

`$\sum_{i=1}^n \frac{\partial Pr(y_i=1|x_{i,1},\dots,x_{i,k})}{\partial x_k} = \sum_{i=1}^n G'(z_i)\times \beta_k$`

---
class: middle

.content-box-green[**R codes to get MME**]

$$
\begin{aligned}
\hat{z}=\hat{\beta}_0+\hat{\beta}_1 x_1+\dots+\hat{\beta}_k x_k
\end{aligned}
$$

```r
#--- get z for all the individuals ---#
z <- predict(probit_lf, type = "link")
```

$$
\begin{aligned}
G'(\beta_0+\beta_1 \bar{x_1}+\dots+\beta_k\bar{x_k})
\end{aligned}
$$
where `$G(z)$` is the cumulative distribution function for the standard normal distribution.

```r
#--- get G'(z) ---#
Gz_indiv <- dnorm(z)
```

$$
\begin{aligned}
G'(\beta_0+\beta_1 \bar{x_1}+\dots+\beta_k\bar{x_k})
\end{aligned}\times \beta_k
$$

```r
#--- mean marignal impact of eduction ---#
mean(Gz_indiv) * probit_lf$coef["educ"]
```

```
##       educ 
## 0.03937009
```

---
class: middle

Fortunately, the `margins` package provides you with a more convenient way of calculating MMEs.

```r
library(margins)

#--- calculate MME based on the probit estimation ---#
mme_lf <- margins(probit_lf, type = "response")

#--- get the summary ---#
summary(mme_lf)
```

```
##    factor     AME     SE       z      p   lower   upper
##       age -0.0159 0.0024 -6.7392 0.0000 -0.0205 -0.0113
##      educ  0.0394 0.0073  5.4186 0.0000  0.0251  0.0536
##     exper  0.0371 0.0052  7.1779 0.0000  0.0270  0.0472
##    exper2 -0.0006 0.0002 -3.2050 0.0014 -0.0009 -0.0002
##   kidsge6  0.0108 0.0132  0.8189 0.4129 -0.0151  0.0367
##   kidslt6 -0.2612 0.0319 -8.1860 0.0000 -0.3237 -0.1986
##  nwifeinc -0.0036 0.0015 -2.4604 0.0139 -0.0065 -0.0007
```

For example, the average marginal effect of `educ` on the probability of "in labor force" is 0.039. That is if education increases by one year, then the probability of being in labor force increases by 4%.

---
class: inverse, center, middle

# Multinomial Choice Model

---
class: middle

# Multinomial Choice

Instead of two options, you are picking one option out of more than two options

+ which carrier?
  * Verizon
  * Sprint
  * AT\&T
  * T-mobile
+ which transportation means to commute?
  * drive
  * Uber
  * bus
  * train
  * bike

---
class: middle

# Multinomial logit model

The most popular model to analyze multinomial choice

+ environmental evaluation

+ tranposrtation

+ marketing

---
class: middle

# Understanding multinomial logit model

.content-box-green[**Choice of train route options**]

+ 10 euros, 30 minutes travel time, one change

+ 20 euros, 20 minutes travel time, one change

+ 22 euros, 22 minutes travel time, no change

.content-box-green[**Associated utility**]

+ `$V_1=\alpha_1 + \beta 10 + \gamma 30 + \rho 1 + v_1$`

+ `$V_2=\alpha_2 + \beta 20 + \gamma 20 + \rho 1 + v_2$`

+ `$V_3=\alpha_3 + \beta 22 + \gamma 22 + \rho 0 + v_3$`

---
class: middle

.content-box-green[**Choice probability**]

Logit model assumes that the probability of choosing an alternative is the following:

+ `$P_1=\frac{e^{V_1}}{e^{V_1}+e^{V_2}+e^{V_3}}$`

+ `$P_2=\frac{e^{V_2}}{e^{V_1}+e^{V_2}+e^{V_3}}$`

+ `$P_3=\frac{e^{V_3}}{e^{V_1}+e^{V_2}+e^{V_3}}$`

.content-box-green[**Notes**]

+ `$0<P_j<1$`, `$^\forall j=1,2,3$`

+ `$\sum_{j=1}^3=1$`

---
class: middle

.content-box-green[**Modeled probability of choices**]

Modeled probability of observing individual `$i$` choosing the option `$i$` chose

`$P_i = \Pi_{j=1}^3 y_{i,j}\times P_j$`

where `$y_{i,j}=1$` if `$i$` chose `$j$`, 0 otherwise.

