Econometrics

.title[
# Econometrics
]
.subtitle[
## Probability: 🎲, 🔮, 🪙
]
.author[
### Florian Oswald
]
.date[
### UniTo ESOMAS </br> 2025-09-17
]

---

---

# Probability

1. Probabilities and RVs
2. PDFs and CDFs of discrete
3. PDFs and CDFs of cont
4. E, Var
5. funcs of RVs
6. standardizing RVs
7. joint and marginal dists
8. conditional dists, independence, covariance, corr
9. LIE
10. correlation and conditional mean
11. KC 2.3
12. random sampling and distributions of resulting statistics

---

# Probabilities

*Probability* is a function that assigns a value in `$[0,1]$` to a *set* (representing *events*).

Consider a fair 6-sided dice. 🎲

* Outcome space: `$\Omega = \{1,2,3,4,5,6\}$`
* Event: a partition of `$\Omega$`.

For example, the events A ("the outcome is even") and B ("the outcome is odd") are:

`$$\begin{align}
A &= \{2,4,6\},\\
B &= \{1,3,5\}.
\end{align}$$`

Having a *fair* dice means that

`$$\Pr(1) = \dots = \Pr(6) = \frac{1}{6}$$`

---

# Probabilities

Given a fair 6-sided dice 🎲, i.e. an outcome space `$\Omega = \{1,2,3,4,5,6\}$`
and events

`$$\begin{align}
A &= \{2,4,6\},\\
B &= \{1,3,5\}.
\end{align}$$`

how does one determine the ***likelihood*** (or the **probability**) of event A occuring?

---

# Discrete Random Variables

A *Discrete Random Variable* is a function mapping outcomes to measurements.

For example, we might call

* `$X$` the number we obtain from throwing the dice once.

* `$X_1,X_2$` the 2 numbers we obtain from throwing the dice two times.

* `$D$` an indicator (either `0`, `1`) for whether a randomly sampled person answers "yes" or "no" when asked whether they have children.

---
class: inverse

# Gambler's Ruin

The dealer tosses a fair coin. If it comes up tails (`T`) you win, if it's heads (`H`)you loose.

Suppose we've seen the following sequence of tosses so far:

```
outcome:  H   H   H   H   H   H   H   H   H   H   ?
toss num: 1.  2.  3.  4.  5.  6.  7.  8.  9.  10. 11.  
```

You have lost 10 times in a row now. Does this increase the probability that you will win the next round?

Which sequence is more likely to occur?

1. `TTTTTTTTTT` (10 `T` in a row)?
2. `HTHTTHTHHT` (random occurence of `T` and `H`)

---

## Gambler's Ruin

The probability of hitting 11 `H` in a row is

$$
\Pr(\text{11 Heads in a Row}) = \frac{1}{2^{11}} = 0.00048 = 0.05\%
$$
--

Ok, but I already had 10 `H`! What's the next toss likely going to be given that miserable history?

$$
\Pr(\text{toss 11 is Heads} | 10 \text{ Heads before})
$$
Let's calculate that probability on the board!

---

# Probability Distributions

---

---

## PDF/PMF and CDF of discrete RVs

* PDF/PMF: Probability Density/Mass Function. Table listing each outcome and the associated probability of observing it.
* CDF: Cumulative Distribution Function. Probability that a given RV takes on a value *less than or equal a certain value* (requires a notion of *ordering* - e.g. what about a dice with 6 different **colors**?)

### For our 6-sided fair dice

```
##   outcome pdf cdf
## 1       1 1/6 1/6
## 2       2 1/6 2/6
## 3       3 1/6 3/6
## 4       4 1/6 4/6
## 5       5 1/6 5/6
## 6       6 1/6 6/6
```

---

## PDF/PMF and CDF of discrete RVs

### For our 6-sided fair dice

<img src="chapter_probability_files/figure-html/unnamed-chunk-2-1.svg" style="display: block; margin: auto;" />
]

.pull-right[
<img src="chapter_probability_files/figure-html/unnamed-chunk-3-1.svg" style="display: block; margin: auto;" />
]

---

## PDF/PMF and CDF of discrete RVs

### For our fair coin

```
##   outcome pdf cdf
## 1    H(0) 1/2 1/2
## 2    T(1) 1/2 2/2
```

---

## PDF/PMF and CDF of discrete RVs

### For our fair coin

<img src="chapter_probability_files/figure-html/unnamed-chunk-5-1.svg" style="display: block; margin: auto;" />
]

.pull-right[
<img src="chapter_probability_files/figure-html/unnamed-chunk-6-1.svg" style="display: block; margin: auto;" />
]

---

## Bernoulli Distribution

* A special discrete RV with 2 outcomes: `0` and `1`, where event `1` ("success") occurs with probability `$p$`.

