Question

Proof that total variation = explained variation + unexplained variation Show that \[ \sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}=\sum_{i=1}^{n}\left(y_{i}-\hat{y}_{i}\right)^{2}+\sum_{i=1}^{n}\left(\hat{y}_{i}-\bar{y}\right)^{2} \] that is Total variation \( = \) Explain

Answer 1

The equation
`\(\sum_(i=1)^(n)\left(y_(i)-\bar{y}\right)^(2)=\sum_(i=1)^(n)\left(y_(i)-\hat{y}_(i)\right)^(2)+\sum_(i=1)^(n)\left(\hat{y}_(i)-\bar{y}\right)^(2)\)`demonstrates that the total variation in a dataset is equal to the sum of the explained variation, attributed to the regression model
`(\(\sum_(i=1)^(n)\left(\hat{y}_(i)-\bar{y}\right)^(2)\))`, and the unexplained variation
`(\(\sum_(i=1)^(n)\left(y_(i)-\hat{y}_(i)\right)^(2)\))`, representing the discrepancies between observed and predicted values. This result highlights the partitioning of the total variation into components attributable to the model and those that remain unaccounted for.

Step-by-step explanation:

Certainly! The equation you've provided is actually an expression of the total sum of squares (total variation) being decomposed into explained sum of squares (explained variation) and unexplained sum of squares (unexplained variation). Let's go through the proof step by step.

1. Total Sum of Squares (Total Variation):*

`\[ \sum_(i=1)^(n)\left(y_(i)-\bar{y}\right)^(2) \]`

This represents the total variation in the dependent variable
`\(y\)`, where
`\(\bar{y}\)` is the mean of
`\(y\)`.

2. Explained Sum of Squares (Explained Variation):

`\[ \sum_(i=1)^(n)\left(\hat{y}_(i)-\bar{y}\right)^(2) \]`

Here,
`\(\hat{y}_(i)\)` represents the predicted (fitted) values of the dependent variable based on the regression model.

3. Unexplained Sum of Squares (Unexplained Variation):

`\[ \sum_(i=1)^(n)\left(y_(i)-\hat{y}_(i)\right)^(2) \]`

This part represents the sum of squares of the residuals, which are the differences between the actual values
`(\(y_(i)\))` and the predicted values
`(\(\hat{y}_(i)\))`.

Now, let's prove that the total variation is equal to the sum of the explained and unexplained variations:

Starting with the total sum of squares:

`\[ \sum_(i=1)^(n)\left(y_(i)-\bar{y}\right)^(2) \]`

This can be expanded as:

`\[ \sum_(i=1)^(n)\left[(y_(i)-\hat{y}_(i)) + (\hat{y}_(i)-\bar{y})\right]^(2) \]`

Expanding the square:

`\[ \sum_(i=1)^(n)\left(y_(i)-\hat{y}_(i)\right)^(2) + 2\sum_(i=1)^(n)\left(y_(i)-\hat{y}_(i)\right)\left(\hat{y}_(i)-\bar{y}\right) + \sum_(i=1)^(n)\left(\hat{y}_(i)-\bar{y}\right)^(2) \]`

Now, notice that the second term (cross-product term) sums to zero when taken over all observations because the residuals
`(\(y_(i)`-
`\hat{y}_(i)\))` sum to zero. Therefore, the equation simplifies to:

`\[ \sum_(i=1)^(n)\left(y_(i)-\bar{y}\right)^(2) = \sum_(i=1)^(n)\left(y_(i)-\hat{y}_(i)\right)^(2) + \sum_(i=1)^(n)\left(\hat{y}_(i)-\bar{y}\right)^(2) \]`

This completes the proof, showing that the total variation is indeed equal to the sum of the explained and unexplained variations.