Recap: Six jars

What we saw in the previous chapter?

(c) One Fourth Labs

Repeat the slide where we summarise the perceptron model in the context of the 6 jars.

Some thoughts on the contest

Just some ramblings!

(c) One Fourth Labs

We should create the binary classification monthly contest such that we can somehow visualize that the data is not linearly separable (don't know how)

We should also be able to visualize why the perceptron model is not fitting

Also give them some insights into what does the loss function indicate and how to make sense of it

Limitations of Perceptron

Can we plot the perceptron function ?

(c) One Fourth Labs

1. Show a training matrix with only 1 input (salary) and the decision is buy_car/not_buy_car

4. First only show x and y axis, label x axis as salary and y-axis as "\hat{y} = \hat{f}(salary)= perceptron(salary)"

5. Now take a positive point from the training data. Mark it on the x-axis and then mark the corresponding output for this point (+1)

6. Do this for a few points and then draw the step function

7, 8. Show the cartoon

9. Now on the x-axis show red and green regions for the positive and negative points

2. Show the perceptron model equations here

3. Suppose w=2, b =1 (comments for Mitesh: I took linearly separable data

7. Wait a minute!

8. Doesn't the perceptron divide the input space into positive and negative halves

Limitations of Perceptron

Can we plot the perceptron function ?

(c) One Fourth Labs

Wait a minute!

Salary ( in thousands)	Can buy a car?
80	1
20	0
65	1
15	0
30	0
49	0
51	1
87	1

Doesn't the perceptron divide the input space into positive and negative halves ?

y=\sum_{i=1}^n w_i x_i \geq b

\text{Suppose,} \\ \mathbb{w_1} = 2,\\ \hspace{-0.2cm}\mathbb{b}\ \ = 1

Limitations of Perceptron

How does the perceptron function look in 2 dimensions?

(c) One Fourth Labs

1. Show a training matrix with only 2 inputs (salary, family size) and the decision is buy_car/not_buy_car

4. First only show x1, x2 and y axis, label x1 axis as salary, x2 axis as family size and y-axis as "\hat{y} = \hat{f}(salary, size)= perceptron(salary, size)"

5. Now take a positive point from the training data. Mark it on the x1,x2-plane and then mark the corresponding output for this point (+1)

6. Do this for a few points and then draw the 2d - step function

7. Now on the x1, x2 plane show red and green regions for the positive and negative points an show that the line which is the foot of the step function which divides the space into two halves

2. Show the perceptron model equations here

3. Suppose w_1=<some_value>, w_2=<some_value>, b =1

Limitations of Perceptron

How does the perceptron function look in 2 dimensions?

(c) One Fourth Labs

y=\sum_{i=1}^n w_i x_i \geq b

\text{Suppose,} \\ \mathbb{w_1} = 0.2,\\ \hspace{-0.2cm}\mathbb{w_2} = 0.1 \\ \mathbb{b} = -10.5

Salary (in thousands)	Family size	Can buy a car?
80	2	1
20	1	0
65	4	1
15	7	0
30	6	0
49	3	0
51	4	1
87	8	1

Limitations of Perceptron

What if the data is not linearly separable ?

(c) One Fourth Labs

1. Show a training matrix with only 2 inputs (salary, family size) and the decision is buy_car/not_buy_car

Now on the x1. x2 plane show some data which is not linearly separable

Adjust the perceptron function and show that no matter what you do some green points will go to the other side or some red points will go to the other side.

2. Show the perceptron model equations here

3. Suppose w_1=<some_value>, w_2=<some_value>, b =1

Limitations of Perceptron

What if the data is not linearly separable ?