For example:

* Tomorrow it will rain with probability `$p$` (it will *not* rain with `$1-p$`).

What kind of RV is flipping a *fair* coin? What about an *unfair* coin?

---

## PDFs and CDFs of continuous Variables

* slightly more complicated because we need calculus and integration.
* The basic idea is the same!

---

# Expected Value and Variance

---

---

## Expected Value and Variance

We write

`$$E(Y) = y_1 p_1 + \dots + y_n p_n$$`

For example for our dice example, where `$X$` is the result of throwing the dice:

`$$\begin{align}
E(X) &= 1 \times \Pr(1) + 2 \times \Pr(2) + \dots + 6 \times \Pr(6) \\
     &=
\end{align}$$`

**Everybody calculate this result now!**

---

# Expected Value and Variance

We write

$$
E(Y) = y_1 p_1 + \dots + y_n p_n
$$

For example for our dice example, where `$X$` is the result of throwing the dice:

`$$\begin{align}
E(X) &= 1 \times \Pr(1) + 2 \times \Pr(2) + \dots + 6 \times \Pr(6) \\
     &= 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + \dots + 6 \times \frac{1}{6} \\
     &= 3.5 \\
\end{align}$$`

* So: We need to know ***all*** weights and values (`$\Pr$` and `$X$`) in order to compute this quantity.

---

# `$E(X)$` is a Theoretical Concept. Mission Impossible?

* Imagine you get this message:

1. Your mission, should you choose to accept it, is to inspect this device : 🔮. Whenever you touch it, it displays a number in `$\{1,\dots,10\}$`. E.g. `4,8,4,9,1,5,10,9,1`...kind of random.
2. We must know the long-run average, i.e. E(🔮). Or something terrible will happen!
3. This message will destroy itself in 10,9,8,...💣

* You **don't know** the theoretical distribution of all the numbers in 🔮 (the `$\Pr$`'s). Time is running ⏰

* What could you do? 🤔

---

---

# `$E(X)$` is a Theoretical Concept. Mission Impossible?

* If we had a huge number of observations from 🔮 (a *sample*), we could come up with a *guess* of E(🔮), based on the data we got. Empirical, like.

``` r
set.seed(12345)  # to ensure reproducibility
Ps = runif(10)
Ps[c(2,3)] = 0
Ps = Ps / sum(Ps) # generate random weights
x = sample(1:10, size = 10000, replace = TRUE, prob = Ps)
xbar = mean(x)
xbar
```

```
## [1] 6.3418
```

* `xbar` = `$\bar{x} = \frac{1}{N} \sum_i^{N} x_i$` is called *Sample Mean*, *Arithmetic Average*, *Sample Average*
* `$\bar{x}$` is an ***estimator*** for `$E(X) = \mu_X$`. (`$E(X) = \mu_X$` by the way.)

---

# Central Tendency - Mean and Median

.pull-left[
`mean(x)`: the average of all values in `x`.
`$$E(X) = \mu_X = \frac{1}{N}\sum_{i=1}^N x_i = \bar{x}$$`

***The second equality 👆 is correct only if...???***

``` r
x <- c(1,2,2,2,2,100)
mean(x)
```

```
## [1] 18.16667
```

``` r
mean(x) == sum(x) / length(x)
```

```
## [1] FALSE
```
]

.pull-right[
`median`: the value `$x_j$` below and above which 50% of the values in `x` lie. `$m$` is the median if

`$$\Pr(X \leq m) \geq 0.5 \text{ and } \Pr(X \geq m) \geq 0.5$$`
    
The median is robust against *outliers*.

``` r
median(x)
```

```
## [1] 2
```
]

---

## Quick Review

1. EV of a bernoulli
2. Continuous RV

---

## Variance

* A measure of *spread* of a distribution.
* The definition of *variance* is

`$$var(Y) = E[(Y-\mu_Y)^2]$$`

if `$Y$` is discrete,

`$$var(Y) = \sum_{i=1}^N (y_i-\mu_Y)^2 \times p_i.$$`
---

## Variance

Consider two `normal distributions` with equal mean at `0`:
]

