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In scaled dot product attention, we scale our outputs by dividing the dot product by the square root of the dimensionality of the matrix:

enter image description here

The reason why is stated that this constrains the distribution of the weights of the output to have a standard deviation of 1.

Quoted from https://www.tensorflow.org/tutorials/text/transformer:

For example, consider that $Q$ and $K$ have a mean of 0 and variance of 1. Their matrix multiplication will have a mean of 0 and variance of $d_k$. Hence, square root of $d_k$ is used for scaling (and not any other number) because the matmul of $Q$ and $K$ should have a mean of 0 and variance of 1, and you get a gentler softmax.

Why does this multiplication have a variance of $d_k$?

If I understand this, I will then understand why dividing by $\sqrt({d_k})$ would normalize to 1.

Trying this experiment on 2x2 arrays I get an output of 1.6 variance:

enter image description here

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It might help to take two small matrices that match the assumptions (mean of zero and variance of one) and just do the matrix multiplication. The dimensionality of K scales Q in the multiplication, scaling the variance simultaneously.

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  • $\begingroup$ Hey, I did do this but it does not come out as expected. I have edited the post $\endgroup$ – Jacob B May 21 at 1:09

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