I understand that the quadratic form of a Normal variable follows chi-square distribution:

if$$x \sim N(\mu,I_p)$$ then $$x'Ax \sim\chi^2(p,\delta)$$ where $\delta = \mu'\Sigma\mu$ is the non-central parameter.

However, as is shown below, I don't know how to prove the independency of these two quadratic form, namely $x'Ax$ and $x'Bx$.




1 Answer 1


I actually have a quite simple idea:

Suppose the rank of $A$ is $r$, I want to show that $x'Ax$ can be expressed with $Y_1,Y_2,...,Y_r$ and $x'Bx$ with $Y_{r+1},...,Y_n$, which supports the independency of two quadratic.

Firstly, we need to discuss the situation where $r=n$. In this case, condition $AB=0$ directly tells us $B$ must be a $0_{n\times n}$ matrix because $A$ is a full rank matrix. Hence $A,B$ are independent. Similarly, we can prove this when $r=n$.

For the following discussion, I will assume $0< r< n$.

  1. Since $A$ is a symmetric matrix, we can decompose it into the form: $A = P'\Lambda P$ where: $$\Lambda = \begin{bmatrix} D_r & 0 \\ 0 &0 \end{bmatrix}, D_r = diag(\lambda_1,\lambda_2,...,\lambda_r)$$

  2. Let's denote $P'BP = H = \begin{bmatrix}H_{11} & H_{12} \\H_{21} & H_{22}\end{bmatrix}$ , where $H_{11}$ is an $r \times r$ matrix. Recall: $$ AB=P'\begin{bmatrix} D_r & 0 \\ 0 &0 \end{bmatrix} P B= 0$$ $P$ is an orthogonal vector, so we can add a $P'P = I_n$ to the tail of it: $$ AB=P'\begin{bmatrix} D_r & 0 \\ 0 &0 \end{bmatrix} P B P'P= P'\begin{bmatrix} D_r & 0 \\ 0 &0 \end{bmatrix} \begin{bmatrix}H_{11} & H_{12} \\H_{21} & H_{22}\end{bmatrix}P=0$$ which yeilds: $$D_rH_{11} = D_rH_{12} = 0$$ note $H_{12} = H_{21} = 0$ because $B$ is a symmetric matrix.
    In general, the discussion in part 2 gives us: $$ P'BP = H = \begin{bmatrix}0 & 0 \\ 0 & H_{22}\end{bmatrix}$$

  3. Finally, let's take a look at $x'Ax$ and $x'Bx$:
    $(1)~ x'Ax = y'P'APy = \Sigma_{i=1}^{r}\lambda_i y_i^2$
    $(2) ~x'Bx = y'P'BPy = y'Hy = (y_{r+1},...,y_n)H_{22}(y_{r+1},...,y_n)'$
    And this proves the question.


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