.content-box-green[**Example**]

`$y_{i,1} = 0, \;\;y_{i,2} = 1,\;\;y_{i,3} = 0$`

`$P_i = \Pi_{j=1}^3 y_{i,j}\times P_j = 0\times P_1 + 1\times P_2 + 0\times P_3$`

---
class: middle

The probability of observing a series of chocies made by all the subjects is
  
`$LL = \Pi_{i=1}^n P_i = \Pi_{i=1}^n \Pi_{j=1}^3 y_{i,j}\times P_j$`

if choices made by the subjects are independent with each other.

.content-box-green[**MLE**]

`$Max_{\beta,\gamma,\rho}\;\; log(LL)$`

---
class: middle

# Interpretation of the coefficients

.content-box-green[**Model in general**]

`$V_{i,j} = \alpha_j + \beta_1 x_{1,i,j} + \dots + \beta_k x_{k,i,j}$`

`$P_{i,j} = \frac{e^{V_{i,j}}}{\sum_{k=1}^J e^{V_{i,k}}}$`

.content-box-green[**Interpretation of the coefficients**]
  
`$\frac{\partial P_{i,j}}{\partial x_{k,i,j}} = \beta_k P_{i,j}(1-P_{i,j})$`
    
+ A marginal change in `$k$`th variable for alternative `$j$` would change the probability of choosing alternative `$j$` by `$\beta_k P_{i,j}(1-P_{i,j})$`
    
+ the sign of the impact is the same as the sign of the coefficient

---
class: middle

.content-box-green[**Implementation in `$R$`**]

You can use `$mlogit$` package to estimate multinomial logit models:

+ format your data in a specific manner

+ convert your data using `$mlogit.data()$`

+ estimate the model using `$mlogit()$`

---
class: middle

```r
#--- library ---#
library(mlogit)

#--- get the travel mode data from the mlogit package ---#
data("TravelMode", package = "AER")

#--- take a look at the data ---#
# first 10 rows
head(TravelMode, 10)
```

```
##    individual  mode choice wait vcost travel gcost income size
## 1           1   air     no   69    59    100    70     35    1
## 2           1 train     no   34    31    372    71     35    1
## 3           1   bus     no   35    25    417    70     35    1
## 4           1   car    yes    0    10    180    30     35    1
## 5           2   air     no   64    58     68    68     30    2
## 6           2 train     no   44    31    354    84     30    2
## 7           2   bus     no   53    25    399    85     30    2
## 8           2   car    yes    0    11    255    50     30    2
## 9           3   air     no   69   115    125   129     40    1
## 10          3 train     no   34    98    892   195     40    1
```

---
class: middle

.content-box-green[**R code: data preparation**]

```r
#--- convert the data ---#
TM <- mlogit.data(TravelMode,
 shape = "long", # what format is the data in?
 choice = "choice", # name of the variable that indicates choice made
 chid.var = "individual", # name of the variable that indicates who made choices
 alt.var = "mode" # the name of the variable that indicates options
)
```

---
class: middle

```r
#--- take a look at the data ---#
# first 10 rows
head(TM, 10)
```

.scroll-box-16[

```
## ~~~~~~~
##  first 10 observations out of 840 
## ~~~~~~~
##    individual  mode choice wait vcost travel gcost income size    idx
## 1           1   air  FALSE   69    59    100    70     35    1  1:air
## 2           1 train  FALSE   34    31    372    71     35    1 1:rain
## 3           1   bus  FALSE   35    25    417    70     35    1  1:bus
## 4           1   car   TRUE    0    10    180    30     35    1  1:car
## 5           2   air  FALSE   64    58     68    68     30    2  2:air
## 6           2 train  FALSE   44    31    354    84     30    2 2:rain
## 7           2   bus  FALSE   53    25    399    85     30    2  2:bus
## 8           2   car   TRUE    0    11    255    50     30    2  2:car
## 9           3   air  FALSE   69   115    125   129     40    1  3:air
## 10          3 train  FALSE   34    98    892   195     40    1 3:rain
## 
## ~~~ indexes ~~~~
##    chid   alt
## 1     1   air
## 2     1 train
## 3     1   bus
## 4     1   car
## 5     2   air
## 6     2 train
## 7     2   bus
## 8     2   car
## 9     3   air
## 10    3 train
## indexes:  1, 2
```
]

---
class: middle

```r
#--- estimate ---#
ml_reg <- mlogit(choice ~ wait + vcost + travel, data = TM)
```

