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IB Mathematics AI AHL Linear transformation MAI Study Notes - New Syllabus

IB Mathematics AI AHL Linear transformation MAI Study Notes

LEARNING OBJECTIVE

  • Linear transformation of a single random variable

Key Concepts: 

  •  Expected value/Variance of linear combinations of n random variables. 
  • unbiased estimation

MAI HL and SL Notes – All topics

LINEAR TRANSFORMATION OF A SINGLE RANDOM VARIABLE

Let $X$ be a random variable, and let $a$ and $b$ be constants. A linear transformation of $X$ is defined as:

$
Y = aX + b
$

 Expected Value of a Linear Transformation

Definition: The expected value (mean) of a linear transformation preserves linearity.

$
E(aX + b) = aE(X) + b
$

The expectation is linear, meaning scaling and shifting behave predictably.

If $\mu_X = E(X)$, then $E(Y) = a\mu_X + b$

 Variance of a Linear Transformation

Definition: The variance of a linear transformation depends only on the scaling factor, not the constant term.

$
\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)
$

Shifting the distribution by $b$ does not affect the spread (variance).

If $\sigma_X^2 = \operatorname{Var}(X)$, then $\operatorname{Var}(Y) = a^2 \sigma_X^2$

Example 

Let \( X \) be a random variable with:

  • \( E(X) = 5 \)

  • \( \text{Var}(X) = 4 \)

Define a new random variable: \( Y = 3X – 7 \)

Find:

  1. \( E(Y) \)

  2. \( \text{Var}(Y) \)

▶️Answer/Explanation

Solution

1. Expected Value:

\( E(Y) = E(3X – 7) = 3E(X) – 7 = 3(5) – 7 = 15 – 7 = \boxed{8} \)

2. Variance:

\( \text{Var}(Y) = \text{Var}(3X – 7) = 3^2 \cdot \text{Var}(X) = 9 \cdot 4 = \boxed{36} \)

LINEAR COMBINATIONS OF N RANDOM VARIABLES

 Expected Value of Linear Combinations of $n$ Random Variables

Let $X_1, X_2, …, X_n$ be random variables with expected values $E(X_i) = \mu_i$, and let $a_1, a_2, …, a_n$ be constants.

Let:

$
Y = a_1X_1 + a_2X_2 + \cdots + a_nX_n
$

Expected Value:

$
E(Y) = E(a_1X_1 + \cdots + a_nX_n) = a_1E(X_1) + \cdots + a_nE(X_n)
$

Linearity of expectation applies even if the variables are dependent.

Variance of Linear Combinations of $n$ Random Variables

If the random variables $X_1, …, X_n$ are independent, then:

$
\operatorname{Var}(Y) = \operatorname{Var}(a_1X_1 + \cdots + a_nX_n) = a_1^2\operatorname{Var}(X_1) + \cdots + a_n^2\operatorname{Var}(X_n)
$

If the variables are not independent, covariances must be included:

$
\operatorname{Var}(Y) = \sum_{i=1}^{n} a_i^2 \operatorname{Var}(X_i) + 2 \sum_{i<j} a_i a_j \operatorname{Cov}(X_i, X_j)
$

Example 

Let \( X \) and \( Y \) be independent random variables with:

  • \( E(X) = 10,\ \text{Var}(X) = 9 \)

  • \( E(Y) = 4,\ \text{Var}(Y) = 1 \)

Define: \( Z = 2X – 3Y \)

Find:

  1. \( E(Z) \)

  2. \( \text{Var}(Z) \)

▶️Answer/Explanation

Solution

1. Expected Value:

\( E(Z) = 2E(X) – 3E(Y) = 2(10) – 3(4) = 20 – 12 = \rm{8} \)

2. Variance:

\( \text{Var}(Z) = 4 \cdot 9 + 9 \cdot 1 = 36 + 9 = \rm{45} \)

UNBIASED ESTIMATORS

An estimator $\hat{\theta}$ of a population parameter $\theta$ is unbiased if:

$
E(\hat{\theta}) = \theta
$

Unbiased estimators give correct population parameter on average.

 Sample Mean as an Unbiased Estimator of Population Mean

Let $X_1, X_2, …, X_n$ be a random sample from a population with mean $\mu$.

Sample Mean:

$
\bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i
$

Expected Value:

$
E(\bar{X}) = \mu
$

Therefore, $\bar{X}$ is an unbiased estimator of the population mean.

 Sample Variance as an Unbiased Estimator of Population Variance

Population Variance:

$
\sigma^2 = \frac{1}{N} \sum_{i=1}^{N} (X_i – \mu)^2
$

Sample Variance (Unbiased):

$
s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (X_i – \bar{X})^2
$

Expected Value:

$
E(s^2) = \sigma^2
$

Hence, $s^2$ is an unbiased estimator of the population variance.

 The denominator $n – 1$ corrects the bias, known as Bessel’s correction.

Example 

A random sample of size \( n = 4 \) is taken with values:

\( x_1 = 8,\quad x_2 = 10,\quad x_3 = 13,\quad x_4 = 11 \)

Find:

  1. The sample mean \( \bar{x} \)
  2. The unbiased sample variance \( s^2 \)
▶️Answer/Explanation

Solution

1. Sample Mean:

\( \bar{x} = \frac{1}{4}(8 + 10 + 13 + 11) = \frac{42}{4} = \rm{10.5} \)

2. Sample Variance:

Use the formula: \( s^2 = \frac{1}{n – 1} \sum_{i=1}^{n} (x_i – \bar{x})^2 \)

  • \( (8 – 10.5)^2 = 6.25 \)
  • \( (10 – 10.5)^2 = 0.25 \)
  • \( (13 – 10.5)^2 = 6.25 \)
  • \( (11 – 10.5)^2 = 0.25 \)

\( s^2 = \frac{1}{3}(6.25 + 0.25 + 6.25 + 0.25) = \frac{13}{3} = \rm{4.\overline{3}} \)

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