Machine Learning Week 1 Quiz 2 (Linear Regression with One Variable) Stanford Coursera

Question 1

Consider the problem of predicting how well a student does in her second year of college/university, given how well she did in her first year.

Specifically, let x be equal to the number of “A” grades (including A-. A and A+ grades) that a student receives in their first year of college (freshmen year). We would like to predict the value of y, which we define as the number of “A” grades they get in their second year (sophomore year).

Here each row is one training example. Recall that in linear regression, our hypothesis is hθ(x)=θ01x, and we use m to denote the number of training examples

xy
54
34
01
43

For the training set given above (note that this training set may also be referenced in other questions in this quiz), what is the value of m? In the box below, please enter your answer (which should be a number between 0 and 10).

Answer:
4

Question 2

Consider the following training set of m=4 training examples:

xy
10.5
21
42
00

Consider the linear regression model hθ(x)=θ01x. What are the values of θ0 and θ1 that you would expect to obtain upon running gradient descent on this model? (Linear regression will be able to fit this data perfectly.)

  • θ0=0.5,θ1=0
  • θ0=0.5,θ1=0.5
  • θ0=1,θ1=0.5
  • θ0=0,θ1=0.5
  • θ0=1,θ1=1

Answer:
θ0=0,θ1=0.5

As J(θ01)=0, y = hθ(x) = θ0 + θ1x. Using any two values in the table, solve for θ0, θ1

Question 3

Suppose we set θ0=−1,θ1=0.5. What is hθ(4)?

Answer:

Setting x = 4, we have hθ(x)=θ01x = -1 + (0.5)(4) = 1

Question 4

Let f be some function so that

f(θ01) outputs a number. For this problem,

f is some arbitrary/unknown smooth function (not necessarily the

cost function of linear regression, so f may have local optima).

Suppose we use gradient descent to try to minimize f(θ01) as a function of θ0 and θ1. Which of the

following statements are true? (Check all that apply.)

  • Even if the learning rate α is very large, every iteration of gradient descent will decrease the value of f(θ01).
  • If the learning rate is too small, then gradient descent may take a very long time to converge.
  • If θ0 and θ1 are initialized at a local minimum, then one iteration will not change their values.
  • If θ0 and θ1 are initialized so that θ01, then by symmetry (because we do simultaneous updates to the two parameters), after one iteration of gradient descent, we will still have θ01.

Answers:

True or FalseStatementExplanation
TrueIf the learning rate is too small, then gradient descent may take a very long time to converge.If the learning rate is small, gradient descent ends up taking an extremely small step on each iteration, and therefor can take a long time to converge
TrueIf θ0 and θ1 are initialized at a local minimum, then one iteration will not change their values.At a local minimum, the derivative (gradient) is zero, so gradient descent will not change the parameters.
FalseEven if the learning rate α is very large, every iteration of gradient descent will decrease the value of f(θ01).If the learning rate is too large, one step of gradient descent can actually vastly “overshoot” and actually increase the value of f(θ01).
FalseIf θ0 and θ1 are initialized so that θ01, then by symmetry (because we do simultaneous updates to the two parameters), after one iteration of gradient descent, we will still have θ01.The updates to θ0 and θ1 are different (even though we’re doing simulaneous updates), so there’s no particular reason to update them to be same after one iteration of gradient descent.

Other Options:

True or FalseStatementExplanation
TrueIf the first few iterations of gradient descent cause f(θ01) to increase rather than decrease, then the most likely cause is that we have set the learning rate to too large a valueif alpha were small enough, then gradient descent should always successfully take a tiny small downhill and decrease f(θ01) at least a little bit. If gradient descent instead increases the objective value, that means alpha is too large (or you have a bug in your code!).
FalseNo matter how θ0 and θ1 are initialized, so long as learning rate is sufficiently small, we can safely expect gradient descent to converge to the same solutionThis is not true, depending on the initial condition, gradient descent may end up at different local optima.
FalseSetting the learning rate to be very small is not harmful, and can only speed up the convergence of gradient descent.If the learning rate is small, gradient descent ends up taking an extremely small step on each iteration, so this would actually slow down (rather than speed up) the convergence of the algorithm.

Question 5

Suppose that for some linear regression problem (say, predicting housing prices as in the lecture), we have some training set, and for our training set we managed to find some θ0, θ1 such that J(θ01)=0.

Which of the statements below must then be true? (Check all that apply.)

  • For this to be true, we must have y(i)=0 for every value of i=1,2,…,m.
  • Gradient descent is likely to get stuck at a local minimum and fail to find the global minimum.
  • For this to be true, we must have θ0=0 and θ1=0 so that hθ(x)=0
  • Our training set can be fit perfectly by a straight line, i.e., all of our training examples lie perfectly on some straight line.

Answers:

True or FalseStatementExplanation
FalseFor this to be true, we must have y(i)=0 for every value of i=1,2,…,m.So long as all of our training examples lie on a straight line, we will be able to find θ0 and θ1) so that J(θ01)=0. It is not necessary that y(i) for all our examples.
FalseGradient descent is likely to get stuck at a local minimum and fail to find the global minimum.none
FalseFor this to be true, we must have θ0=0 and θ1=0 so that hθ(x)=0If J(θ01)=0 that means the line defined by the equation “y = θ0 + θ1x” perfectly fits all of our data. There’s no particular reason to expect that the values of θ0 and θ1 that achieve this are both 0 (unless y(i)=0 for all of our training examples).
TrueOur training set can be fit perfectly by a straight line, i.e., all of our training examples lie perfectly on some straight line.None

Other Options:

True or FalseStatementExplanation
FalseWe can perfectly predict the value of y even for new examples that we have not yet seen. (e.g., we can perfectly predict prices of even new houses that we have not yet seen.)None
FalseThis is not possible: By the definition of J(θ01), it is not possible for there to exist θ0 and θ1 so that J(θ01)=0None
TrueFor these values of θ0 and θ1 that satisfy J(θ01)=0, we have that hθ(x(i))=y(i) for every training example (x(i),y(i))None

Leave a Comment

Your email address will not be published. Required fields are marked *