Choice variance

Task number: 2742

Let the vector \(\mathbf x=(x_1,…,x_n)\in \mathbb R^n\) represents \(n\) observations, i.e. choices. The variance is defined as \(\displaystyle \sigma^2=\sum_{i=1}^n (x_i-\bar x)^2 \), where \(\displaystyle \bar x=\frac 1n \sum_{i=1}^n x_i \) is the average.

Show that \(\sigma^2\) is a quadratic form on \(\mathbb R^n\) and determine the matrix of this form.

  • Resolution

    We use the well known fact

    \(\displaystyle \sigma^2=\sum_{i=1}^n (x_i-\bar x)^2= \sum_{i=1}^n (x_i^2-\bar x^2) \),


    \(\displaystyle \sum_{i=1}^n (x_i^2-\bar x^2)= \sum_{i=1}^n x_i^2- n\bar x^2= \sum_{i=1}^n x_i^2- n\left(\frac 1n \sum_{i=1}^n x_i \right)^2= \sum_{i=1}^n x_i^2- \frac 1n \sum_{i=1}^n\sum_{j=1}^n x_ix_j= \)
    \( \mathbf x^T\mathbf x -\frac1n \mathbf x^T\mathbf J_n\mathbf x= \mathbf x^T\left(\mathbf I_n-\frac 1n \mathbf J_n\right)\mathbf x \),

    where the matrix \(\mathbf J_n\) contains only ones.

    Note that the above relation can also be derived as:

    \(\sum_{i=1}^n (x_i-\bar x)^2=\sum_{i=1}^n (x_i^2-2x_i\bar x+ \bar x^2)= \sum_{i=1}^n x_i^2 -2\bar x \sum_{i=1}^n x_i+ n\bar x^2= \)
    \(\sum_{i=1}^n x_i^2 -n\bar x^2= \sum_{i=1}^n (x_i^2-\bar x^2)\).

  • Result

    The form can be expressed as the product \(\sigma^2=\mathbf x^T\left(\mathbf I_n-\frac 1n \mathbf J_n\right)\mathbf x\).

    The matrix of the form is \(\begin{pmatrix} \frac{n-1}{n} & -\frac1n & … &-\frac1n \\ -\frac1n & \frac{n-1}{n} & \ddots & \vdots \\ \vdots &\ddots &\ddots &-\frac1n\\[3pt] -\frac1n & … & -\frac1n & \frac{n-1}{n} \\ \end{pmatrix} \)

Difficulty level: Hard task
Proving or derivation task
Cs translation
Send comment on task by email