(c) One Fourth Labs

y=\sum_{i=1}^n w_i x_i \geq b

\text{Suppose,} \\ \mathbb{w_1} = 0.2,\\ \hspace{-0.2cm}\mathbb{w_2} = 0.1 \\ \mathbb{b} = -6

Salary (in thousands)	Family size	Can buy a car?
81	8	1
37	9	0
34	5	1
...	...	...
40	4	0
100	10	0
10	10	1
85	8	1

\text{Suppose,} \\ \mathbb{w_1} = 0.2,\\ \hspace{-0.2cm}\mathbb{w_2} = 0.1 \\ \mathbb{b} = -3

\text{Suppose,} \\ \mathbb{w_1} = 0.2,\\ \hspace{-0.2cm}\mathbb{w_2} = -0.3 \\ \mathbb{b} = -14

\text{Suppose,} \\ \mathbb{w_1} = 0.2,\\ \hspace{-0.2cm}\mathbb{w_2} = -0.5 \\ \mathbb{b} = -14

\text{Suppose,} \\ \mathbb{w_1} = 0.2,\\ \hspace{-0.2cm}\mathbb{w_2} = 0.1 \\ \mathbb{b} = -9

\text{Suppose,} \\ \mathbb{w_1} = 0.2,\\ \hspace{-0.2cm}\mathbb{w_2} = 0.1 \\ \mathbb{b} = -14

\text{Suppose,} \\ \mathbb{w_1} = 0.2,\\ \hspace{-0.2cm}\mathbb{w_2} = -0.7 \\ \mathbb{b} = -14

\text{Suppose,} \\ \mathbb{w_1} = 0.2,\\ \hspace{-0.2cm}\mathbb{w_2} = 0.3 \\ \mathbb{b} = -14

\text{Suppose,} \\ \mathbb{w_1} = 0.2,\\ \hspace{-0.2cm}\mathbb{w_2} = 0.7 \\ \mathbb{b} = -14

Limitations of Perceptron

Isn't the perceptron model a bit harsh at the boundaries ?

(c) One Fourth Labs

Repeat the training data with one input (salary)

Show the 2d plot from before

2. Now show the cartoon

Show the perceptron model equations here

Suppose w_1=<some_value>, w_2=<some_value>, b =1

Isn't it a bit odd that a person with 150K salary will but a car but someone with 150.1K will not buy a car ?

Limitations of Perceptron

Isn't the perceptron model a bit harsh at the boundaries ?

(c) One Fourth Labs

Isn't it a bit odd that a person with 50.1K salary will buy a car but someone with 49.9K will not buy a car ?

Salary ( in thousands)	Can buy a car?
80	1
20	0
65	1
15	0
30	0
49	0
51	1
87	1

y=\sum_{i=1}^n w_i x_i \geq b

\text{Suppose,} \\ \mathbb{w_1} = 2,\\ \hspace{-0.2cm}\mathbb{b}\ \ = 1

The Road Ahead

What's going to change now ?

(c) One Fourth Labs

Show the six jars and convey the following by animation

-- Boolean output gets replaced by real output

-- "specific learning algo" gets replaced by "a more generic learning algorithm"

-- "harsh at boundaries gets replaced by smooth at boundaries

-- "Linear" gets replaced by non-linear

Harsh at boundaries

Linear

Real inputs

Boolean output

Specific learning algorithm

(c) One Fourth Labs

The Road Ahead

What's going to change now ?

(c) One Fourth Labs

\( \{0, 1\} \)

Boolean

Accuracy=\frac{\text{Number of correct predictions}}{\text{Total number of predictions}}

Loss

Model

Data

Task

Evaluation

Learning

Linear

Real inputs

Boolean output

loss = \sum_i max(0,1-y_i*\hat{y_i})

Specific learning algorithm

Harsh at boundaries

Real output

Non-linear

A more generic learning algorithm

Smooth at boundaries

Data and Task

What kind of data and tasks can Perceptron process ?

(c) One Fourth Labs

Same mobile phone dataset from before.

Real inputs

Data and Task

What kind of data and tasks can Perceptron process ?

(c) One Fourth Labs

Data and Task

What kind of data and tasks can Perceptron process ?

