# The Vanishing Gradient Problem

October 9th, 2016While neural networks are sometimes intimidating structures, the mechanism for making them work is surprisingly simple: stochastic gradient descent. For each of the parameters in our network (such as weights or biases), all we have to do is calculate the derivative of the loss with respect to the parameter, and nudge it a little bit in the opposite direction.

##### Disappearing gradients

Stochastic gradient descent seems simple enough, but in many networks we might begin to notice something odd: the weights closer to the end of the network change a lot more than those at the beginning. And the deeper the network, the less and less the beginning layers change. This is problematic, because our weights are initialized randomly. If they're barely moving, they're never going to reach the right values, or it'll take them years.

I trained a simple fully connected network to classify MNIST images to illustrate this point. Here, we can see how the gradients change over time for a network with one input layer and two hidden layers:

Notice how the first layer's gradients are much lower that the third layer's, which means those weights are changing by a much smaller amount. If we add more layers, the difference only gets more dramatic. The whole rest of the network is affected by what comes out of the first layer, so if those first weights are totally wrong, our network is not going to perform well.

Here's an iPython notebook implementing this simple network in Tensorflow and plotting the gradients. Feel free to play with the number and size of layers to see how bad it can get.

##### Why do gradients vanish?

Let’s imagine we have a 3-layer network, initialized with some set of weights and activations. For simplicity, let’s envision that each layer has one node.

At each step, the weight first is multiplied by the input at that step (in the following diagram, these intermediate quantities are called $h_i$ for node $i$). There are also 'bias terms' added at each step, which go with the weights but aren't explicitly drawn on each arrow.

Then, the node outputs a function of this product. This result is the output of the node, which is denoted as $z_i$ for node $i$ in the following diagram:

In practice, this function might be a sigmoid, inverse tangent, or ReLU function, but for now we’ll just refer to it as $f$. The final output $\hat{y}$ might be calculated using a different function $g$, such as a softmax.

At the end of this network we end up with a loss, or a measure the difference between what we expected to see and what our network actually outputted. This is commonly calculated with something like a cross-entropy function. For now, all we need to know is that the loss will be a function of $\hat{y}$, the output of the network.

To train the weights of this network, we can use gradient descent. The basic process of gradient descent is 1) calculate the gradient of the loss with respect to each weight and 2) shift the weight in the opposite direction. This process be written like:

Let's try this for the weight $w_3$. To apply this rule, the quantity we need to calulate is that last term, $\frac{\partial Loss}{\partial f(w_3)}$. We can use the chain rule to calculate this partial derivate. Because the loss is a function of $\hat{y}$, which is a function of $z_3$, which is a function of $w_3$, we can express the gradient like this:

We can just keep chain-ruling in the same way to get this quantity for $w_2$ and $w_1$:

Notice that the first two terms are always the same! What's changing are those middle terms (highlighted in yellow in the above image). The farther back the weight, the more of those middle terms we have. So what are those terms anyway? You might realize they're all of the general form $\frac{\partial f(z_{i+1})}{\partial f(z_{i})}$. Let's look at a specific one and try to expand it out:

We know that $z_3$ is just a $f(h_3)$, so we can use the chain rule to first take the derivative of the outer function, $f$, at $h_3$. Then, the second step of the chain rule is to multiply that by the the derivative of the inner function, $h_3$, with respect to $z_2$. We know $h_3$ is just $w_2 \cdot z_2 + b_2$, so the derivative of this with respect to $z_2$ is just $w_2$. This makes our final expression $f'(h_3) \cdot w_2$.

And in fact, this generalizes to all the partial derivates that look like $\frac{\partial f(z_{i+1})}{\partial f(z_{i})}$:

They're all made up of a product of those two terms: an $f'(h)$ and a $w$. So... what are those two terms again? $f$ is the function we use at each step: some common ones are ReLU, sigmoid, or tanh. $f'$ is the derivative of this function. The derivatives of these commonly used functions though, are pretty much always below 1. The second term, $w$, is a weight in the network. Often, weights are initialized from a standard normal distribution, which results in values that are usually below 1.

Here's the thing you may have realized: multiplying two small numbers less than 1 results in an even smaller fraction.

As you saw above, the gradients for earlier weights in the network contain increasingly more of these $\frac{\partial f(z_{i+1})}{\partial f(z_{i})}$ terms, which we now know are usually very small fractions. In a large network with $n$ layers, there are $n$ - $i$ of these terms:

In other words, the number of these terms increases the larger the network gets (the bigger $n$ is) and the farther back in the network the weight is (the smaller $i$ is). Because these are usually small fractions, ss the number of terms increases, the value of the gradient decreases. And this is the vanishing gradient problem in action.

So what are some things people do to combat this problem?

##### 1. Activation Functions

Remember that our vanishing gradient was arising from multiplying lots of $f’(h) \cdot w$ terms. This gives us some insight into why certain functions $f$ (called activation functions) might work better than others for combatting this problem.

For example, while the derivative of a sigmoid function is < 0.25 everywhere, making each term even smaller, the derivative of the ReLU function is one at every point above zero, creating a more stable network. This is also one of the reasons why the inverse tangent activation function is sometimes preferred over the sigmoid.

