# Normalization of conditional probabilities

Let $$p_i$$ be a vector, such that $$p_i\geq 0$$ and $$\sum_i p_i=1$$. Similarly $$\pi_\alpha$$ is a vector satisfying $$\pi_\alpha\geq 0$$ and $$\sum_\alpha \pi_\alpha = 1$$. The two are related via $$\pi_\alpha = \sum_i W_{\alpha i}p_i$$, where $$W_{\alpha i}\geq 0$$. Can we make a mathematically rigorous statement that $$\sum_\alpha W_{\alpha i}=1$$?

Obviously, this is always the case when $$p_i,\pi_\alpha$$ are understood as probabilities and $$W_{\alpha i}$$ as conditional probabilities, in which case $$W_{\alpha i}\geq 0$$ is just the normalization of the conditional probability. Thus the question can be restated as: do we deal here with a simple mathematical fact or is there an additional implicit condition inherent in interpreting $$W_{\alpha i}$$ as probabilities.

Remark: This might be a trivial question due to some glitch in my head, so a clarifying comment could suffice.

• There is no reason why this should hold. Commented Sep 28, 2022 at 10:37

Not necessarily. Here is a counter-example: consider the vectors $$p = \begin{bmatrix} 0.5 \\ 0.5 \end{bmatrix}$$

and

$$\pi = \begin{bmatrix} 0.5 \\ 0.5 \end{bmatrix}$$

Then the matrix

$$W = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$$

verifies $$Wp = \pi$$, while the elements in each of its columns do not sum to 1.

• Good example +1. I should have mentioned in the OP that $p_i$ can be arbitrary (provided that they are non-negative and sum to unity.) Commented Sep 28, 2022 at 12:17

The answer by @CamilleGontier has pushed me in the right direction: what is implicit in the question (but was not explicitly stated in the OP) is that it should work for an arbitrary vector $$p_i$$ (as long as it satisfies conditions $$p_i>0$$ and $$\sum_i p_i=1$$.) We can then consider a set of linearly independent vectors which have only one non-zero component: $$p_i^{(i_0)}=\delta_{i,i_0}$$ Then for each such vector we have corresponding $$\pi_\alpha^{(i_0)}=W_{\alpha i_0},$$ and the normalization condition for $$\pi_\alpha$$ automatically translates into the condition for matrix $$W$$ : $$\sum_\alpha\pi_\alpha^{(i_0)} = \sum_\alpha W_{\alpha i_0}=1$$