NPTEL Introduction to Machine Learning Assignment 11 Answer

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NPTEL Introduction to Machine Learning Assignment

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1 point

What are the necessary conditions for
P(xn)=∑1kπkP(xn|θk)P(xn)=∑1kπkP(xn|θk)
to be a density? (Select all that apply)

0≤πk≤10≤πk≤1

k≤∑kπkk≤∑kπk

P(xi,xj|θk)=P(xi|θk)P(xj|θk)P(xi,xj|θk)=P(xi|θk)P(xj|θk)

∑iπi=1∑iπi=1

Ans   – C,D

Based on the density estimation of a GMM given below, answer questions 2−4.Based on the density estimation of a GMM given below, answer questions 2−4.

1 point

What is value of k?

 3

 4

 5

 6

Ans   – C

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1 point

What is the minimum value of k‘≠kk′≠k, where kk is from previous question, for which you will get a very similar density estimation?

 3

 4

 5

 6

Ans   – C

2 points

Assume equal πiπi for each gaussian model after convergence as in Q2. What would (approximately) be πiπi’s for the model you’ll get with kk′ as in Q3?

 [0.33, 0.33, 0.17, 0.17]

 [0.2, 0.2, 0.6]

 [0.25, 0.25, 0.5]

 [0.2, 0.2, 0.2, 0.4]

Ans   -B

For a set of points (given in orange), the density estimation of a GMM is given below. Based on this, answer questions 5 and 6.For a set of points (given in orange), the density estimation of a GMM is given below. Based on this, answer questions 5 and 6.

1 point

What is the problem evident in the image?

πiπi’s are too big

 The clusters are not sampled from a gaussian distribution.

 The GMM has not converged yet.

 There is no problem

Ans   – A

1 point

What can be done to get a better fit?

 Increase k

 Use a better initialisation

 Learn for more iterations

 There is no problem

Ans   – B

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1 point

What does soft clustering mean in GMMs?

 There may be samples that are outside of any cluster boundary.

 The updates during maximum likelihood are taken in small steps, to guarantee convergence.

 It restricts the underlying distribution to be gaussian.

 Samples are assigned probabilities of belonging to a cluster.

Ans   – C

1 point

What is the update for πkπk in EM algorithm for GMM?

π(m)k=∑Nn=1γ(znk)|v(m−1)N−1πk(m)=∑n=1Nγ(znk)|v(m−1)N−1

π(m)k=∑Nn=1γ(znk)|v(m)Nπk(m)=∑n=1Nγ(znk)|v(m)N

π(m)k=∑Nn=1γ(znk)|v(m−1)Nπk(m)=∑n=1Nγ(znk)|v(m−1)N

π(m)k=∑Nn=1γ(znk)|v(m)N−1πk(m)=∑n=1Nγ(znk)|v(m)N−1

Ans   – A

1 point

KNN is a special case of GMM with the following properties: (Select all that apply)

γi=i(2πϵ)1/2e−12ϵγi=i(2πϵ)1/2e−12ϵ

Covariance = ϵIϵI

μi=μj∀i,jμi=μj∀i,j

πk=1k

Ans   – All

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