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Lecture 5 Discrete Random Variables: Definition and Probability Mass Function

Lecture 5 Discrete Random Variables: Definition and Probability Mass Function. Last Time Reliability Problems Definitions of Discrete Random Variables Probability Mass Functions Families of DRVs Reading Assignment : Sections 2.1-2.3. Probability & Stochastic Processes

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Lecture 5 Discrete Random Variables: Definition and Probability Mass Function

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  1. Lecture 5 Discrete Random Variables: Definition and Probability Mass Function Last Time Reliability Problems Definitions of Discrete Random Variables Probability Mass Functions Families of DRVs Reading Assignment: Sections 2.1-2.3 Probability & Stochastic Processes Yates & Goodman (2nd Edition) NTUEE SCC_03_2008

  2. Lecture 5: DRV: CDF, Functions, Exp. Values Today • Families of DRVs (Cont.) • Cumulative Distribution Function (CDF) • Averages • Functions of DRV Tomorrow • Functions of DRV (Cont.) • Expected Value of a DRV Reading Assignment: Sections 2.4-2.7 Probability & Stochastic Processes Yates & Goodman (2nd Edition) NTUEE SCC_03_2008

  3. Lecture 5: Next Week • Discrete Random Variables • Variance and Standard Deviation • Conditional Probability Mass Function • Continuous Random Variables (CRVs) • CDF • Probability Density Functions (PDF) • Expected Values • Families of CRVs Reading Assignment: Sections 2.8-3.4 Probability & Stochastic Processes Yates & Goodman (2nd Edition) NTUEE SCC_03_2008 5- 3

  4. What have you learned about DRV? • Example: Taiwan’s Presidential Election of 3/22 • D.R.V. • Poll • Gambling Probability & Stochastic Processes Yates & Goodman (2nd Edition) NTUEE SCC_03_2008 5- 4

  5. What have you learned? • Example Alex, Ben, and Tim are three prisoners, one of whom is scheduled to die. If a jailer told Alex that which one of Ben and Tim is going to be freed, will this change the probability of Alex dying? • Let A, B, T, and J be the event that Alex, Ben, and Tim will die and the event that a jailer told Alex that Tim will be freed. Then P(A|J) = P(J|A)P(A)/[P(J|A)P(A) + P(J|B)P(B) + P(J|T)P(T)] = (1/2)(1/3)/[(1/2)(1/3) + 1(1/3) + 0(1/3)] = 1/3. Q: Relation to independence? • Poll • Gambling Probability & Stochastic Processes Yates & Goodman (2nd Edition) NTUEE SCC_03_2008 5- 5

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  8. Example 2.A2 Q: How are X and Y related to events? Q:How would you associate probability to X and Y? Probability & Stochastic Processes Yates & Goodman (2nd Edition) NTUEE SCC_03_2008 Toss a die Q: Could you define two random variables X and Y? 4- 8

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