Deep Learning Specialization
Neural Networks and Deep Learning
Supervised Learning with Neural Networks
Examples:
Input  Output  Application  Algorithm 

Home features  Price  Real Estate  Standard NN 
Ad, user info  Click on ad? (0/1)  Online Advertising  Standard NN 
Image  Object (1,…,1000)  Photo tagging  Convolutional NN 
Audio  Text transcript  Speech recognition  Recurrent NN 
English  Chinese  Machine translation  Recurrent NN 
Image, Radar info  Position of other cars  Autonomous driving  Hybrid 
Logistic Regression as a Neural Network
\[w,b \quad\rightarrow\quad z=w^Tx+b \quad\rightarrow\quad a=\sigma(z) \quad\rightarrow\quad L(a,y)\]Given $x$,
\[\hat{y}=a=\sigma(w^Tx+b)\]with $\sigma(z)=\frac{1}{1+e^{z}}$ and $\sigma’(z)=\sigma(z)(1\sigma(z))$
Loss (error) function is:
\[L(a,y)=(y\log a + (1y)\log(1a))\]Cost function is:
\[J(w,b)=\frac{1}{m}\sum\limits_{i=1}^mL(a^{(i)},y^{(i)})\]Derivatives:
\[\frac{\partial L}{\partial w_i} = \frac{\partial L}{\partial a}\frac{\partial a}{\partial z}\frac{\partial z}{\partial w_i}=(\frac{y}{a}+\frac{1y}{1a})~\cdot~a(1a)~\cdot~x_i\]Neural Network
The first layer $a^{[0] (i)}=x^{(i)}$ is the input layer, the last layer $a^{[L] (i)}=\hat{y}^{(i)}$ is the output layer and the layers between them are hidden layers.
For $a^{[l] (i)}$ and $a_k^{[l+1] (i)}$ we have parameters $w_k^{[l+1]}$ and $b_k^{[l+1]}$ such that
\[z_k^{[l+1] (i)} = (w_k^{[l+1]})^T a^{[l] (i)} + b_k^{[l+1]}\] \[a_k^{[l+1] (i)} = g^{[l+1]}(z_k^{[l+1] (i)})\]where $g(z)$ is the activation function.
So by noting
\[W^{[l]}=\begin{bmatrix} (w_1^{[l]})^T\\ (w_2^{[l]})^T\\ \vdots\\ (w_{n^{[l]}}^{[l]})^T \end{bmatrix}, \quad b^{[l]}=\begin{bmatrix} b_1^{[l]}\\ b_2^{[l]}\\ \vdots\\ b_{n^{[l]}}^{[l]} \end{bmatrix}, \quad z^{[l] (i)}=\begin{bmatrix} z_1^{[l] (i)}\\ z_2^{[l] (i)}\\ \vdots\\ z_{n^{[l]}}^{[l] (i)} \end{bmatrix} \quad a^{[l] (i)}=\begin{bmatrix} a_1^{[l] (i)}\\ a_2^{[l] (i)}\\ \vdots\\ a_{n^{[l]}}^{[l] (i)} \end{bmatrix}\]we have:
\[z^{[l+1] (i)} = W^{[l+1]} a^{[l] (i)} + b^{[l+1]}\] \[a^{[l+1] (i)} = g^{[l+1]}(z^{[l+1] (i)})\]We can continue vectorizing by noting:
\[X=\begin{bmatrix} x^{(1)} & x^{(2)} & \cdots & x^{(m)} \end{bmatrix}\] \[Z^{[l]}=\begin{bmatrix} z^{[l] (1)} & z^{[l] (2)} & \cdots & z^{[l] (m)} \end{bmatrix}\] \[A^{[l]}=\begin{bmatrix} a^{[l] (1)} & a^{[l] (2)} & \cdots & a^{[l] (m)} \end{bmatrix}\]then (in fact the vector $b^{[l+1]}$ is added to each column)
\[Z^{[l+1]} = W^{[l+1]} A^{[l]} + b^{[l+1]}\] \[A^{[l+1]} = g^{[l+1]}(Z^{[l+1]})\]where
 $Z^{[l+1]}$ is of size $(n^{[l+1]},m)$
 $W^{[l+1]}$ is of size $(n^{[l+1]},n^{[l]})$
 $A^{[l]}$ is of size $(n^{[l]},m)$
 $b^{[l+1]}$ is of size $(n^{[l+1]},1)$
 $m$ is the number of samples
 $n^{[l]}$ is the number of neural units in lth layer
It’s a forward propagation.
Activation functions

