1
$\begingroup$

I have a very simple algorithm, and I need to calculate the derivative of an error function, and it gets a bit messy with chain rule.

I have a question whether I'm doing this correctly. To be more precise, my question is why is this correct. I'm taking this from an oxford lecture so I assume it isn't wrong, but I would derive it differently.

The equations:

We have equation for states: $$s_t = \theta_s\phi(s_{t-1})+\theta_x x_t$$

Where $\theta_s$ is the weight associated with $s$ the states, $\theta_x$ is the weight associated with $x$ our input, and $\phi$ is some differentiable function.

We also have an equation for the output of the machine $$y_t = \theta_y\phi(s_t)$$

Furthermore we have error functions for each time step:

$E_t = \frac{1}{2}(y_t-x_t)^2$ and a total error function: $E = \sum_{t=1}^{n}E_t$

So it's fairly simple. we have inputs and we want our outputs to be similar. Our outputs are dependent on states, and the states are a function of the previous state and current input.

The derivative:

I want to find $$\frac{\partial}{\partial \theta_s}E = \sum_{t=1}^{n}\frac{\partial}{\partial \theta_s}E_t$$

From chain rule (according to the lecture), $$\frac{\partial}{\partial \theta_s}E_t = \frac{\partial E_t}{\partial y_t}\frac{\partial y_t}{\partial s_t}\sum_{k=1}^{t}\frac{\partial s_t}{\partial s_k}\frac{\partial s_k}{\partial \theta_s}$$

My question is why this is true

Why not $$\frac{\partial}{\partial \theta_s}E_t = \frac{\partial E_t}{\partial y_t} \frac{\partial y_t}{\partial s_t} \frac{\partial s_1}{\partial \theta_s} \prod_{k=2}^{t}\frac{\partial s_k}{\partial s_{k-1}}$$

It seems like there are many other ways to represent this derivative with chain rule. are they all equal? Why is the form the professor chose correct? and not what i proposed?

$\endgroup$
0
$\begingroup$

This whole thing is a major abuse of notation. The $s_i$ are not a set of independent variables, but you and your professor are writing partial derivatives as if they were, without clear indications of what's being kept constant. I suspect that this contributed to your confusion.

The most direct way to sort this out is to write out $s_t$ as $s_t=\theta_s\phi\left(\theta_s\phi(\ldots)+\theta_xx_{t-1}\right)+\theta_xx_t$ and just differentiate that with respect to $\theta_s$; then you see that you get a sum. Your product is just one of the terms in that sum, namely the $k=1$ term that corresponds to the innermost instance of $\theta_s$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.