.pull-right[
<img src="chapter_probability_files/figure-html/unnamed-chunk-11-1.svg" style="display: block; margin: auto;" />

Compute with:

``` r
var(x)
```
]

---

## Example of other Spread Measures

``` r
# % catholic in 47 french-speaking
# swiss cantons in 1888
plot(swiss$Catholic,rep(1,nrow(swiss)),pch = 3,
     cex = 2,xlab = "% Catholic",yaxt = "n",ylab = "") 
```

<img src="chapter_probability_files/figure-html/unnamed-chunk-13-1.svg" style="display: block; margin: auto;" />
]

| Measure            | `R`              | Result             |
|:---------:|:-------------------:|:---------------------:|
| Variance           | `var(swiss$Catholic)`   | 1739.29   |
| Standard Deviation | `sd(swiss$Catholic)`    | 41.7    |
| IQR                | `IQR(swiss$Catholic)`   | 87.93   |
| Minimum            | `min(swiss$Catholic)`   | 2.15   |
| Maximum            | `max(swiss$Catholic)`   | 100   |
| Range              | `range(swiss$Catholic)` | 2.15, 100 |

]

---

# Computing Variance by hand

1. Compute the variance of our 6-sided dice!
2. compute the variance of `$X \sim \text{Bernoulli}(p)$`!

---

# Two Random Variables <br> `$(X,Y)$`, (🎲, 🎲), (🎲, 🪙), (🪙, 🪙)

---

---

# Two Random Variables: Example

.pull-left[
<img src="chapter_probability_files/figure-html/x-y-corr-1.svg" style="display: block; margin: auto;" />
]

* Here, `$X$` and `$Y$` are ***joint normally*** distributed.
* We would write $$(X, Y) \sim 
\mathcal{N}\biggl(
\begin{bmatrix}
\mu_X \\
\mu_Y
\end{bmatrix},
\,
\Sigma
\biggr)$$`
 
where `$\Sigma$` is a *matrix*

`$$\begin{bmatrix}
\sigma_X^2 & \rho \, \sigma_X \sigma_Y \\
\rho \, \sigma_X \sigma_Y & \sigma_Y^2
\end{bmatrix}$$`

]

Taking as example the data in this plot, the concepts *covariance* and *correlation* relate to the following type of question:

# missing

joint and marginal dists
8. conditional dists, independence, covariance, corr
9. LIE
10. correlation and conditional mean

create/get a table and illustrate

---

# Tabulating Data

`table(x)` is a useful function that counts the occurence of each unique value in `x`:

``` r
table(gapminder$continent)
```

```
## 
##   Africa Americas     Asia   Europe  Oceania 
##     2907     2052     2679     2223      684
```

The same can be achieved using the `count` function (from `dplyr`)

``` r
gapminder %>% count(continent)
```

```
##   continent    n
## 1    Africa 2907
## 2  Americas 2052
## 3      Asia 2679
## 4    Europe 2223
## 5   Oceania  684
```

---

# Tabulating Data

Given two variables, `table` produces a contingency table:

``` r
gapminder_new <- gapminder %>%
  filter(year == 2015) %>%
  mutate(fertility_above_2 = (fertility > 2.1)) # dummy variable for fertility rate above replacement rate
```

``` r
table(gapminder_new$fertility_above_2)
```

```
## 
## FALSE  TRUE 
##    80   104
```
]

``` r
table(gapminder_new$fertility_above_2,gapminder_new$continent)
```

```
##        
##         Africa Americas Asia Europe Oceania
##   FALSE      2       15   20     39       4
##   TRUE      49       20   27      0       8
```
]

With `prop.table`, we can get proportions:

``` r
# proportions by row
prop.table(table(gapminder_new$fertility_above_2,gapminder_new$continent), margin = 1)
# proportions by column
prop.table(table(gapminder_new$fertility_above_2,gapminder_new$continent), margin = 2) 
```

* ⚠️ To obtain `table`s or `crosstable`s with `NA`s, use the `useNA = "always"` or `useNA = "ifany"`

---

# Tabulating Data

Again the `count` function can get you there as well:

``` r
gapminder_new %>%
  count(continent, fertility_above_2)
```

```
##    continent fertility_above_2  n
## 1     Africa             FALSE  2
## 2     Africa              TRUE 49
## 3   Americas             FALSE 15
## 4   Americas              TRUE 20
## 5   Americas                NA  1
## 6       Asia             FALSE 20
## 7       Asia              TRUE 27
## 8     Europe             FALSE 39
## 9    Oceania             FALSE  4
## 10   Oceania              TRUE  8
```

