Help me understand why this summation index is not j

  • #1
docnet
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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?



Screenshot 2024-02-12 at 9.58.44 PM.png

Screenshot 2024-02-12 at 9.58.50 PM.png
 
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  • #2
In the theory of Markov chains, don't they do the matrix-vector operation back-to-front? Just to be different.
 
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  • #3
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.. 😭 :bow::headbang:
 
  • #4
Obvious misprint.
 
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  • #5
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.
 
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  • #6
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.
 
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  • #7
Thank you ! you saved me from an existential crisis for now. :eek::oldsurprised:
 
  • #8
docnet said:
Thank you ! you saved me from an existential crisis for now. :eek::oldsurprised:
It all hinges on the "balance condition". ##P## is ##\pi##-self-adjoint iff that condition holds.
 
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  • #9
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.
 
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  • #10
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.
 
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  • #11
The way I would do this is simply.
$$\langle f,Pg \rangle = \sum_i f(i)\Big(\sum_jP_{ij}g(j)\Big)\pi_i = \sum_{i,j} f(i)g(j)\pi_iP_{ij}$$$$= \sum_{i,j} f(i)g(j)\pi_jP_{ji} =\sum_{i,j} P_{ji} f(i)g(j)\pi_j = \langle Pf, g \rangle$$
 
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  • #12
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##.)
 
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