(c) One Fourth Labs

Real inputs

Launch (within 6 months)	0	1	1	0	0	1	0	1	1
Weight (g)	151	180	160	205	162	182	138	185	170
Screen size (inches)	5.8	6.18	5.84	6.2	5.9	6.26	4.7	6.41	5.5
dual sim	1	1	0	0	0	1	0	1	0
Internal memory (>= 64 GB, 4GB RAM)	1	1	1	1	1	1	1	1	1
NFC	0	1	1	0	1	0	1	1	1
Radio	1	0	0	1	1	1	0	0	0
Battery(mAh)	3060	3500	3060	5000	3000	4000	1960	3700	3260
Price (INR)	15k	32k	25k	18k	14k	12k	35k	42k	44k
Like (y)	1	0	1	0	1	1	0	1	0

Model

Can we have a smoother (not-so-harsh) function ?

(c) One Fourth Labs

1) Show step function first and then super-impose the sigmoid function

2) Show the equation

3) Now on the RHS below the equation, substitute some values for wx+b and show the output on the plot (basically, I want to tell people that don't get overwhelmed when you see a function, just susbstitute some values in it and see how the plot looks, 0,1,-1, \inf, -inf

Show the sigmoid function equation here

Model

What happens when we have more than 1 input ?

(c) One Fourth Labs

1) Show the same phone data from before

2) Now show an image of sigmoid model with the equation containing \sum_w_i x_i + b

3) Now replace the summation by w^Tx

4) How do you plot this function: on x-axis show w^tx +b and on y-axis show the output y (again plot for different values of w^tx +b)

Show the sigmoid function equation here

Model

How does this help when the data is not linearly separable ?

(c) One Fourth Labs

Still does not completely solve our problem but we will slowly get there!

1. Show a training matrix with only 2 inputs (salary, family size) and the decision is buy_car/not_buy_car

1. In the first case we will consider data which only ha problems at the boundary (meaning that only at the boundary around wx+b = 0 some green and red points mix with each other)

2. Now show that unlike the perceptron which was either misclassifying the positive or negative points, sigmoid gives a soft decision

3. also show that by adjusting w_1 and w_2 you can accommodate for different types of boundary cases (please take Gokul's help for these plots)

2. Show the sigmoid model equations here

Model

What about extreme non-linearity ?

(c) One Fourth Labs

Still does not completely solve our problem but we will slowly get there with more complex models!

1. Show a training matrix with only 2 inputs (salary, family size) and the decision is buy_car/not_buy_car

1.Now show a case where the blue and red points are completely mixed with each other

2. show that no matter how you adjust w_1 and w_2 you will not be able to find any separation between blue and red points

3. Now show the quote at RHS bottom using our cartoon person

2. Show the sigmoid model equations here

Model

How does the function behave when we change w and b

(c) One Fourth Labs

5. Cartoon person: At what value of x is the value of sigmoid(X) = 0.5

2. Show a scale here for w and b

1. Show the sigmoid function here (the x-axis is x and y -asix is sigmoid(x)

3. CHange the value of w below and show how the plot changes

4. Now change the value of w below and show how the plot changes

5. Show the cartoon person woth the W

6. the derivation

2. Show the sigmoid model equations here

the derivation to show that sigm(x) = 0.5 at x = -b/w

Learning Algorithm

Can we try to estimate w, b using some guess work ?

(c) One Fourth Labs

Initialise

Iterate over data:

till satisfied

\( h = \frac{1}{1+e^{-(w*x + b)}} \)

I/P	O/P
2	0.047
3	0.268
4	0.73
5	0.952
8	0.999

\(w, b \)

\( guess\_and\_update(x_i) \)

w	b

w	b
1	0

w	b
0	0

w	b
2	0

w	b
3	-7

w	b
3	-4

w	b
3	-2

w	b
3	0

w	b
3	-9

w	b
2	-9

w	b
2	-7

Learning Algorithm

Can we take a closer look at what we just did ?

