Linear Regression

English Version tensorflow
1

https://d2l.ai/chapter_linear-regression/linear-regression.html

2

Q3. I am very appreciated that someone can correct me if i’m wrong. This is my answer:

  1. Assume that y=Xw+b+ϵ,p(ϵ)=1/2e^(−|ϵ|)=1/2e^(−|y−y^|) . So , P(y|x)=1/2e^(−|y−y^|) .
    Negative log-likelihood: LL(y|x)=−log(P(y|x))=−log(∏p(y(i)|x(i)))=∑log2+|y(i)−y^(i)|=∑log2+|y(i)−X(i)w−b|

  2. ∇{w} LL(y|x)=X.T*(Xw−y/|Xw−y|)=0 . From the equation, we get that: (1) : w=(X.TX)^−1X.Ty and (2) w≠X^−1y

  3. I will update soon.

hcmut

3

Honestly, I don’t know what exactly the answer on each question. So, I just tried to write the code based on the questions and the topicห that I understand. And this is the code that I wrote.
import tensorflow as tf
import numpy as np
import matplotlib.pyplot as plt

#u can change the values of x and y as u may like
#alp or learning rate also has the impact to the function, so adjust it properly

y = tf.random.normal([12], 0, 1, tf.float32)
x = np.arange(12)

*def minibatch(y,x,alp):

   w = 1
   b = 0
   z = np.array([])
   while True:
    out = np.sum(x*(w*x + b - y))
    out2 = np.sum(w*x + b - y)
    w1 = w - (alp/len(x))*out
    b1 = b - (alp/len(x))*out2

    if np.abs((w1-w)/w1) < 0.00001 and np.abs((b1-b)/b1) < 0.00001:  
        return w,b,z

    else:
        #print(b)
        z = np.append(z,b)
        w = w1
        b = b1*

Then we will use the out put that we get to plot on the graph by w is weight and z is the array used for storing b values.

w,b,z= minibatch(y,x, 0.01)
plt.plot(x,y, ‘bx’)
plt.plot(x, w*x+b)
plt.grid()
plt.show()

hopefully, this can help you.