A Theory of Probability Homework 4 Solutions

Statistics 100A Homework 4 Solutions

First, note that we have a finite number of trials, n = 4 (although the game goes on forever, we are only concerned with the first 4 balls.) Each trial is a Bernoulli trial that is, each trial has only one of two outcomes: white and not white. Define a success as the event that a white ball is drawn. Then the probability of success p is p = 12 . Since each ball is replaced after it is drawn, we...

Discrete Time Markov Chains : Ergodicity Theory

Lecture 8: Discrete Time Markov Chains: Ergodicity Theory Announcements: 1. We handed out HW2 solutions and your homeworks in Friday’s recitation. I am handing out a few extras today. Please make sure you get these! 2. Remember that I now have office hours both: Wednesday at 3 p.m. and Thursday at 4 p.m. Please show up and ask questions about the lecture notes, not just the homework! No one cam...

Effects of Annotations and Homework on Learning Achievement: An Empirical Study of Scratch Programming Pedagogy

In Taiwan elementary schools, Scratch programming has been taught for more than four years. Previous studies have shown that personal annotations is a useful learning method that improve learning performance. An annotation-based Scratch programming (ASP) system provides for the creation, share, and review of annotations and homework solutions in the interface of Scratch programming. In addition...

Computational Complexity - Homework I (Solutions)

Homework 4 Solutions

k=0 E|X|IAk = E|X|IA <∞ , namely, EXIAn is absolutely summable. 2. Q(A) ≥ 0 and Q(Ω) = 1, with countable additivity of Q shown in part (a). 3. Per our assumption EX = EY . If EX = EY = 0 then both X = 0 a.s. and Y = 0 a.s. so we are done. Otherwise, the probability measures QX(A) = EXIA/EX and QY (A) = EY IA/EY of part (b) agree on the π-system A, hence they must agree on F = σ(A). Considering ...