---
class: middle

.scroll-box-24[

```r
summary(ml_reg)
```

```
## 
## Call:
## mlogit(formula = choice ~ wait + vcost + travel, data = TM, method = "nr")
## 
## Frequencies of alternatives:choice
## air train bus car 
## 0.27619 0.30000 0.14286 0.28095 
## 
## nr method
## 5 iterations, 0h:0m:0s 
## g'(-H)^-1g = 0.000192 
## successive function values within tolerance limits 
## 
## Coefficients :
## Estimate Std. Error z-value Pr(>|z|) 
## (Intercept):train -0.78666667 0.60260733 -1.3054 0.19174 
## (Intercept):bus -1.43363372 0.68071345 -2.1061 0.03520 * 
## (Intercept):car -4.73985647 0.86753178 -5.4636 4.665e-08 ***
## wait -0.09688675 0.01034202 -9.3683 < 2.2e-16 ***
## vcost -0.01391160 0.00665133 -2.0916 0.03648 * 
## travel -0.00399468 0.00084915 -4.7043 2.547e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Log-Likelihood: -192.89
## McFadden R^2: 0.32024 
## Likelihood ratio test : chisq = 181.74 (p.value = < 2.22e-16)
```
]

---
class: middle
  
# Understanding the results

```r
summary(ml_reg)$coef
```

```
## (Intercept):train   (Intercept):bus   (Intercept):car              wait             vcost            travel 
##      -0.786666672      -1.433633718      -4.739856473      -0.096886747      -0.013911604      -0.003994681 
## attr(,"names.sup.coef")
## character(0)
## attr(,"fixed")
## (Intercept):train   (Intercept):bus   (Intercept):car              wait             vcost            travel 
##             FALSE             FALSE             FALSE             FALSE             FALSE             FALSE 
## attr(,"sup")
## character(0)
```

+ intercept for `$air$` is dropped `$(air$` is the base)
  * train:(intercept) is `$-0.786$` means that train is less likely to be chosen if all the other .blue[included] variables are the same

+ the greater the travel time, the less likely the option is chosen

---
class: inverse, center, middle

# Count data (Poisson)

---
class: middle

# Count data (Poisson)

Count variables take non-negative discrete integers `$(0,1,\dots,)$`

+ the number of times individuals get arrested in a year
+ the number of cars owned by a family
+ the number of kids in a family

---
class: middle

# Poisson regression

By far the most popular choice to analyze count variables is Poisson regression

+ The outcome (count) variable is assumed to be Poisson distributed
+ The mean of the Poisson distribution is assumed to be a function of some variables you believe matter

--

.content-box-green[**Poisson distribution**]

Poisson distribution is a discrete probability distribution that describes the probability of the number of events that occur in a fixed interval of time and/or space

$$
\begin{aligned}
Prob(y|\lambda)=\frac{\lambda^y e^{-\lambda}}{y!}, \;\; \mbox{where} \;\; \lambda=E[y]
\end{aligned}
$$

---
class: middle
<img src="data:image/png;base64,#discrete_choice_x_files/figure-html/poisson-1.png" width="60%" style="display: block; margin: auto;" />

.content-box-green[**Poisson regression**]

We try to learn what and how variables affect the expected value (the expected number of events conditional on independent variables).

---
class: middle

.content-box-green[**Expected number of events conditional on independent variables**]

$$
\begin{aligned}
E[y|\mathbf{x}] = G(\beta_0+\beta_1 x_1+\dots + \beta_k x_k)
\end{aligned}
$$

+ This is exactly the same modeling framework we used
  - Linear model: `$G(z)=z$`
  - Probit model: `$G(z)=\Phi(z)$`

--

.content-box-green[**A popular choice of **]$G()$
$$
\begin{aligned}
  G(z) = exp(z)
\end{aligned}
$$

This ensures that the expected value conditional on `$\mathbf{x}$` is always positive

---
class: middle

.content-box-green[**The number of events for two individuals**]

+ Individual 1: `$y=3$` (own three cars)
+ Individual 2: `$y=1$` (own one car)

--

.content-box-green[**Expected number of events observed**]
+ Individual 1: `$\lambda_1= exp(z_1)$`
+ Individual 2: `$\lambda_2= exp(z_2)$`

---
 
.content-box-green[**Probability of observing the number of events we observed**]

$$
\begin{aligned}
\mbox{Individual 1}:& Prob(y=3|\mathbf{x_1})=\frac{\lambda_1^3 e^{-\lambda_1}}{3!}  \\
\mbox{Individual 2}:& Prob(y=1|\mathbf{x_2})= \frac{\lambda_2^1 e^{-\lambda_2}}{1!}
\end{aligned}
$$

--

.content-box-green[**Probability of observing a series of events by all individuals**]

The probability that we observe the collection of choices made by them (if their events are independent)

$$
\begin{aligned}
L=Prob(y_1=3|\mathbf{x_1})\times Prob(y_2=1|\mathbf{x_2}) = \frac{\lambda_1^3 e^{-\lambda_1}}{3!}\times\frac{\lambda_2^1 e^{-\lambda_2}}{1!}
\end{aligned}
$$

which we call llikelihood function.