(c) One Fourth Labs

Initialise

Iterate over data:

till satisfied

\(w, b \)

\( guess\_and\_update(x_i) \)

w	b

w	b
1	0

w	b
0	0

w	b
2	0

w	b
3	-7

w	b
3	-4

w	b
3	-2

w	b
3	0

w	b
3	-9

w	b
2	-9

\( w = w + \Delta w \\ b = b + \Delta b \)

w	b
2	-7

1 = 0 + \(\Delta\)1

0 = 0 + \( \Delta\)0

2 = 1 + \(\Delta\)1

0 = 0 + \( \Delta\)0

3 = 2 + \(\Delta\)1

0 = 0 + \( \Delta\)0

3 = 3 + \(\Delta\)0

-2 = 0 + \( \Delta\)-2

3 = 3 + \(\Delta\)0

-4 = -2 + \( \Delta\)-2

3 = 3 + \(\Delta\)0

-7 = -4 + \( \Delta\)-3

3 = 3 + \(\Delta\)0

-9 = -7 + \( \Delta\)-2

2 = 3 + \(\Delta\)-1

-9 = -7 + \( \Delta\)-2

2 = 3 + \(\Delta\)-1

-7 = -9 + \( \Delta\)2

w = 0

b = 0

\( w = w + \Delta w \)

\( b = b + \Delta b \)

\( \Delta w = some\_guess \)

\( \Delta b = some\_guess \)

Learning Algorithm

Can we connect this to the loss function ?

(c) One Fourth Labs

Initialise

Iterate over data:

till satisfied

\(w, b \)

\( guess\_and\_update(x_i) \)

\( w = w + \Delta w \\ b = b + \Delta b \)

1 = 0 + \(\Delta\)1

0 = 0 + \( \Delta\)0

2 = 1 + \(\Delta\)1

0 = 0 + \( \Delta\)0

3 = 2 + \(\Delta\)1

0 = 0 + \( \Delta\)0

3 = 3 + \(\Delta\)0

-2 = 0 + \( \Delta\)-2

3 = 3 + \(\Delta\)0

-4 = -2 + \( \Delta\)-2

3 = 3 + \(\Delta\)0

-7 = -4 + \( \Delta\)-3

3 = 3 + \(\Delta\)0

-9 = -7 + \( \Delta\)-2

2 = 3 + \(\Delta\)-1

-9 = -7 + \( \Delta\)-2

2 = 3 + \(\Delta\)-1

-7 = -9 + \( \Delta\)2

w = 0

b = 0

\( w = w + \Delta w \)

\( b = b + \Delta b \)

w	b	Loss

w	b	Loss
0	0	0.1609

w	b	Loss
1	0	0.1064

w	b	Loss
2	0	0.1210

w	b	Loss
3	0	0.1217

w	b	Loss
3	-2	0.1215

w	b	Loss
3	-4	0.1198

w	b	Loss
3	-7	0.1081

w	b	Loss
3	-9	0.0209

w	b	Loss
2	-9	0.0636

w	b	Loss
2	-7	0.000

Indeed we were using the loss function to guide us in finding \( \Delta w\) and \(\Delta b\)

Learning Algorithm

Can we visualise this better ?

(c) One Fourth Labs

Initialise

Iterate over data:

till satisfied

\(w, b \)

\( guess\_and\_update(x_i) \)

\( w = w + \Delta w \\ b = b + \Delta b \)

1 = 0 + \(\Delta\)1

0 = 0 + \( \Delta\)0

2 = 1 + \(\Delta\)1

0 = 0 + \( \Delta\)0

3 = 2 + \(\Delta\)1

0 = 0 + \( \Delta\)0

3 = 3 + \(\Delta\)0

-2 = 0 + \( \Delta\)-2

3 = 3 + \(\Delta\)0

-4 = -2 + \( \Delta\)-2

3 = 3 + \(\Delta\)0

-7 = -4 + \( \Delta\)-3

3 = 3 + \(\Delta\)0

-9 = -7 + \( \Delta\)-2

2 = 3 + \(\Delta\)-1

-9 = -7 + \( \Delta\)-2

2 = 3 + \(\Delta\)-1

-7 = -9 + \( \Delta\)2

w = 0

b = 0

\( w = w + \Delta w \)