sigmoid
\[g(z) = \frac{1}{1+e^{z}},\quad g'(z)=g(z)(1g(z))\] 
tanh (better than sigmoid)
\[g(z) = \tanh(z) = \frac{e^ze^{z}}{e^z+e^{z}},\quad g'(z)=1g^2(z)\] 
ReLU (REctified Linear Unit) (Recommanded, faster than tanh)
\[g(z) = \max(0,z),\quad g'(z)\begin{cases}0 & z<0\\ 1 & z \geq 0\end{cases}\] 
Leaky ReLU (i.e. $k=0.01$)
\[g(z) = \max(kz,z),\quad g'(z)\begin{cases}k & z<0\\ 1 & z \geq 0\end{cases}\]
Gradient descent
For the cost function $J(W^{[l]},b^{[l]})=\frac{1}{m}\sum\limits_{i=1}^m L(\hat{y}^{(i)},y^{(i)})$, we note
\[dV := \frac{\partial J}{\partial V}\]and we can update variables like $V = V  \alpha dV$
To calculate the gradient, since
\[z_k^{[l+1] (i)} = (w_k^{[l+1]})^T a^{[l] (i)} + b_k^{[l+1]}\] \[a_k^{[l+1] (i)} = g^{[l+1]}(z_k^{[l+1] (i)})\]We have
\[\begin{align*} dz_k^{[l+1] (i)}&=\frac{\partial J}{\partial a_k^{[l+1] (i)}} \frac{\partial A^{[l+1]}}{\partial z_k^{[l+1] (i)}} = da_k^{[l+1] (i)} {g^{[l+1]}}'(z_k^{[l+1] (i)}) \\ dw_k^{[l+1]}&=\frac{\partial J}{\partial z_k^{[l+1] (i)}} \frac{\partial z_k^{[l+1] (i)}}{\partial w_k^{[l+1]}} = dz_k^{[l+1] (i)} a^{[l] (i)} \\ db_k^{[l+1]}&=\frac{\partial J}{\partial z_k^{[l+1] (i)}} \frac{\partial z_k^{[l+1] (i)}}{\partial b_k^{[l+1]}} = dz_k^{[l+1] (i)} \\ da^{[l] (i)}&=\sum\limits_k \frac{\partial J}{\partial z_k^{[l+1] (i)}} \frac{\partial z_k^{[l+1] (i)}}{\partial a^{[l] (i)}} = \sum\limits_k dz_k^{[l+1] (i)} w_k^{[l+1]} \end{align*}\]Since
\[W^{[l]}=\begin{bmatrix} (w_1^{[l]})^T\\ (w_2^{[l]})^T\\ \vdots\\ (w_{n^{[l]}}^{[l]})^T \end{bmatrix}, \quad b^{[l]}=\begin{bmatrix} b_1^{[l]}\\ b_2^{[l]}\\ \vdots\\ b_{n^{[l]}}^{[l]} \end{bmatrix}, \quad z^{[l] (i)}=\begin{bmatrix} z_1^{[l] (i)}\\ z_2^{[l] (i)}\\ \vdots\\ z_{n^{[l]}}^{[l] (i)} \end{bmatrix} \quad a^{[l] (i)}=\begin{bmatrix} a_1^{[l] (i)}\\ a_2^{[l] (i)}\\ \vdots\\ a_{n^{[l]}}^{[l] (i)} \end{bmatrix}\]we now have, for the ith example,
\[\begin{align*} dz^{[l+1] (i)}&= da^{[l+1] (i)}~\circ~{g^{[l+1]}}'(z^{[l+1] (i)}) \\ dW^{[l+1]}&= dz^{[l+1] (i)} (a^{[l] (i)})^T \\ db^{[l+1]}&= dz^{[l+1] (i)} \\ da^{[l] (i)}&=(W^{[l+1]})^Tdz^{[l+1] (i)} \end{align*}\]By continuing vectorizing with the help of:
\[X=\begin{bmatrix} x^{(1)} & x^{(2)} & \cdots & x^{(m)} \end{bmatrix}\] \[Z^{[l]}=\begin{bmatrix} z^{[l] (1)} & z^{[l] (2)} & \cdots & z^{[l] (m)} \end{bmatrix}\] \[A^{[l]}=\begin{bmatrix} a^{[l] (1)} & a^{[l] (2)} & \cdots & a^{[l] (m)} \end{bmatrix}\]we have,
\[\begin{align*} dZ^{[l+1]}&= dA^{[l+1]}~\circ~{g^{[l+1]}}'(Z^{[l+1]}) \\ dW^{[l+1]}&= \frac{1}{m}dZ^{[l+1]}(A^{[l+1]})^T \\ db^{[l+1]}&= \frac{1}{m}dZ^{[l+1]} [1]_{(m,1)} \\ dA^{[l]}&= (W^{[l+1]})^TdZ^{[l+1]} \end{align*}\]Here $\circ$ means elementwise multiplication, $[1]_{(m,1)}$ means ones((m,1))
and we calculate the average over m examples for $dW, db$.
It’s a backpropagation.
Initialization
We need to initialize all the parameters randomly.
It’s recommanded to initialize $W$ with small value, for example, like [0.01,0.01] and initialize $b$ with zero. We initialize $W$ with small value to make sure that $Z$ will not be too big.
If we initialize all of them with zero, then in each layer all the neural units will act in the same way.
Hyperparameters
The learning rate $\alpha$, the number of iterations, the number of hidden layers and the size of each hidden layer, the choice of activation function, momentum term, mini batch size, regularization parameters, etc. These are all hyperparameters.
Gradient checking
It’s for DEBUG not for training
It doesn’t work with dropout !
 Reshape all the parameters $W^{[l]}, b^{[l]}$ into a big vector $\theta$.
 Reshape all the parameters $aW^{[l]}, ab^{[l]}$ into a big vector $d\theta$.
 Check whether $d\theta$ is the gradient of $J(\theta)$ ?
 Calculate $d\theta_{\text{approx}}[i] = \frac{J(\cdots,\theta_i+\varepsilon,\cdots)  J(\cdots,\theta_i\varepsilon,\cdots)}{2\varepsilon}$
 Compare $d\theta_{\text{approx}}$ and $d\theta$
If we take $\varepsilon=10^{7}$, the L2 error is about $10^{7}$, it’s good.
If grad check fails, try to identify the location of error.
Hyperparameter tuning, Regularization and Optimization
Basic Recipe for Machine Learning:
 High Bias ? (Training data performance)
 Bigger Network (normally will not hurt variance)
