Comments (7)
wq2012
commented on July 28, 2024
@AnzCol Hi Aonan, what are your thoughts here?
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AnzCol
commented on July 28, 2024
Hi,
The output for GRU are those m_s (see Fig.2). We use a running average of
m_{1,2,...,s} to estimate the observation x_s (see the inline equation
below Eq.(11)).
Thanks,
Aonan
…On Thu, Mar 28, 2019 at 11:31 AM Quan Wang ***@***.***> wrote:
@AnzCol <https://github.com/AnzCol> Hi Aonan, what are your thoughts here?
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wq2012
commented on July 28, 2024
@AnzCol I think what @hbredin means is - what if we simply define m_t=x_t, will it still work? Did we have such experiments (my impression is no)?
Personally I don't think it's going to work well.
My understanding is that (@AnzCol please correct me if I'm wrong), the training process forces m_t for each speaker to better fall into a normal distribution. But this is not guaranteed in the distributions of x_t. The power of GRU here is that, to transform the distributions of speaker embeddings into a more clusterable distribution, by learning from the training dataset.
@hbredin Does this explanation make sense to you?
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hbredin
commented on July 28, 2024
@AnzCol I think what @hbredin means is - what if we simply define
m_t=x_t, will it still work? Did we have such experiments (my impression is no)?
This is what I meant, indeed.
My understanding is that (@AnzCol please correct me if I'm wrong), the training process forces
m_tfor each speaker to better fall into a normal distribution. But this is not guaranteed in the distributions ofx_t. The power of GRU here is that, to transform the distributions of speaker embeddings into a more clusterable distribution, by learning from the training dataset.
Except you are still using raw x_t in Equation 11, so the distribution of speaker embeddings is not changed. Or did I miss something?
@hbredin Does this explanation make sense to you?
Not quite sure -- I think I have to think a bit more about this...
I would really like to see an ablative study with m_t = x_t :-)
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zan12
commented on July 28, 2024
Thanks for clarifying your question. Let me make a complement according to
my experience. I personally did some unpublished experiments when m_t=x_t.
Under best hyper-parameter settings, there was a performance drop around 3%
comparing to the current case when m_t is only used to calculate means.
Quan's explanation is very intuitive indeed. Another simple explanation is
that use empirical mean seems to be more statistically stable.
Hervé BREDIN <[email protected]> 于2019年3月28日周四 下午3:41写道：
… @AnzCol <https://github.com/AnzCol> I think what @hbredin
<https://github.com/hbredin> means is - what if we simply define m_t=x_t,
will it still work? Did we have such experiments (my impression is no)?
This is what I meant, indeed.
My understanding is that ***@***.*** <https://github.com/AnzCol> please
correct me if I'm wrong), the training process forces m_t for each
speaker to better fall into a normal distribution. But this is not
guaranteed in the distributions of x_t. The power of GRU here is that, to *transform
the distributions of speaker embeddings into a more clusterable
distribution*, by learning from the training dataset.
Except you are still using raw x_t in Equation 11, so the distribution of
speaker embeddings is not changed. Or did I miss something?
@hbredin <https://github.com/hbredin> Does this explanation make sense to
you?
Not quite sure -- I think I have to think a bit more about this...
I would really like to see an ablative study with m_t = x_t :-)
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wq2012
commented on July 28, 2024
@hbredin Not sure whether this example makes sense: consider two clusters, their distributions of x_t largely overlap with each other, but their distributions of m_t are better separated. Eq. 11 regularizes that m_t should not disjoint too much from x_t.
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hbredin
commented on July 28, 2024
Closing as I got the answers I was looking for :-)
Thanks @AnzCol and @wq2012 !
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