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IIT M FOUNDATION ET1 EXAM QPF1 S1 03 Sep - 03 Sep 2023

MLF End Term18 questions50 marks7 MSQ · 6 SA · 5 MCQFoundation / QPF1
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Q1MCQ0 marks
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  1. Useful Data has been mentioned above.
  2. This data attachment is just for a reference & not for an evaluation.
Q2MSQ2 marksselect all that apply
Which of the following statements accurately describes the key difference between classification and regression models?
  1. In classification, the output variable is continuous and real-valued, while inregression, it is categorical.
  2. Regression models are used for predicting probabilities, whereas classificationmodels focus on predicting absolute values.
  3. Classification models aim to find decision boundaries to separate data intoclasses, while regression models seek to predict a numeric value.
  4. In regression, the commonly used loss function is mean squared error, whilein classification, 0 -1 loss is typically employed.
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Based on the above data, answer the given subquestions. question figure
Q3SA2 marks
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Q5MCQ3 marks
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  1. b1 − b2 + b3 = 0
  2. b1 + b2 + b3 = 0
  3. 2b1 − b2 + b3 = 0
  4. 2b1 + b2 + b3 = 0
Q6MCQ3 marks
A consumer wants to maximize his utility subject to some constraints. He consumes two goods g1 and g2 and the utility function is the sum of g1 and g2. His budget is Rs.100. The per unit price of goods g1 and g2 are Rs.3 and Rs.4 respectively. Chose the correct optimization problem.
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Q7MSQ4 marksselect all that apply
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  1. A − I (where I is the identity matrix of the same size) is also orthogonal.
  2. Both AB and BA are orthogonal.
  3. An and Bn are orthogonal for every n > 1.
  4. A and B are also symmetric matrices.
Q8MSQ4 marksselect all that apply
Which of the following statements is/are true for a real symmetric matrix A?
  1. AT A and AAT have the same rank.
  2. AT A and AAT have the same eigenvalues
  3. AT A and AAT have the same eigenvectors.
  4. AT A and AAT have the same singular values.
Q9MSQ3 marksselect all that apply
Which of the following matrices is/are Unitary but not Hermitian?
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Q10MSQ3 marksselect all that apply
Which among the following is true for PCA?
  1. The first principal component represents the direction along which thevariance of the dataset is maximized.
  2. The first principal component represents the direction along which thevariance of the dataset is minimized.
  3. The first principal component is the eigenvector corresponding to the smallesteigenvalue of the covariance matrix.
  4. The first principal component is the eigenvector corresponding to the largesteigenvalue of the covariance matrix.
Q11MSQ3 marksselect all that apply
Which of the following statements regarding gradient descent is/are correct?
  1. If the step size is chosen as 1, we may “not always” arrive as close to theoptimal solution even after many number of iterations.
  2. Solution obtained for optimization problem is the local minimum of itsobjective function.
  3. Solution obtained for optimization problem is the global minimum of itsobjective function.
  4. Gradient descent converges to the global optimum in the case of convexfunctions.
  5. Gradient descent does not converge to the global optimum in the case ofconvex functions.
Q12MSQ3 marksselect all that apply
Suppose a fair coin is tossed twice. Let A denote the event that the first toss is a tail. Let B denote the event that the second toss is a tail. Let C denote the event that out of the first two tosses, either both are heads or both are tails. Which among the following options are true? Select all that apply.
  1. A, B and C are independent.
  2. A, B and C are dependent.
  3. A, B and C are pairwise independent.
  4. P(A) = P(B) = P(C) = 1/2
Q13SA2 marks
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Q14SA3 marks
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Q15SA3 marks
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Q16SA3 marks
The average life (in years) of an electric bulb follows exponential distribution with parameter λ = 0.4. Using Weak Law of large numbers, find a lower bound on the probability that the mean life of a random sample of 50 such bulbs falls between 1.5 and 3.5 years. Enter your answer correct to three decimal places.
Q17MCQ4 marks
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Q18MCQ4 marks
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Answer key

