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03. Random Variable

Probability Mass Function Expectation 郭俊利 2009/03/16. 03. Random Variable. 1.7 ~ 2.6. Outline. Review Conditional probability Random variable PMF Conditional PMF. Example 1.

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03. Random Variable

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  1. Probability Mass Function Expectation 郭俊利2009/03/16 03. Random Variable

  2. 1.7 ~ 2.6 Outline • Review • Conditional probability • Random variable • PMF • Conditional PMF

  3. Example 1 • From the set of integers {1, 2, 3,. . . , 100000} a number is selected at random. What is the probability that the sum of its digits is 11? All – (a digit over 11) – (a digit over 10) = H511 – C51 – C52 2!

  4. Example 2 • First throw an unbiased die, then throw as many unbiased coins as the point shown on the die. • What is the probability of obtaining k heads? • If 3 heads are obtained, what is the probability that the die showed n?

  5. 36 Random Variable • The random variable is a real-valued function of the outcome of the experiment. Discrete General = Continuous p(x) (a) (b) 6 5 4 3 2 1 The sum of two dices is x, what is p(x) ? p(x) is uniform ? x 1 2 3 4 … 7 … 11 12

  6. upper upper lower lower Expected Value (1/2) • Example: • The expectation of throwing a dice is 3.5 • The answer to a question is 80% correctly, the grade may be 80. Expectation E[X] = Σx pX(x) p(x) > 0 x ∈ X

  7. Expected Value (2/2) • E[a] = a • E[aX] = aE[x] • E[aX + b] = aE[x] + b • E[g(X)] = Σ g(x) px(x) • E[X] = Σ E[Xi] = np (p is uniform!) i = 1 ~ n

  8. Example 3 • The shooting average of A is 2/3 The shooting average of B is 3/4 The shooting average of C is 4/5 (1) P (at least one hit) = (2) P (one hit) = P (two hits) = P (three hits) = (3) (A hit | one hit) = (4) E (how many hits) =

  9. PMF • Probability Mass Function { X = 1, if a head, 0, if a tail. Its PMF is pX(k) = p, if k = 1, 1 – p, if k = 0. {

  10. Example 4 • Let X be a random variable that takes values from 0 to 9 equal likely. (1) Find the PMF of Y = X mod 3 (2) Find the PMF of Z = 5 mod X+1 • A family has 5 natural children and has adopted 2 girls. Find the PMF of the number of girls out of the 7 children.

  11. Example 5 (variance) (1) Find a and E[X]. (2) What is the PMF of Z = (X – E[X])2? (3) From above, find the variance of X. (4)var(X) = Σx (x – E[X])2 pX(x). { p(x) = x2 / a, if |x| < 4 and x ∈ Z 0, otherwise.

  12. Important PMF • Bernoulli • pX(k) = p, 1-p • Binomial • pX(k) = Cnk pk (1 – p)n – k • Geometric • pX(k) = pk (1 – p)n – k • Poisson • pX(k) = e–λλk / k! • e = 2.7183…

  13. 15 5! Poisson • Poisson is a good model for Binomial • When p is very veryvery small and n is veryvery very large, then λ = np • The probability of a wrong words is 0.1% and there are 1000 words in the document, what is the probability of 5 words found? C100050.0015(0.999)95 ≒ e–1

  14. 1 52 Joint PMF • Joint PMF • pX, Y(x, y) = P(X = x, Y = y) • Marginal PMF pX(x) = Σy pX, Y(x, y) pY(y) = Σx pX, Y(x, y) • Example • pX, Y(x, y) = 1 / 52 pX(x) = 13 = 1 / 4

  15. Example 6 X: –3 < x < 5 Y: –2 < y – x < 2 • Joint PMF • pX, Y(x, y) = • Marginal PMF pX(x) = pY(y) = • The averages of X and Y Question X and Y ∈ Z

  16. Example 7 • A perfect coin is tossed n times. Let Yn denote the number of heads obtained minus the number of tails. Find the probability distribution of Yn and its mean.

  17. Conditional PMF • pX|A (x) = P(X = x | A) • pX(x) = Σi P(Ai) pX|Ai (x) • pX, Y(x, y) = pY(y) pX|Y (x|y) • E[X] = Σi P(Ai) E[X | Ai]

  18. Example 8 • You maybe ask 0, 1 or 2 questions equal likely, I answer 75% correctly. X: the number of questions; Y: the number of wrong answers. Constructing joint PMF pX, Y(x, y) to find the probability of… • One question is asked and is answered wrong. • At least one wrong answer.

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