 Train longer
 NN architecture search
 High variance ? (Dev set performance)
 More data (normally will not hurt bias)
 Regularization
 NN architecture search
Regularization
L2 regularization
Logistic regression
\[J(w,b)=\frac{1}{m}\sum\limits_{i=1}^mL(\hat{y}^{(i)},y^{(i)})+\frac{\lambda}{2m}\w\_2^2\]The part of $b$ is omit and if we use $L^1$ norm then $w$ will be sparse.
Neural nerwork
\[J(W^{[l]},b^{[l]})=\frac{1}{m}\sum\limits_{i=1}^mL(\hat{y}^{(i)},y^{(i)})+\frac{\lambda}{2m}\sum\limits_{l=1}^L\W^{[l]}\_F^2\]where the Frobenius norm is defined as:
\[\A\_F^2 = \sum\limits_i \sum\limits_j A_{ij}^2\]So now the gradient of $W^{[l]}$ is:
\[dW^{[l]} = \frac{1}{m}dZ^{[l]}(A^{[l]})^T + \frac{\lambda}{m}W^{[l]}\]Inverted Dropout regularization
For example, for the lth layer we keep each unit with a probability $p$, so we can change $A^{[l]}$ as:
\[A^{[l]} = A^{[l]}~\circ~(\text{rand}((n^{[l]}, m)) < p)\] \[A^{[l]} = \frac{1}{p}A^{[l]}\]For example, there are 50 units and $p=0.8$. So in average there are 10 units shut off. So $A^{[l]}$ is reduced by 20% (20% element of A is zero).
But since $Z^{[l+1]} = W^{[l+1]} A^{[l]} + b^{[l+1]}$, in order to preserve the expected value of $A^{[l]}$, we need to divide it by $p$.
Attention, at test time we do not use dropout.
It’s recommanded for set different keep probability $p$ for each layer, if the size of $A^{[l]}$ is big, we can set $p$ small (i.e. 0.5).
Also, remember that dropout is to prevent overfitting, so if there is no overfitting, we do not have to apply this. For computer vision, we always don’t have enough data so overfitting is often an issue.
Other methods
Data augmentation
For example, we can rotate/flip/cut images to make new data.
Early stopping
Plot training error or $J$ and dev set error, if dev set error increase while training error decrease, we stop.
Optimization
Normalizing training sets
Reduce mean to zero ($x=x\mu$) and variance ($x=x/sigma$) to one. Attention ! Need to use the same parameters to normalize dev/test set.
Do this all the times since there is no harm and we are not sure if we do not need it.
Vanishing/Exploding gradients
When neural network is deep, sometimes the gradient is too small/big.
For example, if all hidden layers is of same size, if $g(z)=z$, $b^{[l]}=0$, $W^{[l]}=W$ then
\[\hat{y}=W^{[L]}W^{L1}X\]It can be very large or small according to its eigenvalues.
Weight initialization for deep networks
A partial solution for gradient vanishing/exploding problems
We know that $W^{[L]}$ is of size $(n^{[l]},n^{[l1]})$, to initialize it (for tanh activation) (Xavier initialization):
\[W^{[L]} = \sqrt{\frac{1}{n^{[l1]}}} \text{rand}((n^{[l]},n^{[l1]}))\]or
\[W^{[L]} = \sqrt{\frac{2}{n^{[l1]}+n^{[l]}}} \text{rand}((n^{[l]},n^{[l1]}))\]If we use ReLU activation functions,
\[W^{[L]} = \sqrt{\frac{2}{n^{[l1]}}} \text{rand}((n^{[l]},n^{[l1]}))\]Minibatch gradient descent
Split the training dataset $X$ into minibatches $X^{{t}}$, and use only a minibatch to calculate the gradient. By this way, the algorithme start to make progress even before finishing the entire giant training set.
Typically the minibatch size is 64, 128, 256 or 512.
Gradient descent with momentum
Exponentially weighted averages
Given data $\theta_1, \theta_2, \cdots$, we can estimate as
\[v_0=0,\quad v_t=\beta v_{t1}+(1\beta)\theta_t\]$v_t$ is an approximate average over $\frac{1}{1\beta}$ days.
In fact,
\[v_t=(1\beta)\theta_t+\beta v_{t1}=(1\beta)\theta_t+(1\beta)\beta\theta_{t1}+\cdots+(1\beta)\beta^{ts}\theta_{s}+\cdots+(1\beta)\beta^{t1}\theta_1\]and we have $\lim\limits_{\varepsilon\rightarrow 0}(1\varepsilon)^{\frac{1}{\varepsilon}}=\frac{1}{e}$, so $\beta^{\frac{1}{1\beta}}\approx\frac{1}{e}$
Bias correction
We can see that $v_t$ is a linear combination of $\theta_i$ and the sum of coefficients is $1\beta^t$. When $t$ is small, $v_t$ is too small and it’s not accurate enough. So we can correct the bias as following:
\[\frac{v_t}{1\beta^t}=(1\beta)\sum\limits_{s=0}^{t1}\beta^s\theta_{ts}\]Momentum
Idea is to average the gradient to prevent oscillation. Then the algorithme may be accelerated.
$V_{dW} = 0, V_{db} = 0$
On iteration $t$:
 Compute $dW$, $db$ using minibatch

