Context:

Let's look at a simple example of why vanishing and exploding gradients occur in RNNs.

Consider a univariate version of RNN with the following update rules

𝑧
(
𝑡
)
=
𝑢
𝑥
(
𝑡
)
+
𝑤
ℎ
(
𝑡
−
1
)

ℎ
(
𝑡
)
=
𝜙
(
𝑧
(
𝑡
)
)

To keep things simple, let us assume
𝜙
is the identity function, i.e.,
𝜙
(
𝑖
)
=
𝑖

Consider we have the a final loss
𝐿
, and computed the derivative of
∂
𝐿
∂
ℎ
𝑇
for some
𝑡
=
𝑇

Using the update rules, the value of
∂
ℎ
𝑇
∂
ℎ
1
comes out to be
𝑤
(
𝑎
𝑇
+
𝑏
)

Main Question:

What is the value of a?

Question

Context:

Let's look at a simple example of why vanishing and exploding gradients occur in RNNs.

Consider a univariate version of RNN with the following update rules

𝑧
(
𝑡
)
=
𝑢
𝑥
(
𝑡
)
+
𝑤
ℎ
(
𝑡
−
1
)

ℎ
(
𝑡
)
=
𝜙
(
𝑧
(
𝑡
)
)

To keep things simple, let us assume 
𝜙
 is the identity function, i.e., 
𝜙
(
𝑖
)
=
𝑖

Consider we have the a final loss 
𝐿
, and computed the derivative of 
∂
𝐿
∂
ℎ
𝑇
 for some 
𝑡
=
𝑇

Using the update rules, the value of 
∂
ℎ
𝑇
∂
ℎ
1
  comes out to be 
𝑤
(
𝑎
𝑇
+
𝑏
)

Main Question:

What is the value of a?

BlackTom AI · Accepted Answer

We start by restating the problem setup and what is being asked.
Context: In a simple RNN with univariate x and identity activation, z(t) = u x(t) + w h(t−1) and h(t) = φ(z(t)). Since φ is the identity, h(t) = z(t) for all t. We consider a final loss L and the derivative ∂L/∂h_T for time T. It is given that this derivative takes the form ∂L/∂h_T = w (a_T + b), where a_T and b are terms that accumulate from the backpropagation through time.
Main idea: When we backpropagate the gradient from time T down to time 1 through the recurrence, each step multiplies by w due to the dependence h(t) on h(t−1) via z(t) and the identity activation. Specifically, in a chain of T to 1 steps, the gradient with respect to h_1 will accumulate a factor of w raised to the number of recurrences traversed.
Now, let us examine how many times the hidden state is fed back into itself from t = T down to t = 2 (i.e., how many backpropagation multiplications by w occur).
- At time T, the gradient contribution starts with ∂L/∂h_T.
- When propagating one step back, we multiply by w to get the contribution to ∂L/∂h_{T−1}.
- This process repeats for each step until we reach h_1, which involves propagating through (T − 1) transitions in total.
Therefore, the cumulative factor contributed by the recurrent connections is w^(T−1). The linear combination that results in ∂L/∂h_T being expressed as w (a_T + b) must account for this (T − 1) count in the exponent structure of a_T, with a_T capturing the sum of terms that scale with w^(T−1) and b capturing any bias-like offset.
Thus, the parameter a that appears in the expression for ∂L/∂h_T is tied to the number of backpropagation steps, which is (T − 1). In other words, a = T − 1.
If you compare with the general form for backpropagating through time in a linear recurrent unit, you typically see gradient terms accumulate as powers of w, and the exponent count equals the number of elapsed time steps, i.e., T−1, confirming why a equals T−1.
In the given setup, there are no alternative options to evaluate here, but the key takeaway is that the depth of temporal recurrence dictates the (T−1) factor in the gradient expression, hence a = T − 1.

类似问题

What problem does LSTM solve that basic RNNs struggle with?

What distinguishes RNNs from feedforward neural networks in terms of processing sequences?

Which of the following best describes the primary purpose of Long Short-Term Memory (LSTM) networks in deep learning?

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