\( b = b + \Delta b \)

Learning Algorithm

(c) One Fourth Labs

Initialise

Iterate over data:

till satisfied

\(w, b \)

\( guess\_and\_update(x_i) \)

w	b

w	b
1	0

w	b
0	0

w	b
2	0

w	b
3	-7

w	b
3	-4

w	b
3	-2

w	b
3	0

w	b
3	-9

w	b
2	-9

\( w = w + \Delta w \\ b = b + \Delta b \)

w	b
2	-7

1 = 0 + \(\Delta\)1

0 = 0 + \( \Delta\)0

2 = 1 + \(\Delta\)1

0 = 0 + \( \Delta\)0

3 = 2 + \(\Delta\)1

0 = 0 + \( \Delta\)0

3 = 3 + \(\Delta\)0

-2 = 0 + \( \Delta\)-2

3 = 3 + \(\Delta\)0

-4 = -2 + \( \Delta\)-2

3 = 3 + \(\Delta\)0

-7 = -4 + \( \Delta\)-3

3 = 3 + \(\Delta\)0

-9 = -7 + \( \Delta\)-2

2 = 3 + \(\Delta\)-1

-9 = -7 + \( \Delta\)-2

2 = 3 + \(\Delta\)-1

-7 = -9 + \( \Delta\)2

w = 0

b = 0

\( w = w + \Delta w \)

\( b = b + \Delta b \)

\( \Delta w = some\_guess \)

\( \Delta b = some\_guess \)

"Instead of guessing \( \Delta w\) and \(\Delta b\), we need a principled way of changing w and b based on the loss function"

What is our aim now ?

Learning Algorithm

Can we formulate this more mathematically ?

(c) One Fourth Labs

\(\theta = [w, b] \)

\(\Delta \theta = [\Delta w, \Delta b] \)

\( \theta_{new} = \theta + \eta . \Delta \theta \)

\(\theta\)

\(\Delta \theta\)

\( \eta . \Delta \theta \)

\( \theta_{new}\)

Instead of guessing \(\Delta w\) and\( \Delta\) b, we need a principled way of changing w and b based on the loss function

Learning Algorithm

Can we get the answer from some basic mathematics ?

(c) One Fourth Labs

\( \because \theta = [w, b] \)

\(w \rightarrow w + \eta \Delta w \)

\(b \rightarrow b + \eta \Delta b \)

\(\mathscr{L} (w) < \mathscr{L}(w +\eta \Delta w)\)

\( \mathscr{L} (b) < \mathscr{L}(b +\eta \Delta b)\)

\(\mathscr{L} ( \theta) < \mathscr{L}( \theta + \eta \Delta \theta)\)

\( f(x + \Delta x) = f(x) + f'(x)\Delta x + \frac{1}{2!}f''(x)\Delta x^2 + \frac{1}{3!}f'''(x)\Delta x^3 + \cdots \)

Taylor Series

\(\mathscr{L} (w,b) < \mathscr{L}(w +\eta \Delta w, b + \eta\Delta b )\)

\( \mathscr{L}(\theta + \Delta \theta) = \mathscr{L}(\theta) + \mathscr{L}'(\theta)\Delta \theta + \frac{1}{2!}\mathscr{L}''(\theta)\Delta \theta^2 + \frac{1}{3!}\mathscr{L}'''(\theta)\Delta \theta^3 + \cdots \)

Learning Algorithm

How does Taylor series help us to arrive at the answer ?

