8
$\begingroup$

It seems reasonable to state that column vectors $\mathbf{x}$ are the most frequently seen standard notation, often using $\mathbf{x}^\intercal$ to denote a row vector (transposed column vector). However, in the study of stochastic processes and probability theory, it appears more standard to use a row vector to represent a probability vector. Why is this?

Maybe it simply just makes more sense to read left to right as forward in 'time': $$\mathbf{x}_n=\mathbf{x}_0P_1P_2\cdots P_n$$ where $P_i$ is the matrix of transition probabilities from the $(i-1)^\text{th}$ to the $i^\text{th}$ step, $\mathbf{x}_0$ is the initial distribution (a row vector), and $\mathbf{x}_n$ is the distribution after $n$ transitions. Maybe this reasoning is somewhat ethnocentric $-$ if a particular language reads right to left, the opposite might be true.

Is there any interesting history on this? Are there stories of 'battles' over standardizing the notation? Are there any other mathematical reasons for choosing row vectors over column vectors?


Also, the transition probabilities in the matrix $P$ can be written $$p_{ij}=\mathbb{P}(\text{next state is $j$ }|\text{ current state is } i).$$ Note that the conditional probability notation is always 'backwards in time', unless you are asking a question about the past conditioned on knowledge of the future. Even then, it is still 'backwards in the mental timeline' $-$ if the left slot is for past knowledge, and the right is for what we want to know about in the future. I'm getting a bit off-topic from the real question at this point though.

$\endgroup$
1
  • 1
    $\begingroup$ I use row vectors because of the left to right progression as you said, and because $M_{i,j}$ is "from $i$ to $j$" which also reads more naturally to me. $\endgroup$ – DanielV Mar 4 '15 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.