--

.content-box-green[**Log-likelihood function**]

$$
\begin{aligned}
LL=log(L)&= log(\frac{\lambda_1^3 e^{-\lambda_1}}{3!}) + log(\frac{\lambda_2^1 e^{-\lambda_2}}{1!})
\end{aligned}
$$

(Remember `$\lambda_i=exp(\beta_0+\beta_1 x_{i,1}+\dots + \beta_k x_{i,k})$`)

--

$$
\begin{aligned}
Max_{\beta_1,\dots,\beta_k}\;\; & LL
\end{aligned}
$$

---
class: middle

# Implementation in R with an example

The number of times a man is arrested during 1986:
$$
\begin{aligned}
Pr(narr86|\mathbf{x})= G(z)
\end{aligned}
$$

where

$$
\begin{aligned}
z = & \beta_0+\beta_1 pcnv+ \beta_2 tottime + \beta_3 qemp86 + \beta_4 inc86\\
& + \beta_5 black + \beta_6 hispan
\end{aligned}
$$

+ `$narr86$`: \# of times arrested in 1986
+ `$pcnv$`: proportion of prior conviction
+ `$tottime$`: time in prison since 18
+ `$qemp86$`: \# quarters employed in 1986
+ `$inc86$`: legal income in 1986 (in $\$100$)

---
class: middle

R code: importing the data

```r
#--- import the data ---#
data <- read.dta13("CRIME1.dta")

#--- take a look ---#
dplyr::select(data, narr86, pcnv, qemp86, inc86) %>%
  head()
```

```
##   narr86 pcnv qemp86 inc86
## 1      0 0.38      0   0.0
## 2      2 0.44      1   0.8
## 3      1 0.33      0   0.0
## 4      2 0.25      2   8.8
## 5      1 0.00      2   8.1
## 6      0 1.00      4  97.6
```

---
class: middle

.content-box-green[**R code: Poisson model estimation using glm()**]

```r
pois_lf <- glm(

#--- formula ---#
  narr86 ~ pcnv + tottime + qemp86 + inc86 + black + hispan,

#--- data ---#
  data = data,

#--- models ---#
  family = poisson(link = "log")
)
```

.content-box-green[**`family` option**]

+ `poisson()`: tells `$R$` that your dependent variable is Poisson distributed
+ `link = "log"`: tells `$R$` that you want to use `$exp()$` (the inverse of `$log()$`) as `$G()$` in `$E(y=1|\mathbf{x})=G(z)$`

---
class: middle

.left4[

```r
msummary(
  pois_lf,
  # keep these options as they are
  stars = TRUE,
  gof_omit = "IC|Log|Adj|F|Pseudo|Within"
)
```
]

.right6[

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style="overflow-wrap:break-word;"><td class="cl-63702a23"></td><td class="cl-63702a22">(0.064)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a55">pcnv</td><td class="cl-63702a54">-0.432***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a23"></td><td class="cl-63702a22">(0.085)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a5f">tottime</td><td class="cl-63702a5e">-0.001</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a23"></td><td class="cl-63702a22">(0.006)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a69">qemp86</td><td class="cl-63702a68">-0.010</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a23"></td><td class="cl-63702a22">(0.029)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a2d">inc86</td><td class="cl-63702a2c">-0.008***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a23"></td><td class="cl-63702a22">(0.001)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a2d">black</td><td class="cl-63702a2c">0.644***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a23"></td><td class="cl-63702a22">(0.074)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a37">hispan</td><td class="cl-63702a36">0.473***</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a23"></td><td class="cl-63702a22">(0.074)</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a41">Num.Obs.</td><td class="cl-63702a40">2725</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-63702a4a">RMSE</td><td class="cl-63702a4b">1.02</td></tr></tbody><tfoot><tr style="overflow-wrap:break-word;"><td colspan="2"class="cl-63702a6a">+ p &lt; 0.1, * p &lt; 0.05, ** p &lt; 0.01, *** p &lt; 0.001</td></tr></tfoot></table></div></template>
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---
class: middle

# Calculate average marginal effects

Just like the binomial regressions we saw earlier, we can use `margins::margins()` function to get the average marginal effects of covariates.

```r
pois_marginal_e <- margins(pois_lf, type = "response")
```