  1. Q1 (MCQ, 0 marks): A
  2. Q2 (MSQ, 2 marks): C, D
  3. Q3 (SA, 2 marks): 0
  4. Q4 (SA, 1 marks): 7.44 to 7.52
  5. Q5 (MCQ, 3 marks): C
  6. Q6 (MCQ, 3 marks): A
  7. Q7 (MSQ, 4 marks): B, C
  8. Q8 (MSQ, 4 marks): A, B, D
  9. Q9 (MSQ, 3 marks): A, D
  10. Q10 (MSQ, 3 marks): A, D
  11. Q11 (MSQ, 3 marks): A, B, D
  12. Q12 (MSQ, 3 marks): B, C, D
  13. Q13 (SA, 2 marks): 225
  14. Q14 (SA, 3 marks): 1
  15. Q15 (SA, 3 marks): 40
  16. Q16 (SA, 3 marks): 0.870 to 0.880
  17. Q17 (MCQ, 4 marks): C
  18. Q18 (MCQ, 4 marks): A

Concepts this paper tests

Auto-generated summary from the source site. Handy as a checklist; the lessons above explain each idea from scratch.

Here’s a structured summary of core topics, concepts, principles, and equations for the exam, followed by Mermaid knowledge graphs to visualize relationships.


Core Topics & Concepts

1. Probability & Statistics

Key Distributions (Discrete & Continuous)

DistributionPMF/PDFCDFMean (E[X]E[X])Variance (Var(X)\text{Var}(X))
Uniform (Discrete)fX(k)=1n,k=a,a+1,…,bf_X(k) = \frac{1}{n}, \quad k = a, a+1, \dots, bFX(x)=k−a+1n,a≤x<bF_X(x) = \frac{k-a+1}{n}, \quad a \leq x < ba+b2\frac{a+b}{2}n2−112\frac{n^2-1}{12}
BernoullifX(k)=pk(1−p)1−k,k=0,1f_X(k) = p^k (1-p)^{1-k}, \quad k=0,1FX(x)=1−p,0≤x<1F_X(x) = 1-p, \quad 0 \leq x < 1ppp(1−p)p(1-p)
BinomialfX(k)=nCkpk(1−p)n−k,k=0,1,…,nf_X(k) = {}^nC_k p^k (1-p)^{n-k}, \quad k=0,1,\dots,nFX(x)=∑i=0knCipi(1−p)n−iF_X(x) = \sum_{i=0}^k {}^nC_i p^i (1-p)^{n-i}npnpnp(1−p)np(1-p)
GeometricfX(k)=(1−p)k−1p,k=1,2,…f_X(k) = (1-p)^{k-1}p, \quad k=1,2,\dotsFX(x)=1−(1−p)kF_X(x) = 1-(1-p)^k1p\frac{1}{p}1−pp2\frac{1-p}{p^2}
PoissonfX(k)=e−λλkk!,k=0,1,…f_X(k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad k=0,1,\dotsFX(x)=e−λ∑i=0kλii!F_X(x) = e^{-\lambda} \sum_{i=0}^k \frac{\lambda^i}{i!}λ\lambdaλ\lambda
Uniform (Continuous)fX(x)=1b−a,a≤x≤bf_X(x) = \frac{1}{b-a}, \quad a \leq x \leq bFX(x)=x−ab−aF_X(x) = \frac{x-a}{b-a}a+b2\frac{a+b}{2}(b−a)212\frac{(b-a)^2}{12}
ExponentialfX(x)=λe−λx,x≥0f_X(x) = \lambda e^{-\lambda x}, \quad x \geq 0FX(x)=1−e−λxF_X(x) = 1-e^{-\lambda x}1λ\frac{1}{\lambda}1λ2\frac{1}{\lambda^2}
NormalfX(x)=1σ2πe−(x−μ)22σ2f_X(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}No closed formμ\muσ2\sigma^2
GammafX(x)=βαΓ(α)xα−1e−βxf_X(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}-αβ\frac{\alpha}{\beta}αβ2\frac{\alpha}{\beta^2}
BetafX(x)=Γ(α+β)Γ(α)Γ(β)xα−1(1−x)β−1f_X(x) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1} (1-x)^{\beta-1}-αα+β\frac{\alpha}{\alpha+\beta}αβ(α+β)2(α+β+1)\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}

Key Theorems & Inequalities

  1. Markov’s Inequality:
P(X≥c)≤E[X]c,c>0P(X \geq c) \leq \frac{E[X]}{c}, \quad c > 0
  1. Chebyshev’s Inequality:
P(∣X−μ∣≥kσ)≤1k2P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}
  1. Weak Law of Large Numbers (WLLN):
P(∣X‾−μ∣>δ)≤σ2nδ2P(|\overline{X} - \mu| > \delta) \leq \frac{\sigma^2}{n\delta^2}
  1. Central Limit Theorem (CLT):
Y−nμnσ≈Normal(0,1)\frac{Y - n\mu}{\sqrt{n\sigma}} \approx \text{Normal}(0,1)

where Y=X1+X2+⋯+XnY = X_1 + X_2 + \dots + X_n.

