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PDF Editor FAQ

What is the reparameterization trick in variational autoencoders?

In VAEs, you need to sample from a gaussian distribution in the middle of your network. The reparametrization consists of saying that sampling from [math]z \sim [/math][math][/math][math][/math][math]\mathcal{N}(\mu, \sigma)[/math] is equivalent to sampling [math]\varepsilon \sim \mathcal{N}(0, 1)[/math] and setting [math]z = \mu [/math][math][/math][math]+ \sigma \odot \varepsilon[/math]. As epsilon can be seen as an input of the network, you sampled while keeping your sampling operation differentiable!

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I agree with Bart's answer, it doesn't seem as if a normal operating agreement online would be able to address your unique situation. Creating a board of directors is not standard and it shouldn't be handled by yourself. Forming your operating agreement the right way is crucial because it lays the foundation for your entire company, you shouldn't take it lightly. You can always find us over at LawTrades - you can use our service to get low cost legal services from excellent business attorneys without the hassle.

What is the re-parametrization trick in Variational Autoencoders (VAE)?

During forward pass, you have an encoder that maps the input [math]x[/math] to a posterior distribution, typically assumed to be a normal distribution with mean [math]\mu(x)[/math] and standard deviation [math]\sigma(x)[/math]. Then, you sample a point [math]z[/math] from the distribution [math]\mathcal{N}(\mu(x), \sigma(x))[/math], and pass [math]z[/math] through a decoder.Now, when you have to backpropagate, the issue is that you cannot get a gradient of the sampled point w.r.t. [math]\mu(x)[/math] and [math]\sigma(x)[/math], because the sampling operation cannot be differentiated. So, what you do is, you sample a point [math]t[/math] from [math]\mathcal{N}(0, 1)[/math], and construct [math]z[/math] as [math]\mu(x) [/math][math][/math][math]+ t \cdot \sigma(x)[/math]. Now, you can pretend that the sampled value [math]t[/math] is a constant, and differentiate [math]z[/math] to get gradients w.r.t. [math]\mu(x)[/math] and [math]\sigma(x)[/math].To give an example, let’s say [math]\mu(x) = 3[/math], and [math]\sigma(x) = 0.4[/math]. So, without reparameterization, [math]z[/math] is sampled from [math]\mathcal{N}(3, 0.4), say 2.8[/math]. This doesn’t give you a differentiable function from [math]\mu(x)[/math] and [math]\sigma(x)[/math] to the value [math]2.8[/math]. But if you reparametrize, you would sample from [math]\mathcal{N}(0, 1)[/math], to get [math]-0.5[/math], and then write [math]\mu(x) - 0.5 \cdot \sigma(x)[/math], which can now be differentiated w.r.t. [math]\mu(x)[/math] and [math]\sigma(x)[/math]. Note that because the sampling operation now involves [math]\mathcal{N}(0, 1)[/math], which doesn’t have any parameters, you don’t need the derivative of this operation.As you do thousands of forward-backward passes, t takes different values, and in expectation, the gradient of the modified expression is the same as the gradient of the original expression.

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