Basics in statistics

class: center, middle, inverse, title-slide

.title[
# Basics in statistics
]
.author[
### Mahendra Mariadassou, INRAE <br> .small[from original slides by Tristan Mary-Huard]
]
.date[
### Shandong University, Weihai (CN)<br>Summer School 2023
]

---

# About this course...

.blue[Topics]

- Basics in statistics

- Bivariate Analysis

- Sampling

- Estimation

- Confidence Intervals

- Tests

- Simple Linear Regression (hopefully)

---
# About this course...

From **biology** to **biology...** through maths !

.blue[Modeling]

Move from a biological question to a statistical model accounting for  
- the nature of the data at hand,  
- the randomness of the experiment,    
- error measurement (see last sessions).

.blue[Statistical inference]

Get some information about some of the (unknown) **parameters** appearing in the model. One aims at  
- estimating the parameter,  
- give a (restricted) range of possible values for the parameter,  
- decide whether the parameter belongs to a given interval or not.

---
# Overview of the first session(s)

**Describing a population**

`$\star$` How are individuals distributed w.r.t. height ? To age ?

`$\star$` How are individuals distributed between male and female ?

**Study the relationship between descriptors**

`$\star$` How are height and age related ?

`$\star$` Is the height distribution the same for male and female ?

---
class: middle, inverse, center

# Finite Population

---
# Definitions

.blue[Population:] a collection of elements of interest

- the collection of individuals of the French population,  
- the collection of all devices produced in a given factory,  
- the collection of all courses that are proposed in a given university.

The elements are called **individuals**.

.blue[Variables:] a collection of measurements that describe each individual  
- height, gender and number of children (French people),  
- weight and length (device),  
- number of hours and level (course).

A measurement made on an individual results in an **observation**.

---
# Definitions

.blue[Quantitative variables:] the variable is a **quantitative** measurement  
- numeric/continuous: the range of the variable is continuous (height),  
- discrete: only some numeric values are possible (number of child.).

.blue[Qualitative variables:] the variable gives information about class membership  
- nominal: gender, ethnicity,  
- ordinal: class of disease severity.

---
# Example

.pull-left[

```
## # A tibble: 8 × 3
##   G         H     C
##   <chr> <dbl> <dbl>
## 1 M      1.86     1
## 2 F      1.72     0
## 3 M      1.75     3
## 4 M      1.9      2
## 5 F      1.65     1
## 6 F      1.68     0
## 7 M      1.7      4
## 8 M      1.82     2
```
]

.pull-right[
- Size of population : `$n=8$`  
- 3 variables: 
  - gender `$G$` (1=Male, 2=Female),    
  - height `$H$`,  
  - number of children `$C$`
- Individual 7

```
##   G   H C
## 1 M 1.7 4
```
]

Variable `$H$` measured on individual `$i$` results in **measurement** `$h_i$`

---
# Probabilities on finite pop.

Consider a **population** of `$n$` individuals.

Let `$A$` and `$B$` be some subsets of individuals, with respective size `$n_A$` and `$n_B$`.

.def[Definition 1] The probability for an individual drawn at random to belong to subset `$A$`, also called probability of `$A$`, is defined as follows:

$$
P(A) = \frac{\text{total number of favorable cases}}{\text{total number of possible cases}} = \frac{n_A}{n} \ \ .
$$

---
# Union, intersection

.def[Definition 2] Denote `$A\bigcap B$` the subset of individuals belonging to subsets `$A$` and `$B$` at the same time. Assume there are `$n_{AB}$` such items, one has:
`$$P\left(A\bigcap B\right) = \frac{n_{AB}}{n} \ \ .$$`

.def[Definition 3] Denote `$A\bigcup B$` the subset of individuals belonging to either `$A$` or `$B$` (or both). One has:
`$$P\left(A\bigcup B\right) = P(A) + P(B) - P\left(A\bigcap B\right) \ \ .$$`

.question[Quizz 1]

---
# Conditional probability

.def[Definition 4] Denote `$P(B|A)$` the conditional probability of `$B$` knowing `$A$`. One has :
`$$P(B|A) =  \frac{n_{AB}}{n_A} =  \frac{n_{AB}}{n}\times\frac{n}{n_A} =  \frac{P(A\bigcap B)}{P(A)} \ \ .$$`

.def[Definition 5] Two events `$A$` and `$B$` are independent if they satisfy the property
`$$P(A\bigcap B) = P(A)\times P(B) \ \ .$$`

