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 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

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

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

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

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(w) = w

How to do calculations Solution 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” E 60 P(E) = (45 – 30)/60 = 1/4

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 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) The c.d.f. makes sense for discrete as well as continuous random variables.

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

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

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(x) = F(x) – F(x – d) for small d F(a) = ∑x ≤ a f(x) 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 – d) d lim d → 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 if x < 0 F(x) = x/60 if x ∈ [0, 60) 1 if x ≥ 60 c.d.f. F(x) p.d.f. f(x) 1/60 if x < 0 dF(x)/dx = 1/60 if x ∈ (0, 60) if x > 60

Cumulative distribution functions discrete c.d.f. F(x) = P(X ≤ x) p.m.f. f(x) = P(X = x) continuous c.d.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) = if x ∈ (0, 1) 1 if x < 0 or x > 1 f(x) A Uniform(a, b) has p.d.f. f(x) = if x ∈ (a, b) 1/(b - a) if x < a or x > b X a b

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?

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

Some practice The c.d.f. of X conditioned on X > 30 is G(x) = (x – 30)/30 for x in [30, 60) The p.d.f. of X conditioned on X > 30 is 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 7.30. It is now 7.30. What is the probability you have to wait past 7.45?

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

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

Interpretation of the p.d.f. The p.d.f. value f(x) d approximates the probability that X in 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 x d If X is uniform, then f(x) = 1/60 P(x ≤ X < x + d) = d/60

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

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

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

Raindrops again Rain is falling on your head at an average speed of l drops/second. 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 Ei = “raindrop hits in interval i” have probability p = l/n and are independent X = interval of first drop X 1 2 P(X = x) = P(E1c…Ex-1cEx) = (1 – p)x-1p

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

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 f(t) = d lim d → 0 P(t ≤ T < t + d) = l e-lt

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

The exponential random variable The c.d.f. of T is F(a) = ∫0 l e-lt dt = e-lt|0 = 1 – e-la a if a ≥ 0 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

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

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

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