ENGG 2040C: Probability Models and Applications Andrej Bogdanov Spring Random variables part two

Review A discrete random variable X assigns a discrete value to every outcome in the sample space. Probability mass function of X : p(x) = P(X = x) Expected value of X : E[X] = ∑ x x p(x). E[N]E[N] N: number of heads in two coin flips

One die Example from last time F = face value of fair 6-sided die E[F] = =

Two dice S = sum of face values of two fair 6-sided dice Solution s pS(s)pS(s) E[S] = … = 7 We calculate the p.m.f. of S :

Two dice again S = sum of face values of two fair 6-sided dice F1F1 F2F2 S = F 1 + F 2 F 1 = outcome of first die F 2 = outcome of second die

Sum of random variables Let X, Y be two random variables. X + Y is the random variable that assigns value X(  ) + Y(  ) to outcome . X assigns value X(  ) to outcome  Y assigns value Y(  ) to outcome 

Sum of random variables F1F1 F2F2 S = F 1 + F … … … …

Linearity of expectation E[X + Y] = E[X] + E[Y] For every two random variables X and Y

Two dice again S = sum of face values of two fair 6-sided dice Solution 2 E[S] = E[F 1 ] + E[F 2 ]= = 7 F1F1 F2F2 S = F 1 + F 2

Balls We draw 3 balls without replacement from this urn: What is the expected sum of values on the 3 balls? 0

Balls S = B 1 + B 2 + B 3 where B i is the value of i -th ball. E[S] = E[B 1 ] + E[B 2 ] + E[B 3 ] p.m.f of B 1 : -101 x p(x)p(x) E[B 1 ] = -1 (4/9) + 0 (2/9) + 1 (3/9) = -1/9 same for B 2, B 3 E[S] = 3 (-1/9) = -1/3.

Three dice E[N] = E[I 1 ] + E[I 2 ] + E[I 3 ] Solution I1I1 I2I2 I3I3 N = I 1 + I 2 + I 3 E[I 1 ] = 1 (1/6) + 0(5/6) = 1/6 E[I 2 ], E[I 3 ] = 1/6 = 3 (1/6) = 1/2 Ik =Ik = 1 if face value of k th die equals 0 if not N = number of s. Find E[N].

Problem for you to solve Five balls are chosen without replacement from an urn with 8 blue balls and 10 red balls. What is the expected number of blue balls that are chosen? What if the balls are chosen with replacement? (a) (b)

The indicator (Bernoulli) random variable Perform a trial that succeeds with probability p and fails with probability 1 – p p(x)p(x) 1 – pp x p = 0.5 p(x)p(x) p = 0.4 p(x)p(x) E[X] = p If X is Bernoulli(p) then

The binomial random variable Binomial(n, p) : Perform n independent trials, each of which succeeds with probability p. X = number of successes Examples Toss n coins. (# heads) is Binomial(n, ½). Toss n dice. (# s) is Binomial(n, 1/6).

A less obvious example Toss n coins. Let C be the number of consecutive changes ( HT or TH ).  C()C() HTHHHHT 3 THHHHHT HHHHHHH 2 0 Examples: Then C is Binomial(n – 1, ½).

A non-example Draw 10 cards from a 52-card deck. Let N = number of aces among the drawn cards Is N a Binomial(10, 1/13) random variable? No! Different trial outcomes are not independent.

Properties of binomial random variables If X is Binomial(n, p), its p.m.f. is p(k) = P(X = k) = C(n, k) p k (1 - p) n-k We can write X = I 1 + … + I n, where I i is an indicator random variable for the success of the i -th trial E[X] = E[I 1 ] + … + E[I n ] = p + … + p = np. E[X] = np

Probability mass function Binomial(10, 0.5) Binomial(50, 0.5) Binomial(10, 0.3) Binomial(50, 0.3)

Investments You have two investment choices: A: put $25 in one stock B: put $½ in each of 50 unrelated stocks Which do you prefer?

Investments Probability model Each stock doubles in value with probability ½ loses all value with probability ½ Different stocks perform independently

Investments N A = amount on choice A N B = amount on choice B 50 × Bernoulli(½) A: put $50 in one stockB: put $½ in each of 50 stocks Binomial(50, ½) E[NA]E[NA] E[NB]E[NB]

Variance and standard deviation Let  = E[X] be the expected value of X. The variance of X is the quantity Var[X] = E[(X –  ) 2 ] The standard deviation of X is  = √ Var[X] They measure how close X and  are typically.

