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5. Continuous Random Variables

5. Continuous Random Variables. Delivery time. A package is to be delivered between noon and 1pm. When will it arrive?. Delivery time. A probability model. Sample space S 1 = { 0, 1, …, 59 }. equally likely outcomes . Random variable X : minute when package arrives.

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5. Continuous Random Variables

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  1. 5. Continuous Random Variables

  2. Delivery time A package is to be delivered between noon and 1pm. When will it arrive?

  3. Delivery time A probability model Sample space S1 = {0, 1, …, 59} equally likely outcomes Random variable X: minute when package arrives X(0) = 0, X(1) = 1, …, X(59) = 59 • X(w) = w E[X] = 0⋅1/60 + … + 59⋅1/60 = 29.5

  4. Delivery time A more precise probability model S2= {0, , , …, 1, 1 , …, 59 } equally likely outcomes • X: minute when package arrives E[X] = 29.983… 59 1 2 1 60 60 60 60

  5. Taking precision to the limit S = the (continuous) interval[0, 60) equally likely outcomes S1 = {0, 1, …, 59} p = 1/60 S2= {0, , …, 59 } p = 1/3600 59 1 60 60 S= [0, 60) p = 0

  6. Uncountable sample spaces In Lecture 2 we said: “The probability of an event is the sum of the probabilities of its elements” but in S = [0, 60) all elements have probability zero! To specify and calculate probabilites, we have to work with the axioms of probability

  7. The uniform random variable Sample space S =[0, 60) Events of interest: intervals [x, y) ⊆ [0, 60) their intersections, unions, etc. P([x, y)) = (y – x)/60 Probabilities: X(w) = w Random variable:

  8. How to do calculations You walk out of the apartment from 12:30 to 12:45. What is the probability you missed the delivery? Solution Event of interest: E = [30, 45) orE = “30 ≤ X < 45” E 60 0 P(E) = (45 – 30)/60 = 1/4

  9. How to do calculations From 12:08 - 12:12 and 12:54 - 12:57 the doorbell wasn’t working. Event of interest: E = “8 ≤ X < 12” ∪ “54 ≤ X < 57” 60 0 P(E) = P([8, 12)) + P([54, 57)) = 4/60 + 3/60 = 7/60

  10. Cumulative distribution function The probability mass function doesn’t make much sense because P(X = x) = 0 for all x. Instead, we can describe X by its cumulative distribution function (c.d.f.)F: F(x) = P(X ≤ x)

  11. Cumulative distribution functions f(x) = P(X = x) F(x) = P(X ≤ x)

  12. Uniform random variable If X is uniform over [0, 60) then X ≤ x x 60 0 F(x) forx < 0 0 P(X ≤ x) = x/60 forx ∈[0, 60) forx > 60 1 x

  13. Cumulative distribution functions discretec.d.f. F(x) = P(X ≤ x) p.m.f. f(x) = P(X = x) ? continuousc.d.f. F(x) = P(X ≤ x)

  14. Discrete random variables: • p.m.f. f(x) = P(X = x) c.d.f. F(x) = P(X ≤ x) f(x) = F(x) – F(x – d) F(a) = ∑x ≤ a f(x) for small d Continuous random variables: The probability density function (p.d.f.) of a random variable with c.d.f. F(x) is • dF(x) • F(x) – F(x – d) = lim f(x) = • dx • d • d → 0

  15. Discrete random variables: • p.m.f. f(x) = P(X = x) c.d.f. F(x) = P(X ≤ x) F(a) = ∑x ≤ a f(x) Continuous random variables: c.d.f. F(x) = P(X ≤ x) • p.d.f. f(x) = dF(x)/dx F(a) = ∫x ≤ a f(x)dx

  16. Uniform random variable ifx < 0 0 F(x) = x/60 ifx ∈[0, 60) ifx ≥ 60 1 c.d.f. F(x) 1/60 ifx < 0 0 1/60 ifx ∈ (0, 60) dF(x)/dx = ifx > 60 0 p.d.f. f(x)

  17. Cumulative distribution functions discretec.d.f. F(x) = P(X ≤ x) p.m.f. f(x) = P(X = x) continuousc.d.f. F(x) = P(X ≤ x) p.d.f. f(x) = dF(x)/dx

  18. Uniform random variable A random variable X is Uniform(0, 1) if its p.d.f. is 1 ifx ∈ (0, 1) f(x) f(x) = ifx <0orx > 1 0 X A Uniform(a, b) has p.d.f. 1/(b - a) ifx ∈ (a, b) f(x) = b a ifx <aorx > b 0

  19. Some practice A package is to be delivered between noon and 1pm. It is now 12.30 and the package is not in yet. When will it arrive?

