Joint_Prob_for_independent_continuous_variables.qmd

---
title: "Joint Probability for iid Random Variables"
format: gfm
editor: visual
---


# Problem


$$
\begin{align}
& \text{Prove that if} \quad X_1, X_2, \dots, X_n \overset{iid}{\sim} F(X), \quad \text{and}\\
& X_i ~~\text{is continuous variable}, \quad i \in (1, 2, \dots, n), \\
& \implies \mathcal{P}(X_{k_1} \leq X_{k_2} \leq X_{k_3} \leq \cdots \leq X_{k_n}) = \frac{1}{n!}, \\
& \text{and} \quad k_i \in (1, 2, 3, \dots, n), \quad k_i \ne k_j, ~~ \forall ~~ i, j \in  (1, 2, 3, \dots, n), ~~ i \ne j.
\end{align}
$$



\ 
\ 

#### We will prove this relation for a specific case, we will generalize it for other cases.

\
\


## First Way

$$
\begin{aligned}
& \text{if}\quad \langle k_1, k_2, k_3, \dots, k_n\rangle  = \langle 1, 2, 3, \dots, n \rangle \implies \\
& \mathbb{P} = \mathcal{P}(X_{k_1} \leq X_{k_2} \leq X_{k_3}\leq \cdots \leq X_{k_n}) = \mathcal{P}(X_1 \leq X_2 \leq X_3 \leq \cdots \leq X_n), \\
& \mathbb{P} = \int_{-\infty}^{+\infty}\int_{x_1}^{+\infty}\int_{x_2}^{+\infty}\int_{x_3}^{+\infty}\cdots\int_{x_{n-1}}^{+\infty}\prod_{i = 1}^n f(x_i)~dx_ndx_{n-1}dx_{n-2}\cdots dx_2dx_1 = \\
& \int_{-\infty}^{+\infty}f(x_1)\int_{x_1}^{+\infty}f(x_2)\int_{x_2}^{+\infty}f(x_3)\int_{x_3}^{+\infty}f(x_4)\cdots\int_{x_{n-1}}^{+\infty}f(x_n)dx_{n-1}dx_{n-2}\cdots dx_2dx_1 = \\
& \int_{-\infty}^{+\infty}f(x_1)\int_{x_1}^{+\infty}f(x_2)\int_{x_2}^{+\infty}f(x_3)\int_{x_3}^{+\infty}f(x_4)\cdots\int_{x_{n-1}}^{+\infty}\frac{d}{dx_n}F(x_n)dx_{n-1}dx_{n-2}\cdots dx_2dx_1, \\
\end{aligned}
$$




$$
\begin{aligned}
& \text{We know that if}~~X_i ~~\text{is continuous Variable}\implies F(X_i)\sim \text{Unif}(0, 1) \implies \\
& \mathbb{P} = \int_{-\infty}^{+\infty}f(x_1)\int_{x_1}^{+\infty}f(x_2)\cdots
\int_{x_{n-1}}^{+\infty}dF(x_{n})dx_{n-1}\cdots dx_2dx_1 = \\
& \int_{x_{n-1}}^{+\infty}f(x_1)\int_{x_2}^{+\infty}f(x_2)\cdots
\int_{x_{n-2}}^{+\infty}f(x_{n-1})(1 - F(x_{n-1}))dx_{n-1}\cdots dx_2dx_1 = \\
& \int_{x_{n-1}}^{+\infty}f(x_1)\int_{x_2}^{+\infty}f(x_2)\cdots
\int_{x_{n-2}}^{+\infty}\frac{d}{dx_{n-1}}F(x_{n-1})(1 - F(x_{n-1}))dx_{n-1}\cdots dx_2dx_1 = \\
& \int_{-\infty}^{+\infty}f(x_1) \int_{x_1}^{+\infty} f(x_2)\cdots \int_{F(x_{n-2})}^1 (1 - F(x_{n-1}))dF(x_{n-1})\cdots dx_1 =\\
& \int_{-\infty}^{+\infty}f(x_1)\int_{x_1}^{+\infty}\int\cdots\int_{x_{n-3}}^{+\infty}\frac{1}{2} - (F(x_{n-2}-\frac{F(x_{n-2})^2}{2}))dx_{n-2}\cdots dx_1, 
\end{aligned}
$$

##### as before


$$
\begin{aligned}
& \mathbb{P} = \int_{-\infty}^{+\infty}f(x_1)\int_{x_1}^{+\infty}\int\cdots\int_{x_{n-4}}^{+\infty}\frac{1}{3!} - (\frac{F(x_{n-3})}{2}-\frac{F(x_{n-3})^2}{2}+\frac{F(x_{n-3})^3}{3!})dF(x_{n-3})\cdots dx_1 = \\
& \int_{-\infty}^{+\infty}f(x_1)\int_{x_1}^{+\infty}\int\cdots\int_{x_{n-5}}^{+\infty}\frac{1}{4!} - (\frac{F(x_{n-4})}{3!}-\frac{F(x_{n-4})^2}{2\times 2!}+\frac{F(x_{n-3})^3}{3!} - \frac{F(x_{n-4})^4}{4!})dF(x_{n-4})\cdots dx_1, 
\end{aligned}
$$