##### 2. Clipping Gradients

There's a small chance the opposite problem arises: if both $f'$ and $w$ happen to be larger numbers, the further back we go, the larger the gradient becomes. When this is extreme, it's called the *exploding gradient* problem. Pascanu et. al. provide a simple solution for exploding gradients: just scale them down whenever they pass above a certain threshold. See their paper for a more detailed geometrical interpretation of why this is an okay thing to do during stochastic gradient descent!

##### 3. LSTMs

As you might imagine, the vanishing gradient becomes a very important issue the deeper a network gets. One type of network that tends to be very deep are recurrent neural networks (RNNs). RNNs are used to model time-dependent data, like words in a sentence. We feed in words one by one, and the nodes in the network store their state at one timestep and use it to inform the next timestep.

If we think about each timestep as a layer, with weights going from one timestep to the next (this is often referred to as “unraveling” an RNN), we can see that our network will be at least as deep as the number of timesteps. When it comes to sentences, paragraphs, or other timeseries data, these sequences we’re feeding in can be very long, so we face the same problems that a very deep neural network would.

The first word fed into an RNN is equivalent to the first layer in the simple neural network from above. If we’re experiencing a vanishing gradient, the weights at the beginning of the network change less and less, and the RNN becomes worse at modeling long-term dependencies. If we’re predicting words of a sentence, the first word in the sentence might actually be really important context for predicting a word at the end, so we don’t want to lose that information.

LSTMs (Long Short-Term Memory Networks) are a special subset of RNNs that are able to deal with remembering information for much longer periods of time. The idea behind an LSTM is actually really simple! Rather than each hidden node being simply a node with a single activation function, each node is a memory cell that can store other information. Specifically, it maintains its own cell state. Normal RNNs take in their previous hidden state and the current input, and output a new hidden state. An LSTM does the same, except it also takes in its old cell state and outputs its new cell state.

So what’s this magic that goes on inside an LSTM memory cell? Let’s split it into three main steps.

1. We decide what from the previous cell state is worth remembering, and tell the cell state to **forget the stuff we decide is irrelevant**.

2. We **selectively update** the cell state based on the new input we’ve just seen.

3. We **selectively decide what part of the cell state we want to output** as the new hidden state.

This is all achieved by a few simple gates: the **forget gate**, the **input gate**, and the **output gate**.

Let's go through the steps with a specific example: translating the English sentence *When we go to France, you speak English but I speak French* to the French sentence *Quand nous allons à France, tu parles Anglais mais je parle Français*.

**Forget Gate**

In the first step, a function of the previous hidden state and the new input passes through the forget gate, letting us know what is probably irrelevant and can be taken out of our cell state. The forget gate will output values close to 1 for parts of the cell state we wish to completely keep, and zero for values we’d like to totally get rid of.

Let's say we're feeding in the example English sentence from above and see the word "you". Now we might like to forget the "we" that appeared previously, since the next verb will likely be conjugated according to the new subject "you".

**Input Gate**

In the second step, a function of the inputs passes through the input gate and is added to the cell state to update it. Following our scenario from above, we might want to add information to the cell state about the new word “you” we’ve just seen -- for example, the fact that it’s a subject, singular and second person.

**Output Gate**

In the final step, the output gate decides what values from our cell state we are going to add to the hidden state output.

In our example, if we expect the next word will be a verb, we might output the information about the current subject that will important for conjugating the verb -- for example, the fact that "you" is singular and second person.

At the same time, we can continue to hold onto things in the cell state that we think might be important not at the next time step, but at some point much later along -- like the fact that the sentence is set in France. This information might not be relevant to the verb appearing next, but it will be helpful information for the end of the sentence, which is about speaking French. The ability to preserve information in the cell state for long stretches of time is a big part of what makes LSTMs special.

The cell state allows an LSTM to surpass the vanishing gradient problem for two main reasons.

First: Remember how the sigmoid activation function always has a derivative less than 0.25? So when we multiply together all those $f’(x) \cdot w$ terms, our gradient just vanishes away? Well, if we look at the cell state in an LSTM, the only thing it's multiplied by is the output of the forget gate, so we can think of $f$ as the weights for the cell state. In that case, what’s the activation function? Technically, there isn’t one, besides the identity function itself. The derivative of the identity function is, conveniently, always one. **So if $f$ = 1, information from the previous cell state can pass through this step unchanged**.

Second: There is one more step, located in the center of the diagram, where we adjust the cell state. Notice that we adjust the cell state by adding some function of the inputs. When we backpropagate and take the derivative of $C_t$ with respect to $C_{t-1}$, **this added term just disappears**!

So, because the forget gate is essentially the weights *and* activation function for the cell state, and because the LSTM can learn to set that forget gate to one for important things in the cell state, information can pass through unchanged.

Because of their ability to capture long-term dependencies, LSTMs have gained a lot of popularity recently. For some interesting applications, see Google's machine translation technology, recent work on image generation, or the use of LSTMs to try and diagnose patients based on clinical timeseries. To learn more about how LSTMs can be extended to work even better, check out attention and memory networks.

Hopefully, this post has given you some insight into the difficulty of modeling long-term dependencies in data and why LSTMs work so well. Reach out with any questions to hsuresh@mit.edu!