  1. Maximum Likelihood Estimation (MLE):
θ∗=arg⁡max⁡θ∏i=1nfX(xi;θ)\theta^* = \arg \max_\theta \prod_{i=1}^n f_X(x_i; \theta)

2. Linear Algebra

Key Concepts

Key Equations

  1. Condition for Solvability of Ax=bA\mathbf{x} = \mathbf{b}:

    • For A∈Rm×nA \in \mathbb{R}^{m \times n}, b\mathbf{b} must lie in the column space of AA.
    • Example: For the given matrix AA, the condition was 2b1−b2+b3=02b_1 - b_2 + b_3 = 0.
  1. Distance from Origin to Surface:

    • For x2−y2−z2=1x^2 - y^2 - z^2 = 1, the minimum distance is 1 (using Lagrange multipliers or geometric intuition).

3. Optimization & Gradient Descent

Key Concepts

Key Equations

  1. Gradient of Quadratic Function:

    • For f(x1,x2)=8x12+2x1x2−x22f(x_1, x_2) = 8x_1^2 + 2x_1x_2 - x_2^2:
∇f=(16x1+2x2,2x1−2x2)\nabla f = (16x_1 + 2x_2, 2x_1 - 2x_2)
  1. Directional Derivative:

    • Dvf(x)=∇f(x)⋅vD_{\mathbf{v}} f(\mathbf{x}) = \nabla f(\mathbf{x}) \cdot \mathbf{v}.
    • Maximum rate of change: ∥∇f∥\|\nabla f\|.

4. Machine Learning Foundations

Key Concepts

Key Equations

  1. Transformation of Gaussian Variables:

    • If X∼Normal(μ,Σ)\mathbf{X} \sim \text{Normal}(\mu, \Sigma), then Y=AX+b∼Normal(Aμ+b,AΣAT)\mathbf{Y} = A\mathbf{X} + \mathbf{b} \sim \text{Normal}(A\mu + \mathbf{b}, A\Sigma A^T).
    • Independence Condition: For Y1=2X1+2X2+1Y_1 = 2X_1 + 2X_2 + 1 and Y2=3X1−2X2−2Y_2 = 3X_1 - 2X_2 - 2, Y1Y_1 and Y2Y_2 are independent if Cov(Y1,Y2)=0\text{Cov}(Y_1, Y_2) = 0.

      • Computed as a=40a = 40 in the given problem.
  1. Exponential Distribution:

    • X∼Exp(λ)X \sim \text{Exp}(\lambda), E[X]=1λE[X] = \frac{1}{\lambda}, Var(X)=1λ2\text{Var}(X) = \frac{1}{\lambda^2}.
    • Weak Law of Large Numbers (WLLN) Application:

      • For λ=0.4\lambda = 0.4, E[X]=2.5E[X] = 2.5.
      • For n=50n = 50, P(1.5≤X‾≤3.5)≥0.875P(1.5 \leq \overline{X} \leq 3.5) \geq 0.875 (using Chebyshev’s inequality).

5. Multivariable Calculus

Key Concepts

Key Equations

  1. Gradient of f(x,y,z)=x2+2y2−3z2f(x,y,z) = x^2 + 2y^2 - 3z^2:
∇f=(2x,4y,−6z)\nabla f = (2x, 4y, -6z)
  1. Directional Derivative at P0=(1,1,1)P_0 = (1,1,1) in direction v\mathbf{v}:
Dvf=∇f(P0)⋅v=(2,4,−6)⋅(−13,13,−13)=0D_{\mathbf{v}} f = \nabla f(P_0) \cdot \mathbf{v} = (2,4,-6) \cdot \left( -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) = 0
  1. Maximum Rate of Change at P0P_0:
∥∇f(P0)∥=22+42+(−6)2=4+16+36=56≈7.48\|\nabla f(P_0)\| = \sqrt{2^2 + 4^2 + (-6)^2} = \sqrt{4 + 16 + 36} = \sqrt{56} \approx 7.48

Graph 2: Linear Algebra

Graph 3: Optimization

Graph 4: Machine Learning

Graph 5: Multivariable Calculus