Math Problem Statement

1.18. By applying formulas (1.14.6) and (1.9.1) show that L (Xi - X)2 i= 1 can be rewritten in the following three forms which are often useful in computation: a) LX? - nx2 , c) LX? - x(L x;) .

Solution

To show that the sum of squares i=1n(XiXˉ)2\sum_{i=1}^{n} (X_i - \bar{X})^2 can be rewritten in the specified forms, we’ll apply formulas (1.14.6) and (1.9.1). For this explanation, let's denote:

  • XiX_i as the ii-th observation,
  • Xˉ\bar{X} as the mean of XiX_i values, i.e., Xˉ=1ni=1nXi\bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i,
  • nn as the number of observations.

Part (a): Show that

i=1n(XiXˉ)2=i=1nXi2nXˉ2\sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - n \bar{X}^2

Step-by-Step Solution:

  1. Expand the left-hand side: Start with the definition of the sum of squared deviations: i=1n(XiXˉ)2=i=1n(Xi22XiXˉ+Xˉ2).\sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} \left( X_i^2 - 2 X_i \bar{X} + \bar{X}^2 \right).

  2. Separate the terms: Distribute the summation: =i=1nXi22i=1nXiXˉ+i=1nXˉ2.= \sum_{i=1}^{n} X_i^2 - 2 \sum_{i=1}^{n} X_i \bar{X} + \sum_{i=1}^{n} \bar{X}^2.

  3. Simplify each term:

    • The first term, i=1nXi2\sum_{i=1}^{n} X_i^2, remains as it is.
    • The second term becomes 2Xˉi=1nXi-2 \bar{X} \sum_{i=1}^{n} X_i. Since i=1nXi=nXˉ\sum_{i=1}^{n} X_i = n \bar{X}, this term simplifies to 2XˉnXˉ=2nXˉ2-2 \bar{X} \cdot n \bar{X} = -2 n \bar{X}^2.
    • The third term, i=1nXˉ2\sum_{i=1}^{n} \bar{X}^2, simplifies to nXˉ2n \bar{X}^2 because Xˉ2\bar{X}^2 is constant with respect to ii.
  4. Combine terms: i=1n(XiXˉ)2=i=1nXi22nXˉ2+nXˉ2=i=1nXi2nXˉ2.\sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - 2 n \bar{X}^2 + n \bar{X}^2 = \sum_{i=1}^{n} X_i^2 - n \bar{X}^2.

This proves part (a).


Part (c): Show that

i=1n(XiXˉ)2=i=1nXi2Xˉi=1nXi.\sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - \bar{X} \sum_{i=1}^{n} X_i.

  1. Use the result from part (a): From part (a), we know that i=1n(XiXˉ)2=i=1nXi2nXˉ2.\sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - n \bar{X}^2.

  2. Substitute nXˉn \bar{X} with i=1nXi\sum_{i=1}^{n} X_i: Recall that Xˉ=1ni=1nXi\bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i, so nXˉ=i=1nXin \bar{X} = \sum_{i=1}^{n} X_i.

  3. Rewrite nXˉ2n \bar{X}^2: nXˉ2=Xˉi=1nXi.n \bar{X}^2 = \bar{X} \sum_{i=1}^{n} X_i.

  4. Substitute into the formula: i=1n(XiXˉ)2=i=1nXi2Xˉi=1nXi.\sum_{i=1}^{n} (X_i - \bar{X})^2 = \sum_{i=1}^{n} X_i^2 - \bar{X} \sum_{i=1}^{n} X_i.

This completes the proof for part (c).


Let me know if you’d like further details or have any questions about these derivations.


Relative Questions:

  1. What is the geometric interpretation of i=1n(XiXˉ)2\sum_{i=1}^{n} (X_i - \bar{X})^2 in statistics?
  2. How can these rewritten forms be useful for computational efficiency?
  3. What is the meaning of Xˉ\bar{X} in terms of probability if XiX_i represents a sample?
  4. How does this approach change if XiX_i values are weighted?
  5. What role does the variance play in relation to these forms?

Tip: Remember, expanding and simplifying terms step-by-step helps avoid mistakes when dealing with summation formulas.

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Math Problem Analysis

Mathematical Concepts

Statistics
Summation Notation
Mean and Variance

Formulas

Sum of Squares: ∑(Xi - X̄)²
Formula for Mean: X̄ = (1/n)∑Xi

Theorems

Sum of Squares Decomposition

Suitable Grade Level

College Level