1
$\begingroup$

Given two independent normally distributed random variables A and B: $$A \sim \mathcal{N(\mu_A, \Sigma_A)}$$$$B \sim \mathcal{N(\mu_B, \Sigma_B)}$$ is there a way to find normally distributed random variable $C \sim \mathcal{N(\mu_C, \Sigma_C)}$ such that $$f_C(x) = r\sqrt{ f_A(x) f_B(x)}$$ for all $x \in \mathbb{R}^n$ and for some $r > 0$

I tried using

$$f_A(x) = \frac{1}{\sqrt{(2 \pi) ^ n \left| \Sigma_A \right|}} \exp(-\frac{1}{2}(x-\mu_A)^T \Sigma_A ^ {-1} (x - \mu_A))$$

$$f_B(x) = \frac{1}{\sqrt{(2 \pi) ^ n \left| \Sigma_B \right|}} \exp(-\frac{1}{2}(x-\mu_B)^T \Sigma_B ^ {-1} (x - \mu_B))$$

in order to modify $f_C$ to form

$$f_C(x) = \frac{1}{\sqrt{(2 \pi) ^ n \left| \Sigma_C \right|}} \exp(-\frac{1}{2}(x-\mu_C)^T \Sigma_C ^ {-1} (x - \mu_C))$$

without any success.

Thanks for any help.

$\endgroup$

1 Answer 1

1
$\begingroup$

This is possible. Note first that the constants in front of the exponential functions in $f_A$ and $f_B$ don't need to bother us, because we have $r$. This means we can concentrate on what happens in the exponents.

The exponent of $f_A$ can be written as: $$-\frac{1}{2} x^T \Sigma_A^{-1}x + \mu_A \Sigma_A^{-1} x - \frac{1}{2}\mu_A \Sigma_A^{-1} \mu_A$$

The last term is constant with respect to $x$, so we can also put it into the constant $r$. Therefore the exponent of $\sqrt{f_A f_B}$ is [without unnecessary constants]: $$-\frac{1}{2} \frac{1}{2}\left(\Sigma_A^{-1} + \Sigma_B^{-1}\right) + \frac{1}{2}\left(\mu_A \Sigma_A^{-1} + \mu_B \Sigma_B^{-1}\right)x$$

Comparing this with the expanded form of the density of $\mathcal{N}(\mu_C, \Sigma_C)$ you only need to choose: $$\begin{align*}\Sigma_C &= \left(\frac{1}{2}\left(\Sigma_A^{-1} + \Sigma_B^{-1}\right)\right)^{-1} \\ \mu_C &= \frac{1}{2}\left(\mu_A \Sigma_A^{-1} + \mu_B \Sigma_B^{-1}\right)\Sigma_C\end{align*}$$

Note that $\Sigma_C$ is actually well-defined, because the sum of two positive definite matrices is positive definite and therefore invertible.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .