Exa 1

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ A.1}$

Show that an independent joint distribution is the tensor product of its marginals.

Independent joint distribution

Anders B Madsen 13 Apr 2024 09:32

Let $\mu \in \mathcal{P}^{A\times B}$ be an uncorrelated joint distribution. This means, that $\mu$ is the tensor product $\mu = \mu_{\alpha} \otimes \mu_{\beta}$ for some $\mu_{\alpha} \in \mathcal{P}^A$ and $\mu_{\beta} \in \mathcal{P}^B$.
For any $a \in A$, the marginal $\mu_A$ is given by:

(1)

\begin{align} \mu_A(a) = \sum_{b\in B} \mu(a,b) = \sum_{b\in B} \mu_{\alpha}(a)\mu_{\beta}(b) = \mu_{\alpha}(a)\sum_{b\in B} \mu_{\beta}(b) = \mu_{\alpha}(a)\,. \end{align}

The same holds for the other marginal $\mu_B$, so $\mu_{\alpha} = \mu_A$ and $\mu_{\beta} = \mu_B$, which shows that the uncorrelated joint distribution $\mu$ is the tensor product of its marginals.