Why are these two random variables identically distributed?

Question

Let $(X_1, X_2, \ldots, X_i, \ldots, X_k)$ be $k$ independent, normally distributed random variables with means $\mu_i$ and variances $1$. Then the random variable $$ \sum_{i=1}^k X_i^2$$ is distributed according to the noncentral chi-squared distribution with parameters: $k$ and $ \lambda=\sum_{i=1}^k \mu_i^2$.

If $(X_1', X_2', \ldots, X_i', \ldots, X_k')$ are also independent normal distributed with variances $1$, and means $\mu'_i$'s s.t. $\sum_{i=1}^k \mu_i'^2 = \lambda$, why are $$ \sum_{i=1}^k X_i'^2$$ and $$ \sum_{i=1}^k X_i^2$$ identically distributed? Thanks.

Answer 1

The $X_i$ can be written as

$$X_i = \mu_i + Y_i$$

where the $Y_i$ are independent standard normal variates (of mean zero). By virtue of the Pythagorean Theorem in $k$ dimension, the expression

$$\chi^2_{k;\mu_1,\mu_2,\ldots,\mu_k}= X_1^2 + X_2^2 + \cdots + X_k^2$$

is recognizable as the squared distance between the point $(-\mu_1, -\mu_2, \ldots, -\mu_k)$ and $(Y_1, Y_2, \ldots, Y_k)$. Because distances remain unchanged by rotations (and reflections) and the distribution of $(Y_1, Y_2, \ldots, Y_k)$ also remains unchanged by rotations, the distribution of this sum of squares will remain unchanged upon rotating (or reflecting) $(-\mu_1, -\mu_2, \ldots, -\mu_k)$ to any other point. The only quantity that is left invariant under rotations and reflections is the magnitude, whence the distribution of $\chi^2_{k;\mu_1,\mu_2,\ldots,\mu_k}$ depends only on $\|(-\mu_1,-\mu_2, \ldots, -\mu_k)\|$ = $\sqrt{\mu_1^2 + \mu_2^2 + \cdots + \mu_k^2}$. Conventionally, the square of this magnitude is used to index the non-central $\chi^2$ distribution.