##### If we continue in the same way, in the following, we will reach an expression set in the form below in each time of integration;


$$
\mathbb{P} = \int_{-\infty}^{+\infty} f(x_1) \int_{x_1}^{+\infty}f(x_2) \cdots \int_{x_{n-k}}^{+\infty}g(F(x_{n-(k-1)}), k)dF(x_{n-(k-1)})\cdots dx_1 \to 
$$


##### that `g` function is a polynomial function of degree `K`

\


###### form of `g` function


$$
\begin{aligned}
& \int g(x, 1)~dx = x, \\
& \int g(x, 2)~dx = x - \frac{x^2}{2}, \\
& \int g(x, 3)~dx = \frac{x}{2}-\frac{x^2}{2} + \frac{x^3}{3!}, \\
& \int g(x, 4)~dx = \frac{x}{3!} -\frac{x^2}{2 \times 2!} + \frac{x^3}{3!} - \frac{x^4}{4!}, \\
& \int g(x, 5) ~ dx = \frac{x}{4!} -\frac{x^2}{2 \times 3!} + \frac{x^3}{2 \times 3!} - \frac{x^4}{4!} + \frac{x^5}{5!}, \\
& \int g(x, 6) ~ dx = \frac{x}{5!} -\frac{x^2}{2 \times 4!} + \frac{x^3}{3 \times 3!} - \frac{x^4}{2 \times 4!} + \frac{x^5}{5!} - \frac{x^6}{6!}, \\
& \int g(x, 7) ~ dx = \frac{x}{6!} -\frac{x^2}{2 \times 5!} + \frac{x^3}{4! \times 3!} - \frac{x^4}{3! \times 4!} + \frac{x^5}{2 \times 5!} - \frac{x^6}{6!} + \frac{x^7}{7!}, \\
& \vdots \\
& \int g(x, k)~ dx = \sum_{i = 1}^k (-1)^{i + 1} \times \frac{x^i}{i! \times (k - i)!} \implies 
\end{aligned}
$$


\ 

$$
\mathbb{P} = \int_{-\infty}^{+\infty} g(F(x_1), n)~dx_1 = \sum_{i = 1}^n (-1)^{i + 1} \times \frac{F(x_1)^i}{i! \times (k-i)!}|_{-\infty}^{+\infty} 
$$


###### We Know That 


$$
F(+\infty) = 1, \quad F(-\infty) = 0 \implies 
$$

\ 

$$
\begin{aligned}
& \mathbb{P} = \sum_{i = 1}^n (-1)^{i + 1} \frac{1}{i! \times (n-i)!} = \\
& \frac{1}{n!} \times \sum_{i = 1}^n (-1)^{i + 1} \frac{n!}{i! \times (n-i)!} = \\
& \frac{1}{n!} \sum_{i = 1}^n {n \choose i} (-1)^{i + 1} = \\
& \frac{-1}{n!} \sum_{i = 1}^n {n \choose i} (-1)^{i} = \\
& \frac{-1}{n!} \sum_{i = 1}^n {n \choose i} (-1)^{i} \times (1)^{(n - i)} = \\
& \frac{-1}{n!} ((-1 + 1)^n - (-1)^0) = \\
& \frac{\textbf{1}}{\textbf{n!}}
\end{aligned}
$$



***



## Second Way

**Lemma (1):**

\ 
\  

$$
\begin{align}
& \text{if} \quad X_1, X_2, \dots, X_n \overset{iid}{\sim} F(X), \quad \quad \text{(i)} \\
& \text{and}\quad  X_i ~~\text{is continuous variable}, \quad i \in (1, 2, \dots, n), \\
& \implies \mathcal{P}(X_i = X_{(1)}) =  \\
& \mathcal{P}(X_1 \leq X_{r_1} \leq X_{r_2} \leq \cdots \leq X_{r_{n-1}}) = \frac{1}{n}, \quad \quad \text{(ii)}\\
& \text{and} \quad r_i \in (2, 3, \dots, n), \quad r_i \ne r_j, ~~ \forall ~~ i, j \in  (2, 3, \dots, n-1), ~~ i \ne j, \\
&\text{Such that} ~~ X_{(1)} ~~ \text{Is The lowest value in the sample}.
\end{align}
$$