$V_{dW} = \beta V_{dW}+(1\beta) dW$
$V_{db} = \beta V_{db}+(1\beta) db$

$W = W  \alpha V_{dW}$
$b = b  \alpha V_{db}$
Practically, we don’t use bias correction for exponentially weighted averages and $\beta = 0.9$.
RMSprop
$S_{dW} = 0, S_{db} = 0$
On iteration $t$:
 Compute $dW$, $db$ using minibatch

$S_{dW} = \beta S_{dW}+(1\beta) dW^2$
$S_{db} = \beta S_{db}+(1\beta) db^2$

$W = W  \alpha \frac{dW}{\sqrt{S_{dW}}+\varepsilon}$
$b = b  \alpha \frac{db}{\sqrt{S_{db}}+\varepsilon}$
Idea is that : when an element of gradient is too big/small, for example $db$ is too big, then by dividing by $\sqrt{S_{db}}$, $db$ will be smaller. The $\varepsilon$ is to prevent division by zero, normally $\varepsilon=10^{8}$.
By using RMSprop, we can use larger $\alpha$.
Adam
Combine Momentum and RMSprop
Adam : Adaptive Moment Estimation
$V_{dW} = 0, V_{db} = 0, S_{dW} = 0, S_{db} = 0$
On iteration $t$:
 Compute $dW$, $db$ using minibatch

