To prove that [tex]\(\sum_{i=1}^n \left(x_i - \mu\right)^2 = \sum_{i=1}^n x_i^2 - n \mu^2\)[/tex], we will start by expanding the left side and then simplifying it to match the right side.
### Step 1: Define the mean
The mean, [tex]\(\mu\)[/tex], of the [tex]\(n\)[/tex] elements [tex]\(x_1, x_2, \ldots, x_n\)[/tex] is:
[tex]\[
\mu = \frac{1}{n} \sum_{i=1}^n x_i
\][/tex]
### Step 2: Expand the left side
Let's start by expanding the left-hand side of the equation:
[tex]\[
\sum_{i=1}^n (x_i - \mu)^2
\][/tex]
Expanding this, we get:
[tex]\[
\sum_{i=1}^n (x_i - \mu)^2 = \sum_{i=1}^n (x_i^2 - 2x_i \mu + \mu^2)
\][/tex]
### Step 3: Distribute the summation
We can distribute the summation over the terms in the expanded expression:
[tex]\[
\sum_{i=1}^n x_i^2 - \sum_{i=1}^n 2x_i \mu + \sum_{i=1}^n \mu^2
\][/tex]
### Step 4: Simplify each term
Let's simplify each term step-by-step:
1. First term:
[tex]\[
\sum_{i=1}^n x_i^2
\][/tex]
This term remains as is.
2. Second term:
[tex]\[
\sum_{i=1}^n 2x_i \mu = 2\mu \sum_{i=1}^n x_i
\][/tex]
Since [tex]\(\mu\)[/tex] is a constant with respect to the summation, it can be taken outside the summation sign.
3. Third term:
[tex]\[
\sum_{i=1}^n \mu^2
\][/tex]
Since [tex]\(\mu^2\)[/tex] is also a constant with respect to the summation:
[tex]\[
\sum_{i=1}^n \mu^2 = n \mu^2
\][/tex]
### Step 5: Combine the simplified terms
Now, combining the simplified terms, we have:
[tex]\[
\sum_{i=1}^n x_i^2 - 2\mu \sum_{i=1}^n x_i + n \mu^2
\][/tex]
### Step 6: Substitute the expression for the mean
Recall that [tex]\(\mu = \frac{1}{n} \sum_{i=1}^n x_i\)[/tex]. Therefore:
[tex]\[
2\mu \sum_{i=1}^n x_i = 2 \left(\frac{1}{n} \sum_{i=1}^n x_i\right) \sum_{i=1}^n x_i = 2 \frac{1}{n} \left(\sum_{i=1}^n x_i\right)^2 = \frac{2}{n} \left(\sum_{i=1}^n x_i\right) \sum_{i=1}^n x_i = 2 \sum_{i=1}^n x_i
\][/tex]
### Step 7: Final expression
Putting it all together:
[tex]\[
\sum_{i=1}^n (x_i - \mu)^2 = \sum_{i=1}^n x_i^2 - 2\mu \sum_{i=1}^n x_i + n \mu^2 = \sum_{i=1}^n x_i^2 - 2 \sum_{i=1}^n x_i \mu + n \mu^2
\][/tex]
Finally, notice that [tex]\(\sum_{i=1}^n x_i \mu = \sum_{i=1}^n \mu x_i = n \mu \mu = n\mu^2\)[/tex] as [tex]\(\mu = \frac{1}{n} \sum_{i=1}^n x_i\)[/tex]:
[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, we have:
[tex]\[
\sum_{i=1}^n (x_i - \mu)^2 = \sum_{i=1}^n x_i^2 - n \mu^2
\][/tex]
This completes the proof.