  20. Some practice Probability model Arrival time X is Uniform(0, 60) We want the c.d.f. of Xconditioned on X > 30: • P(X ≤ x and X > 30) • P(X ≤ x | X > 30) • = • P(X> 30) • P(30 < X ≤ x) • = • P(X> 30) • (x – 30)/60 • (x – 30)/30 • = • = • 1/2

  21. Some practice The c.d.f. of Xconditioned on X > 30 is for x in [30, 60) G(x) = (x – 30)/30 The p.d.f. of Xconditioned on X > 30 is g(x) = dG(x)/dx = 1/30 for x in [30, 60) and 0 outside. So X conditioned onX > 30 is Uniform(30, 60).

  22. Waiting for a friend Your friend said she’ll show up between 7 and 8 but probably around 7.30. It is now 7.30. What is the probability you have to wait past 7.45?

  23. Waiting for a friend Probability model Let’s assume arrival time X has following p.d.f.: f(x) 1/30 x 60 0 30 We want to calculate • P(X > 45) • P(X > 45| X > 30) • = • P(X> 30)

  24. Waiting for a friend f(x) 1/30 x 60 45 0 30 60 • = 1/2 P(X> 30) = ∫30f(x)dx 60 • = 1/8 P(X> 45) = ∫45f(x)dx • so P(X > 45| X > 30)= (1/8)/(1/2) = 1/4.

  25. Interpretation of the p.d.f. The p.d.f. value f(x) d approximates the probability that Xin an interval of length d around x P(x –d≤ X < x) = f(x) d+ o(d) P(x ≤ X < x + d) = f(x) d+ o(d) Example 1/60 If X is uniform, then f(x) = 1/60 d P(x ≤ X < x + d) = d/60 x

  26. Discrete versus continuous p.m.f. f(x) p.d.f. f(x) ∑x ≤ a f(x) ∫x ≤ a f(x)dx P(X ≤ a) E[X] ∑x x f(x) ∫x x f(x)dx ∑x x2 f(x) ∫x x2 f(x)dx E[X2] Var[X] E[(X – E[X])2] = E[X2] – E[X]2

  27. Uniform random variable A random variable X is Uniform(0, 1) if its p.d.f. is 1 ifx ∈ (0, 1) f(x) = ifx <0orx > 1 0 m f(x) ∫0 f(x)dx 1 = ∫0 dx = 1 1 m +s m –s • = 1/2 • = x2/2|0 E[X] = ∫0 x f(x)dx 1 1 • = 1/3 1 • = x3/3|0 m = E[X] s = √Var[X] E[X2] = ∫0 x2 f(x)dx 1 x Var[X] = 1/3 – (1/2)2 = 1/12 √Var[X]= 1/√12 ≈ 0.289

  28. Uniform random variable A random variable X is Uniform(a, b) if its p.d.f. is 1/(b - a) ifx ∈ (a, b) f(x) = ifx <aorx > b 0 Then E[X] = (a + b)/2 Var[X] = (b – a)2/12

  29. Raindrops again Rain is falling on your head at an average speed of l drops/second. 2 1 How long do we wait until the next drop?

  30. Raindrops again Probability model Time is divided into intervals of length 1/n Events Ei = “raindrop hits in interval i” have probability p = l/n and are independent X = interval of first drop X 2 1 P(X = x) • = P(E1c…Ex-1cEx) • = (1 – p)x-1p

  31. Raindrops again X = interval of first drop X = x x-1 x T n n T = time (in seconds) of first drop (x– 1)/n ≤ T < x/n X = x means that P((x– 1)/n ≤ T < x/n) = P(X = x) • = (1 – p)x-1p • = (1 – l/n)x-1(l/n)

  32. Raindrops again T = time (in seconds) of first drop • P((x–1)/n ≤ T < x/n) = (1 – l/n)x-1(l/n) If we set t = (x – 1)/n and d = 1/n we get • P(t ≤ T < t+ d) = (1 – d l)t/d(dl) • = dl e– (d l+ o(d l)) t/d = l e-lt P(t ≤ T < t + d) lim f(t) = • d • d → 0

  33. The exponential random variable The p.d.f. of an Exponential(l) random variable T is le-lt ifx ≥ 0 f(t) = ifx < 0. 0 l= 1 l= 1 p.d.f. f(t) c.d.f. F(t) = P(T≤t)

  34. The exponential random variable The c.d.f. of T is a F(a) = ∫0 le-ltdt= e-lt|0 = 1 –e-la • if a≥ 0 a What should the expected value of T be? (Hint: Rain falls at l drops/second How many seconds till the first drop?) E[T] = 1/l Var[T] = 1/l2

  35. Poisson vs. exponential N 2 1 T Exponential(l) Poisson(l) number of events within time unit time until first event happens description • 1/l • l expectation • 1/l • l std. deviation

  36. Memoryless property How much time between the second and third drop? 2 1 T Solution We start time when the second drop falls. What happened before is irrelevant. Then T is Exponential(l)

  37. Expected time What is the expected time of the third drop? 2 1 T T1 T2 T3 Solution T = T1 + T2 + T3 T is not exponential but E[T]= E[T1] + E[T2] + E[T3] = 3/l

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