# Finding the distribution of a random variable made of a random vector

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?

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

## 1 Answer

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

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