Note that `count` will display `NA`s only if there are any.

``` r
gapminder_new %>% filter(is.na(fertility)) %>% select(country)
```

```
##     country
## 1 Greenland
```

---

# Joint Distributions

``` r
abs_tab
```

```
##           
##            FALSE TRUE
##   Africa       2   49
##   Americas    15   20
##   Asia        20   27
##   Europe      39    0
##   Oceania      4    8
```
There are 184 countries.
]

``` r
prop_tab
```

```
##           
##            FALSE TRUE
##   Africa    0.01 0.27
##   Americas  0.08 0.11
##   Asia      0.11 0.15
##   Europe    0.21 0.00
##   Oceania   0.02 0.04
```
What *fraction* of countries is in each bin?

* 👉The *joint distribution*.
* `$\Pr(X,Y)$` where X = `continent` and Y = `fertility_above_2`
]

👉 The *conditional distribution* is `$\Pr(X=x|Y=y) = \frac{\Pr(X=x,Y=y)}{\Pr(Y=y)}$`

---

# Computing Conditional Distributions

``` r
prop_tab
```

```
##           
##            FALSE TRUE
##   Africa    0.01 0.27
##   Americas  0.08 0.11
##   Asia      0.11 0.15
##   Europe    0.21 0.00
##   Oceania   0.02 0.04
```
]

.pull-right[
1. Compute the Distribution of countries conditional on low fertility?
2. Compute the Distribution of countries conditional on being in Asia?
3. Compute the marginal distribution of high/low fertility
]

---

# How are x and y related? Covariance and Correlation

---

# Covariance

* The covariance is a measure of __joint variability__ of two variables.
    `$$Cov(x,y) = \frac{1}{N} \sum_{i=1}^N(x_i-\bar{x})(y_i-\bar{y})$$`

* The `cov` function computes the covariance:

``` r
cov(gapminder_new$fertility,gapminder_new$infant_mortality, use = "complete.obs")
```

```
## [1] 24.21146
```

* Difficult to interpret because sensitive to the variables' dispersions from the mean

---

# Correlation

* The correlation is a measure of the strength of the __linear association__ between two variables.
    `$$Cor(x,y) = \frac{Cov(x,y)}{\sqrt{Var(x)}\sqrt{Var(y)}}$$`

* The `cor` function computes the correlation:

``` r
cor(gapminder_new$fertility,gapminder_new$infant_mortality, use = "complete.obs")
```

```
## [1] 0.8286402
```

---

# Correlation

* **Correlation is always between -1 and 1!**

---

# Task 4: Summarising data

1. Compute the mean of GDP in 2011 and assign to object `mean`. You should exclude missing values. (*Hint: read the help for `mean` to remove `NA`s*).

1. Compute the median of GDP in 2011 and assign to object `median`. Again, you should exclude missing values. Is it greater or smaller than the average?

1. Create a density plot of GDP in 2011 using `geom_density`. A density plot is a way of representing the distribution of a numeric variable. Add the following code to your plot to show the median and mean as vertical lines. What do you observe?
`geom_vline(xintercept = as.numeric(mean), colour = "red") +` <br>
    `geom_vline(xintercept = as.numeric(median), colour = "orange")`

1. Compute the correlation between fertility and infant mortality in 2015. To drop `NA`s in either variable set the argument `use` to "pairwise.complete.obs" in your `cor()` function. Is this correlation consistent with the graph you produced in Task 3?

---

class: title-slide-final, middle
background-image: url(../img/logo/esomas.png)
background-size: 250px
background-position: 9% 19%

# That's all for Probability!

|                                                                                                            |                                   |
| :--------------------------------------------------------------------------------------------------------- | :-------------------------------- |
| <a href="https://github.com/floswald/Econometrics-Slides">.ScPored[<i class="fa fa-link fa-fw"></i>] | Slides |
| <a href="https://floswald.github.io">.ScPored[<i class="fa fa-link fa-fw"></i>] | My Homepage |
| <a href="https://scpoecon.github.io/ScPoEconometrics/">.ScPored[<i class="fa fa-github fa-fw"></i>]                          | Book                       |