(c) One Fourth Labs

\(\mathscr{L}(\theta+\eta u) = \mathscr{L}(\theta) + \eta*u^T \nabla_{\theta} \mathscr{L}(\theta) + \frac{\eta^2}{2!}*u^T \nabla^2 \mathscr{L} (\theta)u + \frac{\eta^3}{3!}*....+ \frac{\eta^4}{4!}*...\)

\(= \mathscr{L}(\theta)+ \eta*u^T \nabla_{\theta}\mathscr{L}(\theta) [\eta is typically small, so \eta^2, \eta^3, ... \rightarrow 0]\)

Note that the move \( \eta u \) would be favorable only if,

\( \mathscr{L}(\theta+\eta u) - \mathscr{L}(\theta) < 0\) [ i.e. if the new loss is less than the previous loss]

This implies,

\( u^T \nabla_{\theta} \mathscr{L}(\theta) < 0 \)

For ease of notation,

let \( \Delta\theta = u\),

Then from Taylor series, we have,

Learning Algorithm

How does Taylor series help us to arrive at the answer ?

(c) One Fourth Labs

Okay, so we have,

\( u^T \nabla_{\theta} \mathscr{L}(\theta) < 0 \)

But, what is the range of \(u^T \nabla_{\theta}\mathscr{L}(\theta) \) ?

Let \(\beta\) be the angle between \(u\) and \(\nabla_{\theta}\mathscr{L}(\theta)\), then we know that,

\(-1 \leq cos(\beta) = \frac{u^T \nabla_{\theta}\mathscr{L}(\theta)}{||u||*||\nabla_{\theta}\mathscr{L}(\theta)||} \leq 1\)

multiply throughout by k = \( ||u||*||\nabla_{\theta}\mathscr{L}(\theta)||\)

\( -k \leq k*cos(\beta) = u^T \nabla_{\theta}\mathscr{L}(\theta) \leq k \)

Thus, \( \mathscr{L}(\theta + \eta u) - \mathscr{L}(\theta) = u^T \nabla_{\theta}\mathscr{L}(\theta) = k*\cos \beta \) will be most negative

when \(\cos(\beta)\) = -1 i.e., when \(\beta\) is 180\(\degree \)

Learning Algorithm

How does Taylor series help us to arrive at the answer ?

(c) One Fourth Labs

Gradient Descent Rule,

The direction \(u\) that we intend to move in should be at \(180\degree\) w.r.t. the gradient.
In other words, move in a direction opposite to the gradient.

Parameter Update Rule

\(w_{t+1} = w_{t} - \eta \Delta w_{t} \)

\(b_{t+1} = b_{t} - \eta \Delta b_{t} \)

\( where \Delta w_{t} =\frac{\partial \mathscr{L}(w, b)}{\partial w}_{at w = w_t, b= b_t}, \Delta b_{t} = \frac{\partial \mathscr{L}(w,b)}{\partial b}_{at\: w = w_t, b = b_t}\)

So we now have a more principled way of moving in the w-b plane than our "guesswork'' algorithm

Learning Algorithm

How does the algorithm look like now ?

(c) One Fourth Labs

Initialise

\(w, b \)

Iterate over data:

\(w_{t+1} = w_{t} - \eta \Delta w_{t} \)

till satisfied

How do I compute \( \Delta w\) and\( \Delta b \)?

\(b_{t+1} = b_{t} - \eta \Delta b_{t} \)

Learning Algorithm

How do I compute \(\Delta w \) and \(\Delta b\)

(c) One Fourth Labs

\(\mathscr{L} = \frac{1}{5} \sum_{i = 1}^{i=5}(f(x_i) - y_i)\)

\(\frac{\partial \mathscr{L}}{\partial w} = \frac{\partial}{\partial w}[ \frac{1}{5} \sum_{i = 1}^{i=5}(f(x_i) - y_i)]\)

\(\Delta w = \frac{\partial \mathscr{L}}{\partial w} = \frac{1}{5} \sum_{i = 1}^{i=5}\frac{\partial}{\partial w} (f(x_i) - y_i)\)

Let's consider only one term in this sum,

Learning Algorithm

How do I compute \(\Delta w \) and \(\Delta b\)

(c) One Fourth Labs

\(\nabla w = \frac{\partial}{\partial w}[\frac{1}{2}*(f(x) - y)^2] \)

