Seminar 5¶
+import numpy as np
+import pandas as pd
+
+import scipy.stats as sts
+
+import IPython.display as dp
+import matplotlib.pyplot as plt
+import seaborn as sns
+
+dp.set_matplotlib_formats("retina")
+sns.set(style="whitegrid", font_scale=1.5)
+sns.despine()
+
+%matplotlib inline
+/var/folders/33/j0cl7y453td68qb96j7bqcj4cf41kc/T/ipykernel_1174/3109700056.py:10: DeprecationWarning: `set_matplotlib_formats` is deprecated since IPython 7.23, directly use `matplotlib_inline.backend_inline.set_matplotlib_formats()`
+ dp.set_matplotlib_formats("retina")
+
+plt.rc("figure", figsize=(20, 10))
+plt.rc("font", size=13)
+Recap¶
A certain company has $n + m$ employees, consisting of $n$ women and $m$ men. The company is deciding which employees to promote.
+-
+
- Suppose for this part that the company decides to promote $t$ employees, where $1 \leqslant t \leqslant n + m$, by choosing $t$ random employees (with equal probabilities for each set of $t$ employees). What is the distribution of the number of women who get promoted? +
- Now suppose that instead of having a predetermined number of promotions to give, the company decides independently for each employee, promoting the employee with probability $p$. Find the distributions of the number of women who are promoted, the number of women who are not promoted, and the number of employees who are promoted. +
Solution¶
-
+
- We are interested in the number of women $X$ in the set of $t$ promoted employees sampled from $n$ women and $n$ men. What is the distribution in question? +
-
+
- It is hypergeometric distribution $HGeom(n, m, t)$, so we know the answer +$$ + \mathbb{P}(X = k) = \frac{\begin{pmatrix}n\\k\end{pmatrix}\begin{pmatrix}m\\t-k\end{pmatrix}}{\begin{pmatrix}n+m\\t\end{pmatrix}} +$$ +
- If the company decides independently for each of $n$ women if they will be promoted with equal probabilities $p$, the number $Y$ of promoted women follows which distribution? +
-
+
- It is Binomial distribution $Bi(n, p)$, so we know the answer: +$$ +\mathbb{P}(Y = k) = \begin{pmatrix}n\\k\end{pmatrix} p^k (1 - p)^{n-k} +$$ +
CDFs¶
Reminder: the cumulative distribution function (CDF) is defined as +$$ +F_X(x) = \mathbb{P}(X <x) +$$
+It has the following properties:
+-
+
- $F_X$ is non-decreasing +
- $\lim\limits_{x\to-\infty}F_X(x) = 0$ +
- $\lim\limits_{x\to+\infty}F_X(x) = 1$ +
- $F_X$ if right-continuous +
Example 1¶
Draw PMF and CDF of $Bi(3, 0.7)$.
+my_binomial = sts.binom(n=3, p=0.7) # import scipy.stats as sts
+x = np.arange(-1, 5) # import numpy as np
+y = my_binomial.pmf(x)
+
+fig, ax = plt.subplots(1,2)
+ax[0].stem(x, y, label="PMF")
+for xx, yy in zip(x, y):
+ ax[0].text(xx, yy + 0.02, str(round(yy, 2)), horizontalalignment="center")
+ax[0].set_xlabel("x")
+ax[0].set_ylabel("PMF")
+ax[0].legend();
+
+y = my_binomial.cdf(x)
+
+ax[1].step(x, y, where="post", label="CDF")
+for xx, yy in zip(x, y):
+ ax[1].text(xx, yy + 0.02, str(round(yy, 2)), horizontalalignment="center")
+ax[1].set_xlabel("x")
+ax[1].set_ylabel("CDF")
+ax[1].legend();
+
+Example 2¶
A coin is tossed repeatedly until it lands Heads for the first time. Let $X$ be the number of tosses that landed Tails, and let $p$ be the probability of Heads of the coin. Find the distribution of $X$ and its PMF and CDF.