.question[Quizz 2]

.remark[Remark] If `$A$` and `$B$` are two independent events, then
`$$P(B|A) =  \frac{P(A\bigcap B)}{P(A)} = \frac{P(A)\times P(B)}{P(A)} = P(B) \ \ .$$`

---
# Distribution

The **probability distribution** associates each possible value of a variable with its probability, 
  - **synthesis without loss of information**

`$\star$` Qualitative variable example: Gender

```
## 
##     F     M 
## 0.375 0.625
```

`$\star$` Quantitative variable example: Height

```
## 
##  1.65  1.68   1.7  1.72  1.75  1.82  1.86   1.9 
## 0.125 0.125 0.125 0.125 0.125 0.125 0.125 0.125
```

---
# Graphical representation

For **qualitative/discrete** variables, pie charts and histograms provide an **exhaustive** representation of the distribution.

.pull-left-70[
<img src="01_Basics_files/figure-html/unnamed-chunk-5-1.png" height="520px" style="display: block; margin: auto;" />
]

.pull-right-30[

- What about **quantitative** variables ?

- More on that later

]

---
# Limitations

.def[Remark] The probability distribution is a **synthetic** representation only when the variable has a finite (and small) number of possible values.

`$\star$` Good representation for qualitative/discrete variables,

`$\star$` Poor representation for continuous variables.

One can define extra quantities that lead to higher levels of synthesis at the cost of **a loss of information**.

---
## Expectation (mean value)

Example for variable Height

`$$\begin{align}
E(H) & = \frac{1}{8} (1.86 + 1.72+ 1.75+ 1.90+ 1.65+ 1.68+ 1.70+ 1.82)\\
     & = 1.86\times\frac{1}{8} + 1.72\times\frac{1}{8}+ 1.75\times\frac{1}{8}+ 1.90\times\frac{1}{8}+ 1.65\times\frac{1}{8} + \dots \\
     & \quad \dots + 1.68\times\frac{1}{8} + 1.70\times\frac{1}{8}+ 1.82\times\frac{1}{8}\\
     &= 1.76 
\end{align}$$`
and variable Children

`$$\begin{align}
E(C) &= \frac{1}{8}(1 + 0+ 3+ 2+ 1+ 0+ 4+ 2)\\
     &= 0\times\frac{2}{8} + 1\times\frac{2}{8}+ 2\times\frac{2}{8}+ 3\times\frac{1}{8}+ 4\times\frac{1}{8}\\
     &= 1.625     
\end{align}$$`

---
## Expectation (Cont'd)

.def[General definition] For a variable `$Y$` (case of finite population of size `$N$`):

`$$\begin{equation}
E(Y) = \frac{1}{N} \sum_{i=1}^{N} y_i = \sum_{k=1}^{K} y_k \times P(Y=y_k)
\end{equation}$$`

where `$K$` is the number of possible values for variable `$Y$` in the population.

You can compute the mean from:
- the whole population, with `$\sum_{i=1}^N y_i / N$`
- from the probability distribution, with `$\sum_{k=1}^{K} y_k \times P(Y=y_k)$`

.center[
Both encode the .alert[same] information
]

---
## Variance

Average squared distance between expectation and observation

`$$\begin{align}
V(C) &= \frac{1}{8}\Big\{(1-1.625)^2 + (0-1.625)^2+ (3-1.625)^2+ (2-1.625)^2 \\
     &  \quad +(1-1.625)^2+ (0-1.625)^2+ (4-1.625)^2+ (2-1.625)^2\Big\}\\
     &= \frac{2}{8}(0-1.625)^2+ \frac{2}{8}(1-1.625)^2+ \frac{2}{8}(2-1.625)^2\\
     &  \quad + \frac{1}{8}(3-1.625)^2+ \frac{1}{8}(4-1.625)^2\\
     &= 1.734
\end{align}$$`

---
## Variance

.def[General definition] for a variable `$Y$` (case of finite population of size N):

`$$\begin{equation}
V(Y) = \frac{1}{N}\sum_{i=1}^{N} (y_{i}-E(Y))^2 = \sum_{k=1}^{K} (y_k-E(Y))^2\times P(Y=y_{k})
\end{equation}$$`

.def[Standard deviation], expressed in the .alert[same unit] as the measurements.

$$
\sigma(Y) = \sqrt{V(Y)}.
$$

You can again compute the variance / standard deviation from:
- the whole population 
- from the probability distribution