**Proof:**

\ 
\ 

$$
\begin{aligned}
& A_1:\quad X_1 = X_{(1)}, \\
& A_2:\quad X_2 = X_{(1)}, \\
& A_3:\quad X_3 = X_{(1)}, \\
& \vdots \\
& A_{n-1}:\quad X_{n-1} = X_{(1)}, \\
& A_n:\quad X_n = X_{(1)}, \\
& \text{Based (i)}: \quad \mathcal{P}(A_1) = \mathcal{P}(A_2) = \cdots = \mathcal{P}(A_n), \quad \text{(iii)}\\
& \text{and} \quad \mathcal{P}(A_i \cap A_j) = 0, \quad \forall~~ i, j \in (1, 2, ..., n),~~i \ne j\quad \text{(iv)} \\
& \text{It is quite clear that:} \\
& \bigcup_{i = 1}^n A_i = \Omega, \\
& \Omega: \quad \text{Total Sample Space} \implies \\
& \mathcal{P}(\bigcup_{i = 1}^n A_i) = \mathcal{P}(\Omega) = 1, \\
& \text{Based (iv):}\quad \mathcal{P}(\bigcup_{i = 1}^n A_i) = \sum_{i=1}^n \mathcal{P}(A_i) = 1, \\
& \text{Based (i), (iii), (iv)}: \quad \mathcal{P}(A_i) = \frac{1}{n}, ~~i \in (1, 2, ..., n).
\end{aligned}
$$

\

\ 


$$
\begin{aligned}
& \mathbb{P} = \mathcal{P}(X_1 \leq X_2 \leq \cdots \leq X_n), \\
& \text{Based on the existing relationships in the conditional probability for the Combined Events:} \\
& \mathbb{P} = \mathcal{P}(X_1  = \min (X_1, X_2, X_3, \dots, X_n)) \times \\
& \mathcal{P}(X_2  = \min (X_1, X_2, X_3, \dots, X_n)  | X_1  = \min (X_1, X_2, X_3, \dots, X_n)) \times \\
& \mathcal{P}(X_3 = \min(X_1, X_2, X_3, X_4, ..., X_n) | X_1  = \min (X_1, X_2, X_3, \dots, X_n), X_2  = \min (X_2, X_3, \dots, X_n)) \times \\
& \cdots \times \mathcal{P}(X_{n-1} = \min(X_1, X_2, \dots, X_{n-1}, X_n) | \cdots) \times \mathcal{P}(X_n = \min(X_1, X_2, \dots, X_{n-1}, X_n)| \cdots) = \\
& \mathcal{P}(X_1  = \min (X_1, X_2, X_3, \dots, X_n)) \times \\
& \mathcal{P}(X_2  = \min (X_2, X_3, X_3, \dots, X_n)) \times \\
& \mathcal{P}(X_3  = \min (X_3, X_4, X_3, \dots, X_n)) \times \\
& \vdots \\
& \mathcal{P}(X_{n-1}  = \min (X_{n-1}, X_n)) \times \\
& \mathcal{P}(X_n = \min (X_n))
\end{aligned} 
$$


\ 
\ 

$$
\begin{aligned}
& \text{Based Lemma (1):}\\
& \mathbb{P} = \frac{1}{n} \times \frac{1}{n-1}\times \frac{1}{n-2}\times \cdots \times \frac{1}{2}\times \frac{1}{1} = \frac{1}{n!}
\end{aligned}
$$


***


\ 
\ 


## Experimental proof for n^th^ equal to 3 and 4 for Normal Standard Distribution


```{python}
import numpy as np
from scipy import stats
import pandas as pd
import math

def sim_prob(nsim = int(1e+4), Sid = 12, size_sample = 3):
    rvv = stats.norm.rvs(loc = 0, scale = 1, size = nsim * size_sample, 
    random_state = Sid).reshape(nsim, size_sample)
    def get_pr(x):
        y = x.copy()
        x.sort()
        return np.all(x == y)
    temp = np.apply_along_axis(axis = 1, arr = rvv, func1d  = get_pr)
    return temp.sum() / nsim
```


***




#### get result for  `n = 3`

```{python}
from IPython.display import HTML

nSize = [int(1e+2), int(1e+3), int(1e+4), int(1e+5), int(1e+6)]
temp2 = list(map(lambda x: sim_prob(nsim = x), nSize))
np.set_printoptions(suppress = True)
temp3 = np.stack([nSize, temp2, list(np.repeat(1/math.factorial(3), 5))], axis = 1)
res1 = pd.DataFrame(temp3, columns = ['Nsize_simulate', 'Prob_simulate', 
'exact_prob'])
HTML(res1.to_html(index=False))
```

***


#### get result for  `n = 4`


```{python}
temp2 = list(map(lambda x: sim_prob(nsim = x, size_sample = 4), nSize))
temp3 = np.stack([nSize, temp2, list(np.repeat(1/math.factorial(4), 5))], axis = 1)
res2 = pd.DataFrame(temp3, columns = ['Nsize_simulate', 'Prob_simulate', 
'exact_prob'])
HTML(res2.to_html(index=False))
```


***
***