+Solution 2¶
$$ +\mathbb{P}(X = k) = \mathbb{P}(k \text{ tails}) \cdot \mathbb{P}(1 \text{ heads}) = (1 - p)^k p +$$
+Geometric distribution $X \sim Geom(p)$
+my_geometric = sts.geom(p=0.5)
+x = np.arange(-1, 10)
+y = my_geometric.pmf(x)
+
+fig, ax = plt.subplots(1,2)
+ax[0].stem(x, y, label="PMF")
+for xx, yy in zip(x, y):
+ ax[0].text(xx, yy + 0.02, str(round(yy, 2)), horizontalalignment="center")
+ax[0].set_xlabel("x")
+ax[0].set_ylabel("PMF")
+ax[0].legend();
+
+y = my_geometric.cdf(x)
+
+ax[1].step(x, y, where="post", label="CDF")
+for xx, yy in zip(x, y):
+ ax[1].text(xx, yy + 0.02, str(round(yy, 2)), horizontalalignment="center")
+ax[1].set_xlabel("x")
+ax[1].set_ylabel("CDF")
+ax[1].legend();
+
+Independence of random variables¶
Reminder: events $A$ and $B$ were called independent if +$$ +\mathbb{P}(A \cap B) = \mathbb{P}(A) \mathbb{P}(B) +$$
+Random variables $X$ and $Y$ are called independent if +$$ +\mathbb{P}(X \leqslant x, Y \leqslant y) = \mathbb{P}(X \leqslant x) \mathbb{P}(Y \leqslant y) +$$
+Example 3¶
You roll two dice. $X$ is the random result of first die, $Y$ is the random result of second die, $Z = X+Y$.
+How do we find the distribution of $Z$?
+$$ +\mathbb{P}(Z = k) = \sum_{m} \mathbb{P}(X = m) \mathbb{P}(Y = k - m) +$$
+-
+
- Are $X$ and $Y$ independent? +
Yes
+-
+
- Are $X$ and $Z$ independent? +
No
+Transformations of random variables¶
Random variables transform like functions, i.e. if $Y=\varphi(X)$, then +$$ +Y(\omega) = \varphi(X(\omega)) +$$
+Examples of transformations
+-
+
- Linear transformations of random variables and vectors $Y = aX + b$ +
- Non-linear invertible transformations of random variables $Y = g(X)$ +
- Sums $Y = X_1 + X_2$ +
If $g(\cdot)$ is one-to-one, it simplifies to: +$$ +\mathbb{P}(Y = g(x)) = \mathbb{P}(X = x) +$$
+The general formula: +$$ +\mathbb{P}(g(X) = y) = \sum\limits_{x \text{ such that } g(x) = y} \mathbb{P}(X = x) +$$
+Example 4¶
Let $X$ be a random variable with CDF $F_X$. Find CDF of $Y = a X + b$.
+Solution 4¶
If $a > 0$: +$$ +F_Y(y) = \mathbb{P}(Y \leqslant y) = \mathbb{P}(a X + b \leqslant y) = \mathbb{P}\left(X \leqslant \frac{y - b}{a}\right) +$$
+Example 5¶
Let $X \sim Bin(n, p)$. Find the PDF of $Y = \exp(X)$.
+Solution 5¶
So $g(x) = \exp(x)$, it's one-to-one and the inverse is $g^{-1}(x) = \log x$. +$$ +\mathbb{P}(Y = y) = \mathbb{P}(X = g^{-1}(y)) = \mathbb{P}(X = \log y) +$$
+Example 6¶
Let $X \sim Be(p)$. Find CDF of $Y = 2X$.
+Solution 6¶
$$ +F_Y = \mathbb{P}(Y \leqslant y) = \mathbb{P}(2 X \leqslant y) = \mathbb{P}\left(X \leqslant \tfrac{y}{2}\right) = F_X\left(\tfrac{y}{2}\right) +$$
+my_bernoulli = sts.bernoulli(p=0.6)
+x = np.arange(-2, 4)
+y = my_bernoulli.cdf(x)
+y2 = my_bernoulli.cdf(x / 2)
+
+fig, ax = plt.subplots()
+ax.step(x, y, where="post", label="CDF X")
+ax.step(x, y2, where="post", ls="--", label="CDF 2X")
+for xx, yy, yy2 in zip(x, y, y2):
+ ax.text(xx, yy + 0.02, str(round(yy, 2)), horizontalalignment="center")
+ ax.text(xx, yy2 + 0.02, str(round(yy2, 2)), horizontalalignment="center")
+ax.set_xlabel("x")
+ax.set_ylabel("CDF")
+ax.legend();
+
+Example 7¶
Let $X \sim Be(p)$. Find the distribution of $Y = X^2$.
+Solution 7¶
-
+
- $X \sim Be(p)$ +
- $X \in \{0, 1\}$ +
- $Y = X^2$ +
- $Y \in \{0, 1\}$ +
So we need to define PMF at 0 and 1:
+$$ +\begin{cases} +\mathbb{P}(Y = 0) = \mathbb{P}(X^2 = 0) = \mathbb{P}(X = \sqrt{0}) = \mathbb{P}(X = 0) = 1 - p \\ +\mathbb{P}(Y = 1) = \mathbb{P}(X^2 = 1) = \mathbb{P}(X = \pm \sqrt{1}) = \mathbb{P}(X = -1) + \mathbb{P}(X = 1) = 0 +p = p +\end{cases} +$$
+