.center[
Both encode the .alert[same] information
]

---
.small[
.pull-left[

```
##     Population Children
##  1:     Pop. 2        1
##  2:     Pop. 2        3
##  3:     Pop. 2        2
##  4:     Pop. 2        2
##  5:     Pop. 2        3
##  6:     Pop. 2        4
##  7:     Pop. 2        1
##  8:     Pop. 2        1
##  9:     Pop. 2        2
## 10:     Pop. 2        3
## 11:     Pop. 2        1
## 12:     Pop. 2        3
## 13:     Pop. 3        0
## 14:     Pop. 3        3
## 15:     Pop. 3        4
## 16:     Pop. 3        4
## 17:     Pop. 3        3
## 18:     Pop. 3        4
## 19:     Pop. 3        0
## 20:     Pop. 3        0
## 21:     Pop. 3        0
## 22:     Pop. 3        4
## 23:     Pop. 3        0
## 24:     Pop. 3        3
## 25:     Pop. 3        0
## 26:     Pop. 3        4
## 27:     Pop. 3        3
##     Population Children
```
]
]

.pull-right[
<img src="01_Basics_files/figure-html/unnamed-chunk-7-1.png" style="display: block; margin: auto;" />

.question[Quizz 3]
]

---
## Expectation vs Variance

Number of children per individual in 2 populations:

**Population 2**: `$\{1,3,2,2,3,4,1,1,2,3,1,3\}$`

**Population 3**: `$\{0,3,4,4,3,4,0,0,0,4,0,3,0,4,3\}$`

`$$\begin{equation}
\begin{array}{c||c|c|c|c|}
        & Size  & Mean  & Variance  & St.Dev \\
\hline
Pop.2   & 12    & 2.16  & 0.97      & 0.98   \\
Pop.3   & 15    & 2.13  & 3.18      & 1.78   \\
\hline
\end{array}
\end{equation}$$`

Variance/std deviation measures how .alert[spread out] the population is.

- Here Population 2 is **more homogeneous** than Population 3.

.blue[Expectation versus variance]

- Expectation gives the **location** of the population
- Variance quantifies the **dispersion** from the expectation.

---
# Median

The median `$y_{0.5}$` for distribution of variable `$Y$` is the smallest value such that at least `$50\%$` of the population has a value of `$Y$` lower or equal to `$y_{0.5}$`.

.pull-left[
![](01_Basics_files/figure-html/unnamed-chunk-8-1.png)
]

.pull-right[
.blue[Example of computation] Number of Children in Population 1

- Collect data `$\{1, 0, 3, 2, 1, 0, 4, 2\}$`
- Reorder data from lowest to highest `$\{0, 0, 1, 1, 2, 2, 3, 4\}$`
- Value 1 splits the population into 2 groups of equal size
]

.center[
The .blue[median] splits the population into two parts of (roughly) .alert[equal size].
]

---
### Understanding medians

.pull-left[
Compute the median of:
- P1: `$\{0, 0, 1, 1, 2, 2, 3, 4\}$`

- P2: `$\{0, 0, 1, 1, 1, 2, 2, 3, 4\}$`

- P3: `$\{0, 0, 1, 1, 1, 2, 2, 3, 4, 5, 6\}$`

.question[Quizz 4]
]

.pull-right[
.small[

```r
c(0, 0, 1, 1, 2, 2, 3, 4) |> median()
```

```
## [1] 1.5
```

```r
c(0, 0, 1, 1, 1, 2, 2, 3, 4) |> median()
```

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

```r
c(0, 0, 1, 1, 1, 2, 2, 3, 4, 5, 6) |> median()
```

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

.center[
.alert[Beware], computers try to find the midpoint instead. 
]

---
### Understanding medians (II)

.question[Quizz 5]

.pull-left[
![](01_Basics_files/figure-html/unnamed-chunk-10-1.png)
]

.pull-right[
![](01_Basics_files/figure-html/unnamed-chunk-11-1.png)
]

---
# Quantiles

The .blue[quantile] `$y_{\alpha}$` for distribution of variable `$Y$` is the smallest value such that at least `$\alpha\times100\%$` of the population has a value of `$Y$` lower or equal to `$y_{\alpha}$`. In particular

`$$\begin{equation}
P(Y \leq y_\alpha) \geq \alpha \quad \text{and} \quad P(Y < y_\alpha) \leq \alpha
\end{equation}$$`

The quantile `$y_{\alpha}$`  splits the population into two parts of sizes (roughly equal to) `$\alpha\times100\%$` and `$(1-\alpha)\times100\%$`.

