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🔷 Expectation and Variance of the Hypergeometric Distribution
✅ Random Variable X and Its Range
The hypergeometric distribution is a discrete probability distribution that describes the probability of obtaining a certain number of elements with a specific characteristic (e.g., category or label) when a fixed number of samples are drawn without replacement from a finite population.
The random variable \( X \) denotes the number of elements in the sample that fall under the target category. Its range is given by:
\[ \max(0, n - (N - K)) \leq X \leq \min(n, K) \]
This reflects the feasible range of values based on the size of the population and sample.
For example, if \( N = 20 \), \( K = 5 \), and \( n = 4 \), then:
- \( \min(4, 5) = 4 \)
- \( \max(0, 4 - (20 - 5)) = \max(0, -11) = 0 \)
Thus, \( X \in \{ 0, 1, 2, 3, 4 \} \).
✅ X as a Sum of Indicator Variables
In the hypergeometric distribution, the variable \( X \) can be expressed as a sum of indicator variables as follows:
\[ X = I_1 + I_2 + \cdots + I_n \]
Each \( I_j \) is an indicator variable that equals 1 if the \( j \)th selected item is in the target category, and 0 otherwise.
✅ \(I_j\) Does Not Indicate Time Order
The index \( j \) in \( I_j \) refers to the position in the sample, not the order in which the item was drawn.
For instance, suppose the population consists of 2 strawberry-flavored candies (denoted as S) and 3 lemon-flavored candies (L), i.e., \( (S, S, L, L, L) \), and we sample 3 items without replacement. If we obtain the sample \( (L, S, L) \), then:
- \( I_1 = 0 \): The first item is lemon
- \( I_2 = 1 \): The second item is strawberry
- \( I_3 = 0 \): The third item is lemon
Note that \( I_1, I_2, I_3 \) are not independent, since sampling is done without replacement. The selection of one item affects the probability of selecting others.
✅ Why \(\mathbb{E}[I_j] = \frac{K}{N}\)
Each \( I_j \) represents the chance that a particular sample position contains an element from the target category. This chance equals the proportion of such elements in the population.
Thus, the expectation of each indicator variable is always:
\[ \mathbb{E}[I_j] = \frac{K}{N} \quad \text{for all } j \in \{1, \dots, n\} \]
✅ Verifying \(\mathbb{E}[I_j]\) Using Full Enumeration of (1,1,0,0,0)
※ Note: This example does not represent the distribution of \( X \), but demonstrates that the expected value of each position remains consistent across permutations.
✅ Sample-Based Example: X = I₁ + I₂
Let’s verify the expectation of \( X \) using the values and probabilities of \( I_1 \), \( I_2 \) in real samples:
✅ Generalized Expectation Formula
We have shown the structure \( X = I_1 + I_2 \) for small \( n \), but the same logic applies for any \( n \).
When \( X = I_1 + I_2 + \cdots + I_n \), we compute the expectation as follows:
This result uses the linearity of expectation, which holds even if the variables are dependent.
Since each \( \mathbb{E}[I_j] = \frac{K}{N} \), we get:
✅ Comparing Three Expectation Methods
We now summarize the three methods of computing the expectation:
- Indicator Variables: \( 0.4 + 0.4 = 0.8 \)
- Probability Distribution: \( 0 \cdot 0.3 + 1 \cdot 0.6 + 2 \cdot 0.1 = 0.8 \)
- General Formula: \( 2 \cdot \frac{2}{5} = 0.8 \)
All three methods yield exactly the same result, confirming consistency between structure and numeric values.
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