ENGG 2040C: Probability Models and Applications Andrej Bogdanov Spring 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 X(0) = 0, X(1) = 1, …, X(59) = 59 E[X] = 0 ⋅ 1/60 + … + 59 ⋅ 1/60 = 29.5 X(  ) = 

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

Taking precision to the limit S = the (continuous) interval [0, 60) equally likely outcomes p = 1/60 p = 1/3600 p = 0 S 1 = {0, 1, …, 59} S 2 = {0,, …, 59 } S = [0, 60)

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

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

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) or E = “30 ≤ X < 45” P(E) = (45 – 30)/60 = 1/4 E 0 60

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

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)

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

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

Cumulative distribution functions p.m.f. f(x) = P(X = x) discrete c.d.f. F(x) = P(X ≤ x) continuous c.d.f. F(x) = P(X ≤ x) ?

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) f(x) = F(x) – F(x –  ) for small  Continuous random variables: The probability density function (p.d.f.) of a random variable with c.d.f. F(x) is f(x) = F(x) – F(x –  )  lim  → 0 dF(x) dx =

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: p.d.f. f(x) = dF(x)/dx c.d.f. F(x) = P(X ≤ x) F(a) = ∫ x ≤ a f(x)dx

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

Cumulative distribution functions discrete c.d.f. F(x) = P(X ≤ x) continuous c.d.f. F(x) = P(X ≤ x) p.m.f. f(x) = P(X = x) p.d.f. f(x) = dF(x)/dx

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

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

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

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

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

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

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

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

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

Uniform random variable A random variable X is Uniform(0, 1) if its p.d.f. is f(x) = 0 if x ∈ (0, 1) 1 if x 1 E[X] = ∫ 0 x f(x)dx 1 ∫ 0 f(x)dx 1 = ∫ 0 dx = 1 1 = x 2 /2| 0 1 = 1/2  E[X 2 ] = ∫ 0 x 2 f(x)dx 1 = x 3 /3| 0 1 = 1/3 Var[X] = 1/3 – (1/2) 2 = 1/12  –   +  f(x)f(x) x  = E[X]  = √Var[X] √Var[X] = 1/√12 ≈ 0.289

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

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

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

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

Raindrops again T = time (in seconds) of first drop P((x – 1)/n ≤ T < x/n) = (1 – /n) x-1 ( /n) If we set t = (x – 1)/n and  = 1/n we get P(t ≤ T < t +  ) = (1 –  ) t/  (  ) f(t) =  lim  → 0 P(t ≤ T < t +  ) = e - t =  e – (  + o(  )) t/ 

The exponential random variable The p.d.f. of an Exponential( ) random variable T is f(t) = e - t 0 if x ≥ 0 if x < 0. p.d.f. f(t) = 1 c.d.f. F(t) = P(T ≤ t) = 1

The exponential random variable The c.d.f. of T is F(a) = ∫ 0 e - t dt = e - t | 0 = 1 – e - a a a if a ≥ 0 What should the expected value of T be? (Hint:Rain falls at drops/second How many seconds till the first drop?) E[T] = 1/ Var[T] = 1/ 2

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

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

T Expected time What is the expected time of the third drop? 1 2 T3T3 T2T2 T1T1 Solution T = T 1 + T 2 + T 3 T is not exponential but E[T] = E[T 1 ] + E[T 2 ] + E[T 3 ] = 3/