---
# Expectation versus Median

Number of children per individual in 2 populations:
- Population 2: `$\{1,3,2,2,3,4,1,1,2,3,1,3\}$`
- Population 3: `$\{0,3,4,4,3,4,0,0,0,4,0,3,0,4,3\}$`
- Population 4: `$\{1,3,2,2,3,4,1,1,2,3,1,8\}$`

`$$\begin{array}{c||c|c|c|c|c|c|c|}
        & \text{Size}  & \text{Mean}  & \text{Variance}      & \text{Std Dev.}  & C_{20\%}   & \text{Median}    & C_{80\%}    \\
\hline
Pop.2   & 12    & 2.16  &     0.97      & 0.98      &              &          &               \\
Pop.3   & 15    & 2.13  &     3.18      & 1.78      &              &          &               \\
Pop.4   & 12    & 2.58  &     3.57      & 1.89      &              &          &               \\
\hline
\end{array}$$`

.def[Remark 1:] Outliers have less influence on median than on expectation. More generally, quantiles are **robust** to outliers.

.def[Remark 2:] Usually more than 1 descriptor (mean, variance, median...) is requested to accurately describe a population.

---
### Graphical representation: Continuous variable

Height in Population 1

.pull-left-70[
<img src="01_Basics_files/figure-html/unnamed-chunk-12-1.png" width="1000px" style="display: block; margin: auto;" />
]

.pull-right-30[
- Histogram: values of variable `$Y$` in abscisses, counts in ordinates.

- Box-plot: displays `$H_{25\%}$`, `$H_{50\%}$` and `$H_{75\%}$`, outliers (if any) and the `$1.8\times IQR$` range in ordinates
]

.center[
And **no general convention for boxplots...**
]

---
## Cumulative distribution function

For a quantitative variable `$Y$` with probability distribution `$P$`, the cdf is noted `$F$` and defined as follows:
`$$F(y) = P(Y\leq y).$$`

The definition is `$F(y) = P(Y\leq y)$` and **not** `$F(y) = P(Y < y)$`.

---
## Probability distribution vs CDF

Providing the CDF is equivalent to providing the probability distribution function !

Compute the following:

.pull-left[ 
- `$P(C = 4)$` 
- `$P(C = 2)$` 
- `$P(1 \leq C \leq 3)$`
]
--

.pull-right[
- `$P(C=4)={0.33 = 5/15}$`
- `$P(C=2)={0}$`
- `$P(1\leq C\leq3)={0.26=4/15}$`
]

---
class: middle, inverse, center

# A detailed example

---
class: center
# The summer school dataset

```
## # A tibble: 6 × 5
##   Height Weight   Age Year  Gender
##    <dbl>  <dbl> <int> <chr> <chr> 
## 1   160.   55.4    23 3A    F     
## 2   168.   52.9    23 3A    F     
## 3   165.   50.7    22 3A    F     
## 4   164.   62.6    19 1A    F     
## 5   186.   66.5    20 2A    M     
## 6   176.   59.6    20 1A    M
```

---
## Univariate analysis

.blue[Qualitative variables:]

`$\star$` Gender

```
##          F     M
## Nb   362.0 369.0
## Freq   0.5   0.5
```

`$\star$` Age

```
##          19     20     21     22    23
## Nb   252.00 196.00 139.00 100.00 44.00
## Freq   0.34   0.27   0.19   0.14  0.06
```

---
# Univariate analysis

.blue[Quantitative variables:]

---
## Univariate analysis

.blue[Cumulative distribution function]

---
class: middle, inverse, center

# Infinite population

---
## Case I: qualitative variable

.blue[Coin tossing]
Each coin tossing is a 0/1 experiment:

`$$X = \begin{cases}
1 & \text{if  tail is obtained} \\
0 & \text{otherwise}
\end{cases}$$`

How can we describe the population of **all** coin tossing trials ?

