1. P robability 1 07. Midterm Practice Condition Independence 郭俊利 2009/04/13

2. Probability 2 Concept  Homework  p(n) > 1 Discrete  f ( x ) > 1 Continuous  F( x ) > 1 2.1 ~ 4.1

3. Probability 3 Basic Probability Set P(A ∪ B ∪ C) = P(A) + P(A C ∩ B) + P(A C ∩ B C ∩ C) Condition P(A ∩ B ∩ C) = P(A) P(B|A) P(C|A ∩ B) Independence P(A ∩ B) = P(A) P(B) Problem 1.31 – error bit

4. Probability 4 Expectation E[a] = a E[aX + b] = aE[X] + b E[g(X)] = Σ x g(x) p x (x) E[X] = Σ i E[X i ] = np (p is uniform!) E[X] = Σ i P(A i ) E[X | A i ] E[X] = E[E[X|Y]] var(aX + b) = a 2 var(X)

5. Probability 5 Graph Mean Center (not find f(x) ) Variance E[X 2 ] – E[X] 2 (need f(x) or formula) f(x)f(x) x 1/3 2/3 1 2

6. Probability 6

7. 7 Other E.V. Independence E[XY] = E[X] E[Y] var(X + Y) = var(X) + var(Y) Sum E[T] = E[X] E[N] var(T) = var(X) E[N] + E 2 [X] var(N) T = X 1 + … + X N f(x, y) = f(x) f(y)

8. Probability 8 Conditional Sum of Independence 1 st Let X 1, X 2 and X 3 be independent and identical binomial random variables, that is, P(X i = k) = C n k p k (1 − p) n−k, 0 ≤ k ≤ n. Compute the P(Z = X 1 + X 2 + X 3 ) Compute E[Z], var(Z) E[  ] = np ; var(  ) = np(1-p)

9. Probability 9

10. 10 Joint f X,Y (x,y) = f Y (y) f X|Y (x|y) X, Y are independent, X and Y are in [0, 2]. f XY (x, y) = xy / 4, find E[ f(x, y) ] Double integration or … ∵ f(x, y) = f(x) f(y) ∴ f(x) = x / 2 f(y) = y / 2 or f(x) = x f(y) = y / 4

11. Probability 11

12. Probability 12 Important Random Variable Bernoulli p X (k) = p, 1-p Binomial p X (k) = C n k p k (1 – p) n – k Geometric p X (k) = (1 – p) k-1 p E[X] = p var(X) = p(1-p) E[X] = np var(X) = np(1-p) E[X] = 1/p var(X) = (1-p)/p 2

13. Probability 13 Geometric random number Xiao-Quan plays a game rock-paper-scissors with another. He plays until he loses. Find the expectation of the number of rounds. If Xiao-Quan has won 3 times and drawn 2 times, how many rounds will he expect to play? Problem 2.22, 2.23 Non-memoryless

14. Probability 14 Conditional Sum of Independence 2 nd Suppose that X and Y are independent and identical geometrical random variables with parameter p, that is, P(X = k) = P(Y = k) = q k−1 p, k ≥ 1. Compute P(X = i | X + Y = n), i = 1, 2,..., n − 1. Compute E(X | X + Y = n), var(X | X + Y = n). P(X = i | X + Y = n) = P(X = i ∩ X + Y = n) P(X + Y = n) = P(X = i ∩ Y = n – i) P(X + Y = n) = P(X = i) P(Y = n – i) P(X + Y = n) = 1 / n-1

15. Probability 15 Exponential random number f(x) = λe –λx P(x ≧ a) =∫ a ∞ λe –λx dx = –e –λx | a ∞ = e –λa F(x) = 1 – e –λx ( Geometric F(n) = 1 – (1–p) n ) E[X] = 1 / λ var(X) = 1 / λ 2 (E[X 2 ] = 2 / λ 2 )

16. Probability 16 Exponential Examples The spent time of work is modeled as an exponential random variable. The average time that Xiao-Ming completes the task is 10 hours. What is the probability that Xiao-Ming has done this task early (in advance) ? f(x) = ½ λe –λx, x ≥ 0 ½ λe +λx, x < 0 F(x) = ?

17. Probability 17

18. Probability 18 Condition P(A|B) = P(A∩B) / P(B) p X|A (x) = P({X = x} ∩ A) / P(A) f X|A (x) = f(x) / P(A) Roll a fair die shown k points, what is the probability p(k) given some roll is even number? Xiao-Wang arrivals is a uniform random variable from 7:10 to 7:30. The bus comes at 7:15 and 7:30. What is the waiting time f(x) of Xiao-Wang? Show exponential random number is memoryless.

19. Probability 19

20. Probability 20 Normal random number Standard normal distribution N( – a) = P(Y ≦ – a) = P(Y ≧ a) = 1 – P(Y ≦ a) N( – a) = 1 – N(a) CDF P(X ≦ a) = P(Y ≦ ) = N( ) a – μ σ a – μ σ The grades of the exam is suitable for a normal random variable. The average of grades = 60 and the standard deviation = 20. What is the probability that Xiao-Kuo ’ s grade will be higher over 70?

21. Probability 21 F() and f () Derived Linear Y = aX + b General y = g(x), x = h(y)  y = g(h(y)) f Y (y) = f X ( h(y) ) |h’(y)|

22. Probability 22 Derived Distribution Find the PDF of Z = g(X, Y) = Y/X F Z (z) =0 ≤ z ≤ 1 F Z (z) =z > 1 f Z (z) =

23. Probability 23 Linear Mapping X is an exponential random variable, Y = – λX + 2, Find the PDF of Y f(x) = λe –λx f(y) = e –λ(y-2 / –λ) = e y-2 λ |-λ|

24. Probability 24 Travel Problem Xiao-Hua is driving from Boston to New York 180 miles. His speed is uniformly distributed between 30 and 60 mph. What is the distribution of the duration of the trip? f V (v) = 1 / 30 30 ≤ v ≤ 60 T(v) = 180 / v f T (t) = f V (v) T(t)’ = 6 / t 2

25. Probability 25

26. Probability 26 Convolution W = X + Y P W (w) = Σ P X (x) P Y (w – x) W = |X| + 2Y p(x) = 1/3, if x = –1, 0, 1 p(y) = 1/2, y = 0 1/3, y = 1 1/6, y = 2

27. Probability 27 Maximum W = max {X, Y} (X, Y = [0, 1] uniformly) W = max {X, Y} (X, Y = 0.1, 0.2, …, 1.0 uniformly) P(X ≦ w) = P(Y ≦ w) = w F W (w) = P(X ≦ w) P(Y ≦ w) = w 2 f W (w) = 2w p W (w) = F W (w) – F W (w – 0.1) = w 2 – (w–0.1) 2