$V_{dW} = \beta_1 V_{dW}+(1\beta_1) dW$
$V_{db} = \beta_1 V_{db}+(1\beta_1) db$
$S_{dW} = \beta_2 S_{dW}+(1\beta_2) dW^2$
$S_{db} = \beta_2 S_{db}+(1\beta_2) db^2$
 Bias correction
 $V_{dW} = \frac{V_{dW}}{1\beta_1^t},\quad V_{db} = \frac{V_{db}}{1\beta_1^t}$
 $S_{dW} = \frac{S_{dW}}{1\beta_2^t},\quad S_{db} = \frac{S_{db}}{1\beta_2^t}$

$W = W  \alpha \frac{V_{dW}}{\sqrt{S_{dW}}+\varepsilon}$
$b = b  \alpha \frac{V_{db}}{\sqrt{S_{db}}+\varepsilon}$
Hyperparameters:
 $\alpha$ need to be tune
 $\beta_1 = 0.9$
 $\beta_2 = 0.999$
 $\varepsilon = 10^{8}$
Learning rate decay
We note one epoch is one pass through the data, then we can decrease learning rate $\alpha$ as following:
\[\alpha = \frac{1}{1+\text{decay rate}~*~\text{epoch num}}\alpha_0\]or
 $\alpha = \text{decay rate}^\text{epoch num}\alpha_0$
 $\alpha = \frac{k}{\sqrt{\text{epoch num}}}\alpha_0$
Hyperparameter tuning
Hyperparameters:
 learning rate
 minibatch size, number of hidden units
 etc.
Try random values instead of using a grid to tune. When finding a good point, we can zoom in and sample more densily to tune.
Appropriate scale
 number of layers : 2,3,4, sample on linear scale.
 learning rate $\alpha$ : $10^{4}$ ~ $1$, sample on log scale.
Batch Normalization
In NN, can we normalize $z^{[l]}$ (or $a^{[l]}$) to train $w^{[l]},b^{[l]}$ faster.
\[z_{\text{norm}}^{[l] (i)} = \frac{z^{[l] (i)}  \mu}{\sqrt{\sigma^2+\varepsilon}}\] \[\tilde{z}^{[l] (i)} = \gamma z_{\text{norm}}^{[l] (i)} + \beta\]Here we normalize $z^{[l] (i)}$ then we adjust it to have a specific variance $\gamma^2$ and bias $\beta$.
\[X \xrightarrow{W^{[1]},b^{[1]}} Z^{[1]} \xrightarrow{\gamma^{[1]},\beta^{[1]}} \tilde{Z}^{[1]} \rightarrow a^{[1]} \rightarrow \cdots\]Now, the parameters are $W^{[l]},b^{[l]},\gamma^{[l]},\beta^{[l]}$. We need to learn $\gamma^{[l]},\beta^{[l]}$ and they are of size ($n^{[l]}$,$1$).
In fact, we know that $Z^{[l+1]} = W^{[l+1]} A^{[l]} + b^{[l+1]}$, but since we will normalize $Z^{[l+1]}$, $b^{[l+1]}$ is no longer useful. So we can get rid of it.
Moreover, when using minibatch, we normalize $Z^{[l]}$ by only using the data on the minibatch. When testing, we need to calculate the $\mu, \sigma$ differently.
It’s proposed to calculate $\mu, \sigma$ for each mini batch and then we use exponentially weighted averages to get $\mu, \sigma$ for the entire dataset.
Multiclass classification
Softmax layer
hard max : [1, 0, 0, 0]
soft max : [0.8, 0.1, 0.002, 0.098]
To do multiclass classification, we change the last activation function.
We know that $z^{[L]} = W^{[L]} a^{[L1]} + b^{[L]}$, then we define $a^{[L]}$ as follows:
\[t = \exp(z^{[L]}), \quad \hat{y}=a^{[L]}=\frac{t}{\sum t_i}\]then $a_i^{[L]}$ represents the possibility that the sample belongs to class i.
If we define $a^{[L]}=g^{[L]}(z^{[L]})$, the function $g^{[L]}$ takes a vector as input and the output is also a vector. It’s different from the other activation functions that we have seen.
Loss function
For example, we have
\[y=\begin{bmatrix}0\\1\\0\\0\end{bmatrix} ,\quad \hat{y}=\begin{bmatrix}0.3\\0.2\\0.1\\0.4\end{bmatrix}\]then the loss function
\[L(\hat{y},y)=\sum\limits_j y_j\log\hat{y}_j=\log(0.2)\]Deep learning frameworks
 Caffe / Caffe 2
 CNTK
 DL4J
 Keras
 Lasagne
 mxnet
 PaddlePaddle
 TensorFlow
 Theano
 Torch
How to choose ?
 Ease of programming (development and deployment)
 Running speed
 Truly open (open source with good governance)
Structuring Machine Learning Projects