Let `$n_T$` and `$n_H$` be the number of tails (resp. heads) after `$n$` trials. Then

`$$\begin{equation}
P(X = 1) = \underset{n\rightarrow\infty}{\lim}\frac{n_{T}}{n} \text{ and }  P(X = 0) = \underset{n\rightarrow\infty}{\lim}\frac{n_{H}}{n} \ \ .
\end{equation}$$`

Note that by definition `$n_T+n_H=n \Leftrightarrow n_T/n + n_H/n =1$`
`$$\Rightarrow P(X = 0)+P(X = 1)=1 \ \ .$$`

---
## Example 1: Coin tossing

The population of coin tossing trials (or alternatively the result of a coin tossing trial) can be
.blue[modeled] using the .alert[Bernoulli] distribution:

`$$X \hookrightarrow \mathcal{B}(p).$$`

.blue[Probability distribution]
`$$P(X=x) = p^x(1-p)^{1-x} = \begin{cases} p & \text{if } x = 1 \\ 1 - p & \text{if } x = 0 \end{cases}$$`

.blue[Expectation and variance] 
--
`$$E(X)=p,\quad V(X)=p(1-p)$$`

---
## Example 2: Cumulated coin tossing
Now consider the cumulated  result of `$K$` coin tossing trials:

`$$X \in \{0,...,K\}$$`

One can define

`$$P(X=x)\ =\ \underset{n\rightarrow \infty}{\lim}\frac{n_x}{n}\ \overset{def}{=}\ p_x,\quad \forall x\in\{0,...,K\}.$$`
--

Alternatively,  assuming .blue[independence between trials] and .blue[identical "tail" probabilities], one has

`$$P(X=x)={x \choose K} p^x(1-p)^{K-x}$$`
which corresponds to the .blue[Binomial probability distribution]:

`$$X\hookrightarrow \mathcal{B}(K,p)$$`
--
One easily check that
`$$\sum_xP(X=x)=1,\quad E(X)=Kp,\quad V(X)=Kp(1-p)$$`

---
## Case II: quantitative discrete variable

.blue[Light bulb lifetime]

Denote `$X$` the lifetime of a bulb, measured in hours.
How can we describe the population of **all** bulbs ?

Let `$n_x$` be the number of bulbs that last `$x$` hours over `$n$` bulbs. Then
`$$\begin{equation}
P(X = x) = \underset{n\rightarrow\infty}{\lim}\frac{n_{x}}{n}
\end{equation}$$`

--
As for the previous example, one has:
$$
\sum_{x=1}^{\infty} P(X = x) = 1 \ \ .
$$

---
## Example: Light bulb lifetime

.pull-left[
Assume one observed a sample of bulb lifetimes:

166, 143, 150, 136, 148,
168, 148, 174,...

]

.pull-right[
![](01_Basics_files/figure-html/unnamed-chunk-22-1.png)
]

---
## Example: Light bulb lifetime

The light bulb population distribution can **modeled**
using the Poisson distribution: `$L \hookrightarrow \mathcal{P}(\lambda)$`

.pull-left-70[
<img src="01_Basics_files/figure-html/unnamed-chunk-24-1.png" style="display: block; margin: auto;" />
]

.pull-right-30[
.blue[Probability distribution]

`$$P(X=x)= \begin{cases} \frac{e^{-\lambda}\lambda^x}{x!} & \text{if } x \in \mathbb{N} \\ 0 & \text{Otherwise} \end{cases}$$`
]

---
## Example II: Light bulb lifetime

Expectation and variance of a random variable with Poisson distribution can be theoretically obtained as follows:

.pull-left[
.blue[Expectation]
$$
E(X) = \lambda
$$
]

.col-right[
.blue[Variance]
$$
V(X) = \lambda
$$
]
--

.pull-left[
`$$\begin{align}
E(X) &= \sum_0^{+\infty} x\times P(X=x) \\
&= \sum_1^{+\infty} x\times\frac{e^{-\lambda} \lambda^x}{x!} \\
&= \sum_1^{+\infty}  \frac{e^{-\lambda} \lambda^x}{(x-1)!}\\
&= \lambda \sum_0^{+\infty}  \frac{e^{-\lambda} \lambda^x}{x!} = \lambda
\end{align}$$`
]

.col-right[
`$$\begin{align}
E[X(X-1)] &= \sum_0^{+\infty} x(x-1)\times P(X=x) \\
&= \sum_2^{+\infty} x(x-1)\times\frac{e^{-\lambda} \lambda^x}{x!} \\
&= \sum_2^{+\infty}  \frac{e^{-\lambda} \lambda^x}{(x-2)!}\\
&= \lambda^2 \sum_0^{+\infty}  \frac{e^{-\lambda} \lambda^x}{x!} = \lambda^2
\end{align}$$`
]

---
## Example III: Continuous variable

.blue[Deviance to target]

Denote `$X$` the deviance to target observed for a given trial in a shooting contest.
How can we describe the population of **all** deviances ?

