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IB Mathematics AI SL Concept of discrete random variables and their probability MAI Study Notes - New Syllabus

IB Mathematics AI SL Concept of discrete random variables and their probability MAI Study Notes

LEARNING OBJECTIVE

  • Concept of discrete random variables and their probability distributions.

Key Concepts: 

  • Discrete random variables
  • E (X) for discrete data.

MAI HL and SL Notes – All topics

PROBABILITY DISTRIBUTION OF A RANDOM VARIABLE X

A random variable \( X \) takes on some values in a given domain at random!!! It may be A discrete variable takes on values in a finite or numerable set, while a continuous variable takes on values in some interval(s).

DISCRETE RANDOM VARIABLE

Let \( X \) be a variable which takes on the values
$ 10, 20, 30 $
with probabilities
$ 0.2, 0.3, 0.5 $
respectively. We often use a table:

$\rm  x$$10$$20$$30$
$\text{P(X=x)}$$0.2$$0.3$$0.5$

1. All the probabilities are non-negative numbers; and
2. Their sum is always 1.

Then we say that \( X \) is a discrete random variable.

To express that the probability that \( X=10 \) is 0.2, we write:
$ P(X=10) = 0.2 $
Similarly, \( P(X=20) = 0.3 \) and \( P(X=30) = 0.5 \).

In general, for a discrete random variable \( X \) with:

$ \begin{array}{c|cccc}
X & x_1 & x_2 & x_3 & \cdots \\
\hline
P(X=x) & p_1 & p_2 & p_3 & \cdots \\
\end{array} $

it holds:
1. \( p_i \geq 0 \), for all \( i \).
2. \( \sum p_i = 1 \), i.e., \( p_1 + p_2 + p_3 + \cdots = 1 \).

We write:
$ P(X=x_1) = p_1, \quad P(X=x_2) = p_2, \quad \text{and so on.} $
(We also say that a probability function \( p: x_i \mapsto y_i \) is defined).

◆ THE EXPECTED VALUE \( \mu = E(X) \)

The mean \( \mu \) or otherwise the expected value \( E(X) \) is defined by:
$ E(X) = \sum x_i p_i = x_1 p_1 + x_2 p_2 + x_3 p_3 + \cdots $

For our example:

$ \begin{array}{c|ccc}
X & 10 & 20 & 30 \\
\hline
P(X=x) & 0.2 & 0.3 & 0.5 \\
\end{array} $

the expected value (otherwise the mean) is:
$ E(X) = 10 \times 0.2 + 20 \times 0.3 + 30 \times 0.5 = 23 $

NOTICE: Explanation for \( \mu = E(X) \)

In fact, the mean here is not different than the mean in statistics.

Consider the following ten numbers:
$ 10, 10, 20, 20, 20, 30, 30, 30, 30, 30 $

The probabilities to select 10, 20, or 30 are as in the table above.

The mean in statistics is also:
$ \mu = \frac{10 \times 2 + 20 \times 3 + 30 \times 5}{10} = 10 \times \frac{2}{10} + 20 \times \frac{3}{10} + 30 \times \frac{5}{10} = 23 $

Example

Consider:

$\rm  x$$10$$20$$30$
$\text{P(X=x)}$$a$$b$$0.5$

Given that \( E(X)=23 \), find the values of \( a \) and \( b \).

▶️Answer/Explanation

Solution:

Solution

We use two relations:
1. \( a + b + 0.5 = 1 \quad \Rightarrow \quad a + b = 0.5 \)
2. \( 10a + 20b + 30 \times 0.5 = 23 \quad \Rightarrow \quad 10a + 20b = 8 \)

The solution of the system is \( a = 0.2 \) and \( b = 0.3 \).

The probability distribution applies in many betting games:

Example

Consider again the same table above. But now we select one of the numbers 10, 20, 30 at random.

If we select 10, we earn 6 points.
If we select 20, we earn 1 point.
 If we select 30, we lose 2 points.

What is the expected number of points in one game?

▶️Answer/Explanation

Solution:

Expected profit:

$ \text{Expected profit} = 6 \times 0.2 + 1 \times 0.3 – 2 \times 0.5 = 0.5 $

That is, in each game we earn 0.5 points on average.

◆ MEDIAN-MODE

These measures, known from statistics, are defined analogously:

MODE = The value \( X=a \) of the highest probability.
MEDIAN = The value \( X=m \) where the probability splits in two equal parts (0.5-0.5).

Example

Consider again the probability distribution:

$\rm  x$$10$$20$$30$
$\text{P(X=x)}$$0.2$$0.3$$0.5$

Find Mode and median.

▶️Answer/Explanation

Solution:

Mode

 $P(X = 10) = 0.2$
 $P(X = 20) = 0.3$
 $P(X = 30) = 0.5$

The highest probability is 0.5, which occurs at $x = 30$, so:
$\rm{\text{Mode} = 30}$

Median

Cumulative probabilities
$P(X \leq 10) = 0.2$
$P(X \leq 20) = 0.2 + 0.3 = 0.5$
$P(X \leq 30) = 0.2 + 0.3 + 0.5 = 1.0$

The median is the smallest $x$ such that $P(X \leq x) \geq 0.5$.
Since $P(X \leq 20) = 0.5$, the median is:

$\rm{\text{Median} = 20}$

◆ VARIANCE (Only for HL )

We define:
$ \text{Var}(X) = E(X-\mu)^2 $
that is:
$ \text{Var}(X) = (x_1-\mu)^2 \times p_1 + (x_2-\mu)^2 \times p_2 + (x_3-\mu)^2 \times p_3 + \cdots $

An equivalent definition is:
$ \text{Var}(X) = E(X^2) – \mu^2 $
where:
$ E(X^2) = x_1^2 \times p_1 + x_2^2 \times p_2 + x_3^2 \times p_3 + \cdots $

Example

Consider again the probability distribution:

$\rm  x$$10$$20$$30$
$\text{P(X=x)}$$0.2$$0.3$$0.5$

Using first and second both definition Find $ \text{Var}(X) $

▶️Answer/Explanation

Solution:

We have seen that \( \mu = E(X) = 23 \).

1. Using the first definition:
$ \text{Var}(X) = (10-23)^2 \times 0.2 + (20-23)^2 \times 0.3 + (30-23)^2 \times 0.5 = 61 $

2. Using the second definition:
$ E(X^2) = 10^2 \times 0.2 + 20^2 \times 0.3 + 30^2 \times 0.5 = 590 $
$ \text{Var}(X) = 590 – 23^2 = 61 $

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