Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Material Type: Exam; Class: PROBAB MTH,ELEC ENG; Subject: STATISTICS; University: Iowa State University; Term: Unknown 1989;

Typology: Exams

Pre 2010

1 / 19

Download Study Guide for Final Exam Preparation | STAT 322 and more Exams Statistics in PDF only on Docsity! SUMMARY AND EXAMPLES Preparing final exam: (Dec. 15, W. 9:45 - 11:45 am). Importance of materials: 1. All homework solutions; 2. Lecture notes examples and past exam questions; 3. Relevant textbook content. EE/STAT 322, #25 1 COVERED TOPICS Outline: • Introduction To Probability: Random Experiments and Events, Definition of Probability, Elementary Set Theory, The Axiomatic Approach. • Set Operations, Conditional Probability, Independence • Combined Experiment Bernoulli Trials and its Applications • Concept of A Random Variable: − Distribution Function (CDF), Probability Density Function (PDF), − Mean Values and Moments EE/STAT 322, #25 2 COVERED TOPICS (CONT.) • Correlation Functions − Auto-correlation Functions – Properties, examples, − Crosscorrelation Functions − Relation to stationary, ergodic processes. − Correlation matrix, covariance matrix. • Power Spectral Density (PSD) Relation of PSD to Fourier Transform, Mean-Square values from PSD, White Noise • Central Limit Theorem (CLT) − Facts about Gaussian PDF and MGF − Markov and Chebyshev inequalites. • Linear MMSE estimation, Orthogonality Principle, MSE EE/STAT 322, #25 5 EXAMPLE: MMSE ESTIMATE Let X and Y be two jointly Gaussian distributed random variables. X ∼ N(0, σ2X), and Y ∼ N(0, σ 2 Y ), and their correlation coefficient is ρ = E[XY ] σXσY . Let Ŷ = aX be an estimate of Y using X. 1) Find a such that the MSE J = E[|Y − Ŷ |2] is minimized. 2) Find the minimum MSE Jmin. 3). Show that E[Ŷ (Y − Ŷ )] = 0. Solution: (1) J is minimized if dJda = 0, where J = E[|Y − aX| 2]. dJ da = 0 ⇒E[(Y − aX)(−2X)] = 0 ⇒E[Y X] = aE[X 2]. Since E[Y X] = ρσXσY and E[X 2] = σ2X, we have a = E[Y X]/E[X2] = ρσXσY /σ 2 X = ρσY /σX. EE/STAT 322, #25 6 EXAMPLE: MMSE ESTIMATE (CONT.) (2) Method I: Jmin = E[|Y − aX| 2] = E[Y 2] + E[a2X2] − E[2aXY ] = σ2Y + ( ρ2σ2Y σ2X )σ2X − 2 ρσY σX ρσY σX = σ2Y + ρ 2σ2Y − 2ρ 2σ2Y = (1 − ρ 2)σ2Y . Method II: (orthogonal principle:) Jmin = E[|Y − aX| 2] = E[Y 2] − E[a2X2] = σ2Y − ( ρ2σ2Y σ2X )σ2X = (1 − ρ 2)σ2Y . (3) E[Ŷ (Y − Ŷ )] = E[aX(Y − aX)] = aE[XY ] − a2E[X2] = (ρσYσX )ρσXσY − ( ρ2σ2Y σ2 X )σ2X = ρ2σ2Y − ρ 2σ2Y = 0. EE/STAT 322, #25 7 EXAMPLE: MGF (CONT.) Y and Z are independent RVs, where Y is exponentially distributed with parameter 2, and Z is Gaussian with mean 3 and variance 4. (a) Find the MGF for 2Z + 3; (b) Find the MGF for Y + Z. Solution: fY (y) = 2e −2y, and Z ∼ N(3, 4). MY (s) = 2/(2 − s), and MZ(s) = e µs+σ2s2/2 = e3s+2s 2 . (a) M2Z+3(s) = E[e s(2Z+3)] = e3sE[es2Z] = e3sMZ(2s) = e3se3(2s)+2(2s) 2 = e9s+8s 2 . (b) MY +Z(s) = MY (s)MZ(s) = 2 2−se 3s+2s2. EE/STAT 322, #25 10 EXAMPLE (GAUSSIAN CONDITIONAL DISTRIBUTION) Let X be a Gaussian RV with mean 4 and variance 16. (a) Find the probability P (X < 4) and P (X > 6) Solution: X = 4. σx = 4. P (X < 4) = Q(−(4 − 4)/4) = Q(0) = 0.5. P (X > 6) = Q((6 − 4)/4) = Q(0.5). EE/STAT 322, #25 11 (b) Find the conditional distribution function FX(x|X < 0), and draw the shape of FX(x|X < 0). Solution: P (X < 0) = FX(0) = Q(−(0 − 4)/4) = Q(1). FX(x|X < 0) = P (X≤x,X<0) P (X<0) = { P (X<x) P (X<0) X < 0 P (X<0) P (X<0) = 1 X > 0 , FX(x) = Q(−(x − 4)/4) = Q(1 − x/4). ⇒FX(x|X < 0) = Q(1 − x/4)/Q(1) for X < 0. Thus, FX(x|X < 0) = { Q(1 − x/4)/Q(1) X < 0 1 X > 0 . EE/STAT 322, #25 12 (c) find X − Y , σ2x−y, and E[(X − Y ) 2] = (X − Y )2, Solution: E[X − Y ] = E(X) − E(Y ) = 1 − 2 = −1. σ2x−y = σ 2 x + σ 2 y − 2ρσxσy = 1 + 9 − 2 ∗ 0.5 ∗ 1 ∗ 3 = 10 − 3 = 7. E[(X − Y )2] = (E[X − Y ])2 + σ2x−y = (−1) 2 + 7 = 8. [Hint: Z2 = Z 2 + σ2Z, let Z = X − Y ]. EE/STAT 322, #25 15 EXAMPLE (PSD) Use the Parsevel’s theorem to evaluate the following integral: ∫ ∞ −∞ ( sin(4ω) 4ω · 1 ω2+4 ) dω. Solution: F (ω) = ( sin(4ω) 4ω ) = (sinc(8Tf)) ⇒f(t) = F−1{F (ω)} = 18, |t| ≤ 4. g(t) = F−1{G(ω)} = F−1{ 1 ω2+4 } = 14e −2|t|. Using Parseval’s theorem: ∫ ∞ −∞ f(t)g(t)dt = 12π ∫ ∞ −∞ F (ω)G(−ω)dω. ⇒ ∫ ∞ −∞ ( sin(4ω) 4ω · 1 ω2+4 ) dω = 2π ∫ ∞ −∞ f(t)g(t)dt = 2π ∫ 4 −4 1 8 1 4e −2|t|dt = 2π32 ∫ 4 0 2e−2tdt = π16e −2t|04 = π 16(1 − e −8). EE/STAT 322, #25 16 EXAMPLE (PSD) A stationary process X(t) has the PSD SX(ω) = 1 ω2+1 + 2π4δ(ω). (a) Find RX(τ). (b) Find X̄, X̄2 and σ 2 X. Solution: (a) RX(τ) = F −1{SX(ω)}. F−1{ 2Aβ ω2+β2 + 2π4δ(ω)} = Ae−β|τ | + 4, where A = 12, β = 1. ⇒RX(τ) = 1 2e −|τ | + 4. (b) Since X(t) is stationary, we get RX(∞) = X̄ 2, ⇒X̄2 = 12e −∞ + 4 = 4 ⇒X̄ = ±2. X̄2 = RX(0) = 1 2 + 4 = 4.5. σ2X = X̄ 2 − X̄2 = 4.5 − 4 = 0.5. EE/STAT 322, #25 17