Deviance is assumed to be measured with **absolute** precision:

`$x=3$` means that the deviance is exactly 3.0000000... cm.

Let `$n_x$` be the number of trials that deviated from target with deviance `$x$` in a sample of `$n$` trials. One has:
$$
P(X = x) = \underset{n\rightarrow\infty}{\lim}\frac{n_{x}}{n} = 0 \text{!!!}
$$

.alert[But]

`$\star$` Some values seem more likely than others,

`$\star$` The probability of a range of values is not 0: `$P(x_1<X<x_2)>0$`

---
## A short illustration of density
.blue[Mass density function]

`$\star$` Consider a **homogeneous** stick of length 1m and weight 0.7kg.

.pull-left[
`$$\begin{array}{lll}
\text{Interval (m)} & \text{Density (kg/m)} & \text{Mass (kg)} \\
[0,1]   & 0.7  & 0.7  \\
[0,0.5] & 0.7 &  0.35 \\
[0,0.25] & 0.7 & 0.175  \\
[x,x+dx] & 0.7 & 0.7dx \\
\end{array}$$`
]

.pull-right[
.center[![](Figures/SegmentF-2.png)]
]

`$\star$`  Consider an **inhomogeneous** stick of length 1m and weight 0.7kg.
One needs to provide at each point `$x$` the density `$f(x)$`, that may differ from an `$x$` to another:

`$$\begin{array}{lll}
\text{Interval (m)} & \text{Density (kg/m)} & \text{Mass (kg)} \\
[0,1]   & 0.7  & 0.7  \\
[x,x+dx] & f(x) &  f(x)dx \\
\end{array}$$`

---
## A more formal definition

.blue[Probability density function]
The density function is defined at point `$x$` as

$$
\begin{align}
f(x)dx &= P(x \leq X \leq x+dx) \\
\Rightarrow f(x) &=& \frac{P(x \leq X \leq x+dx)}{dx}.
\end{align}
$$

If `$X$` takes its values in interval `$[a,b]$`, then one can "sum" (integrate) probabilities associated with all possible values of `$X$` to obtain:

$$
\int_a^b f(x)dx = P(a \leq X \leq b) = 1
$$

From this, the expectation and variance for a quantitative continuous variables are defined as:

`$$\begin{align}
E(X) &= \int x \times f(x)dx \\
V(X) &= \int (x-E(X))^2 \times f(x)dx
\end{align}$$`

And in general, for any numeric function `$g$` (such as `$g(x) = x$` for the mean and `$g(x) = (x - E[X])^2$` for the variance)

$$
E\left[ g(X) \right] = \int g(x) \times f(x)dx
$$

The expectations corresponds to

---
## Back to example III: Deviance to target

The population of deviances in a shot trial can be **modeled** using the Normal distribution:
$$
X \hookrightarrow \mathcal{N}(\mu,\sigma^2),\ \text{ with } \mu=0
$$

.pull-left[
<img src="Figures/Normale.png" width="399" />
]

.pull-right[
.blue[Probability density function]
`$$f(x)=\frac{1}{\sqrt{2\pi}\sigma} \exp^{-\frac{(x-\mu)^2}{2\sigma^2}}$$`
.blue[Expectation and variance]
$$ E(X)=\mu \ \ \text{ and } \ \ V(X)=\sigma^2 $$
]

---
## Quantiles for continuous variables

The definition of quantile `$x_\alpha$` does not change: minimum value such that
`$$P(X\leq x_\alpha)\geq \alpha$$`

.blue[Interpretation] in terms of integration

Minimum value `$x_\alpha$` such that
`$$\int_{-\infty}^{x_\alpha}f(x)dx\geq \alpha$$`
--

.blue[Remark] Note that both the pdf and the cdf are characteristic to the probability distribution.