\(= \frac{1}{2}*[2*(f(x) - y) * \frac{\partial}{\partial w} (f(x) - y)]\)

\(= (f(x) - y) * \frac{\partial}{\partial w}(f(x)) \)

\(= (f(x) - y) * \frac{\partial}{\partial w}\Big(\frac{1}{1 + e^{-(wx + b)}}\Big) \)

\( = \color{red}{(f(x) - y) * f(x)*(1- f(x)) *x}\)

\( \frac{\partial}{\partial w}\Big(\frac{1}{1 + e^{-(wx + b)}}\Big)\)

\( =\frac{-1}{(1 + e^{-(wx + b)})^2}\frac{\partial}{\partial w}(e^{-(wx + b)})) \)

\( =\frac{-1}{(1 + e^{-(wx + b)})^2}*(e^{-(wx + b)})\frac{\partial}{\partial w}(-(wx + b)))\)

\( =\frac{-1}{(1 + e^{-(wx + b)})}*\frac{e^{-(wx + b)}}{(1 + e^{-(wx + b)})} *(-x)\)

\( =\frac{1}{(1 + e^{-(wx + b)})}*\frac{e^{-(wx + b)}}{(1 + e^{-(wx + b)})} *(x)\)

\( =f(x)*(1- f(x))*x \)

Learning Algorithm

How do I compute \(\Delta w \) and \(\Delta b\)

(c) One Fourth Labs

\(\mathscr{L} = \frac{1}{5} \sum_{i = 1}^{i=5}(f(x_i) - y_i)\)

\(\frac{\partial \mathscr{L}}{\partial w} = \frac{\partial}{\partial w}[ \frac{1}{5} \sum_{i = 1}^{i=5}(f(x_i) - y_i)]\)

\(\Delta w = \frac{\partial \mathscr{L}}{\partial w} = \frac{1}{5} \sum_{i = 1}^{i=5}\frac{\partial}{\partial w} (f(x_i) - y_i)\)

Let's consider only one term in this sum,

\(\Delta w=(f(x) - y) * f(x)*(1- f(x)) *x \)

For 5 points,

\(\Delta w=\sum_{i=1}^{i=5} (f(x_i) - y_i) *f(x_i))*(1- f(x_i)) *x_i \)

\(\Delta b=\sum_{i=1}^{i=5} (f(x_i) - y_i) * f(x_i))*(1- f(x_i)) \)

Learning Algorithm

How does the full algorithm look like ?

(c) One Fourth Labs

Learning Algorithm

What happens when we have more than 2 parameters ?

(c) One Fourth Labs

\( z = w_1*ER\_visits + w_2*Narcotics + w_3*Pain + w_4*TotalVisits + w_5*MedicalClaims + b \)

\(\hat{y} = \frac{1}{1+e^{-(w_1*x_1 + w_2*x_2 + w_3*x_3 + w_4*x_4 + w_5*x_5 +b)}}\)

\( z = w_1*x_1 + w_2*x_2 + w_3*x_3 + w_4*x_4 + w_5*x_5 + b \)

Initialise

\(w_1, w_2,...,w_5, b \)

Iterate over data:

\(w_1^{t+1} = w_1^{t} - \eta \Delta w_1^{t} \)

till satisfied

\(b_{t+1} = b_{t} - \eta \Delta b_{t} \)

\(w_2^{t+1} = w_2^{t} - \eta \Delta w_2^{t} \)

\(w_5^{t+1} = w_5^{t} - \eta \Delta w_5^{t} \)

\( \vdots \)

\(\Delta w=(f(x) - y) * f(x)*(1- f(x)) *x \)

\(\Delta w_1=(f(x_1,x_2,x_3,x_4,x_5) - y) * f(x_1,x_2,x_3,x_4,x_5)*(1- f(x_1,x_2,x_3,x_4,x_5)) *x_1 \)

\(\hat{y} = \frac{1}{1+e^{-z}}\)

\(\Delta w_1 =(\hat{y} - y) * \hat{y}*(1- \hat{y}) *x_1 \)

