Help me understand why this summation index is not j

The below image is an excerpt from a website about Markov Chains.

In the red boxed which I put in the image, I don't understand why the term ##g(i)## isn't being summed over ##j## instead of ##i##, since the outer sum is over the ##i##th element of the vector ##Pg##, which is the dot product between the ##i##th row of ##P## and ##g##.

I expected ##\langle f,Pg \rangle## to expand as $$\sum_i f(i)(Pg)_i \pi_i = \sum_i f(i)\Big(\sum_jP_{ij}g(j)\Big)\pi_i .$$

But, the website shows $$\langle f,Pg \rangle= \sum_i f(i)\Big(\sum_jP_{ij}g(i)\Big)\pi_i .$$ What am I misunderstanding?

In the theory of Markov chains, don't they do the matrix-vector operation back-to-front? Just to be different.

PeroK said:

In the theory of Markov chains, don't they do the matrix-vector operation back-to-front? Just to be different.

Yes, if you mean left side multiplication, I think for probability distributions. Right side multiplication is done for expected values. The special inner product is defined using ##Pg## so I thought it was the usual right side multiplication... I am also not understanding how in the last line, ##P_{ji}## is switched with ##P_{ij}## as soon as the summation order changed..

Obvious misprint.

PeroK said:

In the theory of Markov chains, don't they do the matrix-vector operation back-to-front? Just to be different.

Even if that is the case (I have not checked), the quoted expressions have typos. Exactly what they should be depends on this definition so I am not going to get into details regarding that.

Having browsed the page: The first equation in the quote should have ##g(j)##, not ##g(i)##. This is also clear from the second step where they do write ##g(j)##.

For the second equation they should not have switched the indices and their target expression should not have switched indices.

I’d say a case of bad proof reading in general.

Thank you ! you saved me from an existential crisis for now.

docnet said:

Thank you ! you saved me from an existential crisis for now.

It all hinges on the "balance condition". ##P## is ##\pi##-self-adjoint iff that condition holds.

PeroK said:

It all hinges on the "balance condition". ##P## is ##\pi##-self-adjoint iff that condition holds.

Yes, but that was used already when writing the first expression in that equation down. The step where they just switch the indices on P is actually wrong and results in a wrong expression that they then interpret as if it were the correct one.

Orodruin said:

Yes, but that was used already when writing the first expression in that equation down. The step where they just switch the indices on P is actually wrong and results in a wrong expression that they then interpret as if it were the correct one.

It is a mess. At this level, I would expect to highlight that the balance condition is precisely the condition that makes ##P## self-adjoint. It's not just useful for the proof, it's the whole basis of the proof.

The way I would do this is simply.
$$\langle f,Pg \rangle = \sum_i f(i)\Big(\sum_jP_{ij}g(j)\Big)\pi_i = \sum_{i,j} f(i)g(j)\pi_iP_{ij}$$$$= \sum_{i,j} f(i)g(j)\pi_jP_{ji} =\sum_{i,j} P_{ji} f(i)g(j)\pi_j = \langle Pf, g \rangle$$

PeroK said:

The way I would do this is simply.
$$\langle f,Pg \rangle = \sum_i f(i)\Big(\sum_jP_{ij}g(j)\Big)\pi_i = \sum_{i,j} f(i)g(j)\pi_iP_{ij}$$$$= \sum_{i,j} f(i)g(j)\pi_jP_{ji} =\sum_{i,j} P_{ji} f(i)g(j)\pi_j = \langle Pf, g \rangle$$

That's sort of what I did as well! I thought about adding an extra step

$$\sum_{i,j} P_{ji} f(i)g(j)\pi_j=
\sum_j g(j)\Big(\sum_iP_{ji}f(i)\Big)\pi_j=
\langle Pf, g \rangle$$

where the change of the order of the summation is allowed because the double sum would converge absolutely, conditional on ##g(j)## and ##f(i)## being bounded in ##\mathbb{R}^n##. (And because ##P_{ij}\leq 1## for all ##i,j,## and ##\sum_j\pi_j=1##.)