Consider linear regression with mini-batch stochastic gradient descent. Let X denote the data matrix and let t denote the target vector. Let d = number of features and let n = number of training examples. Write the weight update rule for mini-batch stochastic gradient descent in terms of the samples in a particular batch B. Assume the batch size is b.
<extra_id_1>Assistant
The weight update rule for mini-batch stochastic gradient descent is given by:
 

 w(i+1) = w(i) - alpha * [ (t - Xw(i))/b ]
 

 where alpha is the learning rate and [ (t - Xw(i))/b ] is the gradient of the loss function evaluated at w(i) in the direction of the samples in batch B.
<extra_id_1>User
Now, write the loss function in vector notation using the matrix B of the batch and the weights of the model w. Include a 1/b term in the loss. Calculate the gradient of the loss w.r.t. w in vector form
 

 Both your answers should be in terms of B, b, w, and t 
<extra_id_1>Assistant
The loss function for mini-batch stochastic gradient descent can be written in vector form as:
 

 loss = (1/b) * (t - Xw)^T * (t - Xw)
 

 The gradient of this loss function w.r.t. w is given by:
 

 gradient = (1/b) * (t - Xw)^T * X
<extra_id_1>User
Can you argue that the gradient above is of the form X^T*a for some vector a? First show it equals B^T*c for some vector c, and then show B^T*c = X^T*a for some vector a. It's a simple argument you don't need to give a proof
Yes, the gradient of the loss function can be written as a product of two matrices, X^T and a vector a. Here's the argument:
 

 1. Let B be the matrix of the batch samples, where each row b_i represents a sample t_i - Xw. Then we can write the gradient as:
 

 gradient = B^T * (t - Xw)
 

 2. We can expand the product B^T * (t - Xw):
 

 gradient = B^T * (t - Xw)
 = (B^T * B) * (t - Xw)
 = (I - B B^T) * (t - Xw)
 

 where I is the identity matrix.
 

 3. The matrix I - B B^T is diagonal, with entries 1 - b_{11}, 1 - b_{22}, ..., 1 - b_{dd}, where b_{ii} is the i-th element of the batch. Therefore, we can write:
 

 gradient = diag(1 - b_{11}, 1 - b_{22}, ..., 1 - b_{dd}) * (t - Xw)
 

 4. The gradient vector a can be found by taking the first element of each row of the diagonal matrix diag(1 - b_{11}, 1 - b_{22}, ..., 1 - b_{dd}):
 

 a = (1 - b_{11}, 1 - b_{22}, ..., 1 - b_{dd})^T
 

 5. Finally, we can show that the gradient X^T * a is equal to the gradient B^T * c for some vector c:
 

 X^T * a = (X^T * (t - Xw)) * a
 = (t - Xw)^T * X * a
 = B^T * (t - Xw)
 

 Therefore, the gradient of the loss function can be written as X^T * a, where a is a vector found by taking the first elements of each row of the diagonal matrix B^T.