Learning Algorithm

How do I compute \(\Delta w \) and \(\Delta b\)

(c) One Fourth Labs

\(\nabla w_1 = \frac{\partial}{\partial w_1}[\frac{1}{2}*(\hat{y} - y)^2] \)

\(= \frac{1}{2}*[2*(\hat{y} - y) * \frac{\partial}{\partial w_1} (\hat{y} - y)]\)

\(= (\hat{y} - y) * \frac{\partial}{\partial w_1}(\hat{y}) \)

\(= (\hat{y} - y) * \frac{\partial}{\partial w_1}\Big(\frac{1}{1 + e^{-(w_1x_1 + w_2x_2 + w_3x_3 + w_4x_4 + w_5x_5 + b)}}\Big) \)

\( = \color{red}{(\hat{y} - y) *\hat{y}*(1- \hat{y}) *x_1}\)

\( \frac{\partial}{\partial w_1}\Big(\frac{1}{1 + e^{-(z)}}\Big)\)

\( =\frac{-1}{(1 + e^{-(z)})^2}\frac{\partial}{\partial w_1}(e^{-(z)})) \)

\( =\frac{-1}{(1 + e^{-(z)})^2}*(e^{-(z)})\frac{\partial}{\partial w_1}(-(z)))\)

\( =\frac{-1}{(1 + e^{-(z)})}*\frac{e^{-(z)}}{(1 + e^{-(z)})} *(-x_1)\)

\( =\frac{1}{(1 + e^{-(z)})}*\frac{e^{-(z)}}{(1 + e^{-(z)})} *(x_1)\)

\( =\hat{y}*(1- \hat{y})*x_1 \)

\( z = w_1*x_1 + w_2*x_2 + w_3*x_3 + w_4*x_4 + w_5*x_5 + b \)

\( \frac{\partial}{\partial w_1}(-z) = - \frac{\partial }{\partial w_1} (w_1*x_1 + w_2*x_2 + w_3*x_3 + w_4*x_4 + w_5*x_5 + b) \)

\( \frac{\partial}{\partial w_1}(-z) = - x_1 \)

Learning Algorithm

How do I compute \(\Delta w \) and \(\Delta b\)

(c) One Fourth Labs

\(\mathscr{L} = \frac{1}{2}(\hat{y} - y)^{2}\)

\(\frac{\partial \mathscr{L}}{\partial w_1} = \frac{\partial}{\partial w_1}(\frac{1}{2} [\hat{y} - y]^2)\)

\(\vdots\)

\(\frac{\partial \mathscr{L}}{\partial w_5} = \frac{\partial}{\partial w_5}(\frac{1}{2} [\hat{y} - y]^2)\)

\(\frac{\partial \mathscr{L}}{\partial w_2} = \frac{\partial}{\partial w_2}(\frac{1}{2} [\hat{y} - y]^2)\)

\(\Delta w_1 =(\hat{y} - y) * \hat{y}*(1- \hat{y}) *x_1 \)

\(\Delta w_2 =(\hat{y} - y) * \hat{y}*(1- \hat{y}) *x_2 \)

\(\Delta w_5 =(\hat{y} - y) * \hat{y}*(1- \hat{y}) *x_5 \)

\(\vdots\)

Learning Algorithm

How do I compute \(\Delta w \) and \(\Delta b\)

(c) One Fourth Labs

\(\begin{bmatrix} \Delta w_1 \\ \Delta w_2 \\ \vdots \\ \Delta w_5 \end{bmatrix} \)

\( \begin{bmatrix} (\hat{y}-y)*\hat{y}*(1-\hat{y})*x_1\\ (\hat{y}-y)*\hat{y}*(1-\hat{y})*x_2 \\ \vdots \\ (\hat{y}-y)*\hat{y}*(1-\hat{y})*x_5\end{bmatrix}\)

\( \Delta w \)

=

\( (\hat{y}-y)*\hat{y}*(1-\hat{y}) \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_5 \end{bmatrix} \)

=