ThmDex – An index of mathematical definitions, results, and conjectures.

Result R5395 on D3932: Probability distribution equality relation

Factorization need not preserve equality in distribution

Formulation 0

Let $X, Y \in \text{Bernoulli}(1/2)$ each be a D3999: Standard Bernoulli random boolean number.

Then

(1)	\begin{equation} X \overset{d}{=} Y \end{equation}
(2)	\begin{equation} 2 X \overset{d}{\neq} X + Y \end{equation}

Proofs

Proof 0

Let $X, Y \in \text{Bernoulli}(1/2)$ each be a D3999: Standard Bernoulli random boolean number.

The first claim is clear. For the second, consider what values each side can attain. The random variable $2 X$ can attain values only in the set $\{ 0, 2 \}$, while the random variable $X + Y$ can attain values in the set $\{ 0, 1, 2 \}$. $\square$