Answer :
Sure, let's prove the given equation step-by-step:
Given:
[tex]\[ \sum_{i=1}^n (x_i - \mu)^2 = \sum_{i=1}^n x_i^2 - n \mu^2 \][/tex]
Proof:
1. Expand the left-hand side (LHS):
Start with the expression [tex]\(\sum_{i=1}^n (x_i - \mu)^2\)[/tex].
[tex]\[ \sum_{i=1}^n (x_i - \mu)^2 \][/tex]
Using the binomial expansion, we can expand this to:
[tex]\[ \sum_{i=1}^n (x_i^2 - 2x_i\mu + \mu^2) \][/tex]
Now, distribute the summation across the terms inside:
[tex]\[ = \sum_{i=1}^n x_i^2 - \sum_{i=1}^n 2x_i\mu + \sum_{i=1}^n \mu^2 \][/tex]
2. Simplify each term:
- [tex]\(\sum_{i=1}^n x_i^2\)[/tex] remains as is.
- For the second term, factor out the constant [tex]\(2\mu\)[/tex]:
[tex]\[ - \sum_{i=1}^n 2x_i\mu = -2\mu \sum_{i=1}^n x_i \][/tex]
- For the third term, [tex]\(\mu\)[/tex] is also a constant and can be treated as such:
[tex]\[ \sum_{i=1}^n \mu^2 = \sum_{i=1}^n \mu \cdot \mu = \mu^2 \sum_{i=1}^n 1 = \mu^2 \cdot n \][/tex]
(since [tex]\(\sum_{i=1}^n 1\)[/tex] simply sums [tex]\(1\)[/tex] from [tex]\(1\)[/tex] to [tex]\(n\)[/tex], which equals [tex]\(n\)[/tex]).
This gives us:
[tex]\[ = \sum_{i=1}^n x_i^2 - 2\mu \sum_{i=1}^n x_i + n\mu^2 \][/tex]
3. Relate to the right-hand side (RHS):
Recall we want to show this equals [tex]\(\sum_{i=1}^n x_i^2 - n \mu^2\)[/tex].
- Notice that we can rewrite the expanded LHS (as calculated above) in terms of the known quantities:
[tex]\[ \sum_{i=1}^n x_i^2 - 2\mu \sum_{i=1}^n x_i + n\mu^2 \][/tex]
- Recognize that the mean [tex]\(\mu\)[/tex] can be expressed as:
[tex]\[ \mu = \frac{1}{n} \sum_{i=1}^n x_i \][/tex]
- Substitute [tex]\(\mu\)[/tex]:
[tex]\[ - 2\mu \sum_{i=1}^n x_i = -2 \left( \frac{1}{n} \sum_{i=1}^n x_i \right) \sum_{i=1}^n x_i = -2 \left( \frac{1}{n} (\sum_{i=1}^n x_i)^2 \right) = -\frac{2}{n} \left( \sum_{i=1}^n x_i \right)^2 \][/tex]
Which simplifies back to:
[tex]\[ -2\mu \sum_{i=1}^n x_i = -2\mu n \mu = -2\mu^2 n \text{ (because } \mu = \frac{\sum_{i=1}^n x_i}{n}) \][/tex]
4. Combine all terms:
Now, the LHS [tex]\(\sum_{i=1}^n x_i^2 - 2\mu \sum_{i=1}^n x_i + n\mu^2\)[/tex] can be seen as:
[tex]\[ \sum_{i=1}^n x_i^2 - 2n\mu^2 + n\mu^2 = \sum_{i=1}^n x_i^2 - n \mu^2 \][/tex]
Hence, this is equal to the RHS:
Therefore:
[tex]\[ \sum_{i=1}^n\left(x_i - \mu\right)^2 = \sum_{i=1}^n x_i^2 - n \mu^2 \][/tex]
And the proof is complete!
Given:
[tex]\[ \sum_{i=1}^n (x_i - \mu)^2 = \sum_{i=1}^n x_i^2 - n \mu^2 \][/tex]
Proof:
1. Expand the left-hand side (LHS):
Start with the expression [tex]\(\sum_{i=1}^n (x_i - \mu)^2\)[/tex].
[tex]\[ \sum_{i=1}^n (x_i - \mu)^2 \][/tex]
Using the binomial expansion, we can expand this to:
[tex]\[ \sum_{i=1}^n (x_i^2 - 2x_i\mu + \mu^2) \][/tex]
Now, distribute the summation across the terms inside:
[tex]\[ = \sum_{i=1}^n x_i^2 - \sum_{i=1}^n 2x_i\mu + \sum_{i=1}^n \mu^2 \][/tex]
2. Simplify each term:
- [tex]\(\sum_{i=1}^n x_i^2\)[/tex] remains as is.
- For the second term, factor out the constant [tex]\(2\mu\)[/tex]:
[tex]\[ - \sum_{i=1}^n 2x_i\mu = -2\mu \sum_{i=1}^n x_i \][/tex]
- For the third term, [tex]\(\mu\)[/tex] is also a constant and can be treated as such:
[tex]\[ \sum_{i=1}^n \mu^2 = \sum_{i=1}^n \mu \cdot \mu = \mu^2 \sum_{i=1}^n 1 = \mu^2 \cdot n \][/tex]
(since [tex]\(\sum_{i=1}^n 1\)[/tex] simply sums [tex]\(1\)[/tex] from [tex]\(1\)[/tex] to [tex]\(n\)[/tex], which equals [tex]\(n\)[/tex]).
This gives us:
[tex]\[ = \sum_{i=1}^n x_i^2 - 2\mu \sum_{i=1}^n x_i + n\mu^2 \][/tex]
3. Relate to the right-hand side (RHS):
Recall we want to show this equals [tex]\(\sum_{i=1}^n x_i^2 - n \mu^2\)[/tex].
- Notice that we can rewrite the expanded LHS (as calculated above) in terms of the known quantities:
[tex]\[ \sum_{i=1}^n x_i^2 - 2\mu \sum_{i=1}^n x_i + n\mu^2 \][/tex]
- Recognize that the mean [tex]\(\mu\)[/tex] can be expressed as:
[tex]\[ \mu = \frac{1}{n} \sum_{i=1}^n x_i \][/tex]
- Substitute [tex]\(\mu\)[/tex]:
[tex]\[ - 2\mu \sum_{i=1}^n x_i = -2 \left( \frac{1}{n} \sum_{i=1}^n x_i \right) \sum_{i=1}^n x_i = -2 \left( \frac{1}{n} (\sum_{i=1}^n x_i)^2 \right) = -\frac{2}{n} \left( \sum_{i=1}^n x_i \right)^2 \][/tex]
Which simplifies back to:
[tex]\[ -2\mu \sum_{i=1}^n x_i = -2\mu n \mu = -2\mu^2 n \text{ (because } \mu = \frac{\sum_{i=1}^n x_i}{n}) \][/tex]
4. Combine all terms:
Now, the LHS [tex]\(\sum_{i=1}^n x_i^2 - 2\mu \sum_{i=1}^n x_i + n\mu^2\)[/tex] can be seen as:
[tex]\[ \sum_{i=1}^n x_i^2 - 2n\mu^2 + n\mu^2 = \sum_{i=1}^n x_i^2 - n \mu^2 \][/tex]
Hence, this is equal to the RHS:
Therefore:
[tex]\[ \sum_{i=1}^n\left(x_i - \mu\right)^2 = \sum_{i=1}^n x_i^2 - n \mu^2 \][/tex]
And the proof is complete!
