Firstly let's explain some signs used later:

  1. $y_1, \ldots, y_2$ are random variables,
  2. $\mathbb{E}(y_i) = \mu_i$,
  3. $\text{Cov}(y_i, y_j) = \sigma_{ij}$,
  4. $Y = (y_1 \ldots y_n)^{T}$,
  5. $\mathbb{E}(Y) = \mu = (\mu_1 \ldots \mu_n)^{T}$,
  6. $\text{Cov}(Y) = \mathbb{E}[(Y- \mu)(Y - \mu)^{T}]$.

Now we can define a random vector $Y$ with normal distribution - $N(\mu, \Sigma)$.
Let's consider a random variable: $$Z = (Y-\mu)^{T} \Sigma^{-1}(Y-\mu).$$ What is the distribution of $Z$? How can it be found?
I know that one method would be to find expected value and covariance but I don't know how. Are there any other possibilities?

  • $\begingroup$ isn't it just a Chi-squared distribution? since Z is as sort of "square" of standard normals? $\endgroup$ – RScrlli Nov 20 '18 at 22:47
  • $\begingroup$ @RamiroScorolli pretty much! Typically you diagonalise the quadratic form and then rewrite it as a sum of independent non-central Chi-squared random variables $\endgroup$ – Nadiels Nov 20 '18 at 23:21

You have that $X$ is distributed $N(\mu,\Sigma)$ so:

$(X-\mu)\sim N(0,\Sigma)$ and

$\Sigma^{-1/2}(X-\mu)\sim N(0,\Sigma^{-1/2}\cdot\Sigma\cdot{\Sigma^{-1/2}}^{'})$

$\Sigma^{-1/2}(X-\mu)\sim N(0,\Sigma^{-1/2}\cdot\Sigma^{-1/2}({\Sigma^{1/2}}^{'}\cdot\Sigma^{1/2})\cdot{\Sigma^{-1/2}}^{'})$ Since $\Sigma^{-1/2}$ is symmetric by construction (use an orthogonal eigendecomposition) this leads to:

$\Sigma^{-1/2}(X-\mu)\sim N(0,I)$ i.e. a standard normal.

Calculating the square of this new random variable gives us: $(\Sigma^{-1/2}(X-\mu))^{'}\cdot \Sigma^{-1/2}(X-\mu)=(X-\mu)^{'}\Sigma^{-1}(X-\mu)$ Which is precisely $Z$
Since it's the square of a Standard Normal, the distribution will be Chi-Squared with k (dimension of the vector $y$) df

  • $\begingroup$ Thank you very much! Your solution is very smart and elegant :) $\endgroup$ – Hendrra Nov 21 '18 at 7:57
  • 1
    $\begingroup$ Your welcome! Glad you found it useful $\endgroup$ – RScrlli Nov 21 '18 at 8:06

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