---
## Illustration: quantiles for the `$\mathcal{N}(0,1)$` distribution

.pull-left[
![](Figures/QuantilesNormaleCentreeReduite_Extrait.png)

]

.col-right[
Compute 
- `$P(X\leq 1.05)$` 
- `$P(X\geq -1)$`
- `$x_{0.75}$`
- `$x_{0.1}$`
- `$P(-1\leq X\leq 1)$`
]

.pull-right[
- `$P(X\leq 1.05)={0.853}$`
- `$P(X\geq -1)={P(X\leq1)=0.841}$`
- `$x_{0.75}={0.67}$`
- `$x_{0.1}={-x_{0.9}=-1.28}$`
- `$P(-1\leq X\leq 1)={P( X\leq 1)-P( X\leq-1)=0.682}$`
]

---
## Finite vs infinite population: sampling

.blue[Sampling in a finite population]
Consider a population of 3 individuals: 1 male and 2 female.

`$\star$` Probability of sampling 1 male ?

$$
P(G_1=m) = 1/3
$$

`$\star$` Probability of sampling 2 males ? .question[Quizz6]

$$
P(G_1=m\cap G_2=m) = P\left(G_2=m | G_1=m\right)P\left(G_1=m\right) = 0
$$

.blue[Conclusion]

`$\star$` `$G_1$` drawn according to the distribution of the population,

`$\star$` `$G_2$` given `$G_1$` **not** drawn according to the distribution of the population. The second draw distribution depends on the first draw: .alert[no independence between samplings]

---
## Finite vs infinite population: sampling

.blue[Sampling in an infinite population]
Consider an infinite population of males and females, with proportions 1/3 and 2/3.

`$\star$` Probability of sampling 1 male  ?

`$$P(G_1=m) = 1/3$$`

`$\star$` Probability to sample 2 males ?

`$$P(G_2=m\cap G_1=m) = P\left(G_2=m | G_1=m\right)P\left(G_1=m\right) = (1/3)^2$$`

--
.blue[Conclusion]
  
`$\star$` `$G_1$` drawn according to the distribution of the population,

`$\star$` `$G_2$` **also** drawn in the distribution of the population.

This leads to the so-called **"i.i.d" assumption**: all observations are  **i**ndependent and **i**dentically **d**istributed.

---
## Exercise 1: Tennis

Assume that a tennisman play games that last

- 3 sets in 60% of the games,
- 4 sets in 25% of the games,
- 5 sets in 15% of the games.

- What is the expectation of number of played set for this player ?
- What is the variance of number of played set for this player ?

.blue[Answer:] `$E(N)=3.55$`,  `$V(N)=0.547$`

---
## Exercise 2: The Pepsi factory

A Pepsi factory machine builds cans of height 115mm, with a standard deviation of 0.4mm. The Pepsi specification requires all cans lower than 114mm or 116mm to be withdrawn.

- Assuming the distribution of the can height is normal, what proportion of cans meets the specification ?
- Assume the machine get disrupted such that produced cans have height 114.5 on average. What proportion of cans meets the specification ?

.blue[Answer:]

- Q1: 0.987
- Q2: 0.894

---
## Exercise 3: properties of expectation and variance

Assume that variable `$X$` as expectation `$\mu$` and variance `$\sigma^2$`.
Let `$a$` and `$b$` be two constants.
Provide the expression of

- `$E(aX+b)$`
- `$V(aX+b)$`

.blue[Answer:]
- `$E(aX+b) = a\mu+b$`
- `$V(aX+b) = a^2\sigma^2$`

Proof in lecture 3.

---
## Exercise 4: Moments

Compute `$E(X^2)$` in the following cases:
- `$X\sim \mathcal{B}(p)$`
- `$X\sim \mathcal{N}(\mu,\sigma^2)$`
- `$X\sim \mathcal{E}(\lambda)\quad\quad$`

.blue[Answer:] 
- `$p$`,  
- `$\sigma^2+\mu^2$`, 
- `$\lambda(\lambda+1)$`

---

## Exercise 5: Dice

Consider `$X$` the value obtained when rolling a fair dice.
Provide the expression of

- `$E(X)$`
- `$E[\|X-E(X)\|]$`

.blue[Answer:]
- `$E(X) = 3.5$`
- `$E[\|X-E(X)\|]=1.5$`

---
## Summary

- In statistics, characterizing a population amounts to providing a .alert[probability distribution] (qualitative/discrete case) or a **probability density function** (continuous case).

- Alternatively, one can provide the **cumulative distribution function**.

- Some "classical" theoretical distributions allow one to **model** different populations.

- These theoretical populations are usually **parametric**, i.e. a few parameters rule the shape of the distribution.

- **Expectation** and **variance** can be expressed in terms of the distribution parameters.