Single number evaluation metric
Use a single number to evaluate the performance of algorithms

Satisficing and Optimizing metric
Given n metrics, we optimize over one metric and give constraints that the other n1 metrics have to satisfy
i.e., we optimize the accuracy and ask the run time to be <100ms

Train/dev/test distribution
Have to make sure that the data in train/dev/test set comes from a same distribution
i.e., if we put data of USA in training set and put data of China in dev set, it will not work :(

Size of the dev/test sets
Before, we can divide the data by 6:2:2 for train/dev/test
But now, if the data size is very huge, we may divide the data by 98 : 1 : 1

When to change dev/test sets and metrics
For cat classification, if algo A has 3% error and algo B has 5% but A show some pornographic image, we can increase the weight of these pornographic in the metric
 Humanlevel performance
avoidable bias is the difference between humanlevel error and training error
variance is the difference between training error and dev error

When the error of algorithm is much worse than humanlevel performance, we focus on (avoidable) bias to reduce error. But if the error of algorithm is at the same level as human, we focus on the variance to reduce the difference of errors over train and dev set.

Humanlevel error as a proxy for Bayes error
Bayes error rate is the lowest possible error rate for any classifier of a random outcome


Error Analysis

Cleaning up incorrectly labeled data

DL algorithms are quite robust to random errors in the training set but not to systematic errors (like classify dogs as cats)

Examine the data to figure out the reason of errors (incorrectly labeled data in dev set or for example classify lions as cats, etc.) and estimate the improvement (if it’s worth enough to fix it)



Mismatched training and dev/test set

Training and testing on different distributions
For example, we’d like to make an app to classify cats and we have 200K images from webpages and 10K images from mobile apps.
Instead of putting all data together and shuffle to prepare train/dev/test set. It’s better to divide like this since we care more about the performance on app:
 Train: 200K web data and 5K app data
 Dev: 2.5K app data
 Test: 2.5K app data

Bias and Variance with mismatched data distributions
For example, training set has a different distribution than dev/test set and humanlevel error is near 0, training error is 1% and dev error is 10%.
In order to find out the reason of the avoidable bias, we can divide a small part of traning data as trainingdev set and we train on the remaining train set then study the error on trainingdev/dev/test set.
If train error is 1%, trainingdev error is 9%, dev error is 10%, the problem is that algorithm generalize not well.
If train error is 1%, trainingdev error is 1.5%, dev error is 10%, then it’s a data mismatched problem, the distribution of train/dev data is not the same.
If train error is 10%, trainingdev error is 11%, dev error is 12%, there is a high bias problem.
If train error is 10%, trainingdev error is 11%, dev error is 20%, we have a high bias problem as well as a mismatched problem.

Addressing data mismatch
We can make artificial data. For example, we have 10K hours audio data and 1 hour car noise, we can make sythesized auto by adding noise into data. BUT ! Attention that the algorithm may overfit to the 1 hour noise even though all car noise seems the same to human.


Learning from multiple tasks

Transfer learning
For example, use the neural network of image recognition to do radiology diagnosis. We only need to change the parameters of the last layer or add some more layers to fit to the new training data.
It’s also called pretraining and finetuning.
When to use transfer learning ?
Transfer from A to B

Task A and B have the same input x

Have a lot more data for task A than B

Low level features from A could be helpful for learning B


Multitask learning
Use a single network to learn do multitask, i.e. y is multidimensional.
When multitask makes sense ?

Training on a set of tasks that could benefit from having shared lowerlevel features.

Amount of data for each task is quite similar. For example, we have 100 tasks and each task has only 1K examples.

We can train a big enough neural network to do well on all the tasks.



Endtoend deep learning
For example, to translate audio to transcript, we can do multiple stages like: audio > feature > phoneme > word > transcript or simply use a neural network to translate from audio directly to transcript.
Endtoend learning may need learge amount of data and sometimes it’s useful to have handdesigned components.