Calculating variance  = E[N A ] y25 2 q(y)q(y) 1 p.m.f of (N A –  ) 2 Var[N A ] = E[(N A –  ) 2 ] x050 p(x)p(x) ½ p.m.f of N A = 25 2  = std. dev. of N A = 25  –   + 

Another formula for variance Var[X] = E[(X –  ) 2 ] = E[X 2 – 2  X +  2 ] = E[X 2 ] + E[–2  X] + E[  2 ] = E[X 2 ] – 2  E[X] +  2 = E[X 2 ] – 2  +  2 = E[X 2 ] –  2 for constant c, E[cX] = cE[X] for constant c, E[c] = c Var[X] = E[X 2 ] – E[X] 2

Variance of binomial random variable Suppose X is Binomial(n, p). Then X = I 1 + … + I n, where I i = 1, if trial i succeeds 0, if trial i fails Var[X] = E[X 2 ] –  2 = E[X 2 ] – (np) 2  = E[X] = np E[X2]E[X2] = E[(I 1 + … + I n ) 2 ] = E[I … + I n 2 + I 1 I 2 + … + I n-1 I n ] = E[I 1 2 ] + … + E[I n 2 ] + E[I 1 I 2 ] + … + E[I n-1 I n ] E[I i 2 ] = E[I i ] = p E[I i I j ] = P(I i = 1 and I j = 1) = P(I i = 1) P(I j = 1) = p 2 = n p = n(n-1) p 2 = np + n(n-1) p 2 – (np) 2

Variance of binomial random variable Suppose X is Binomial(n, p). Var[X] = np + n(n-1) p 2 – (np) 2  = E[X] = np = np – p 2 = np(1-p) Var[X] = np(1-p)  = √ np(1-p). The standard deviation of X is

Investments N A = amount on choice A N B = amount on choice B 50 × Bernoulli(½) A: put $50 in one stockB: put $½ in each of 50 stocks Binomial(50, ½)    = 25  = √ 50 ½ ½ = 3.536…  –   + 

Apples About 10% of the apples on your farm are rotten. You sell 10 apples. How many are rotten? Probability model N umber of rotten apples you sold is Binomial(n = 10, p = 1/10). E[N] = np = 1

Apples You improve productivity; now only 5% apples rot. You can now sell 20 apples and only one will be rotten on average. N is now Binomial(20, 1/20).

Binomial(10, 1/10) Binomial(20, 1/20)

The Poisson random variable A Poisson(  ) random variable has this p.m.f.: p(k) = e -   k /k! k = 0, 1, 2, 3, … Poisson random variables do not occur “naturally” in the sample spaces we have seen. They approximate Binomial(n, p) random variables when  = np is fixed and n is large (so p is small) p Poisson(  ) (k) = lim n → ∞ p Binomial(n,  /n) (k)

Rain is falling on your head at an average speed of 2.8 drops/second. Divide the second evenly in n intervals of length 1/n. Raindrops Let E i be the event “raindrop hits during interval i.” Assuming E 1, …, E n are independent, the number of drops in the second N is a Binomial(n, p) r.v. Since E[N] = 2.8, and E[N] = np, p must equal 2.8/n. 0 1

Raindrops Number of drops N is Binomial(n, 2.8/n) 0 1 As n gets larger, the number of drops within the second “approaches” a Poisson(2.8) random variable:

Expectation and variance of Poisson If X is Binomial(n, p) then E[X] = npVar[X] = np(1-p) When p =  /n, we get E[X] =  Var[X] =  (1-  /n) As n → ∞, E[X] →  and Var[X] → . This suggests When X is Poisson(  ), E[X] =  and Var[X] = .

Problem for you to solve Rain falls on you at an average rate of 3 drops/sec. You walk for 30 sec from MTR to bus stop. When 100 drops hit you, your hair gets wet. What is the probability your hair got wet?

Problem for you to solve Solution On average, 90 drops fall in 30 seconds. So we model the number of drops N you receive as a Poisson(90) random variable. Using the online Poisson calculator at or the poissonpmf(n, L) function in 13L07.py we getonline P(N > 100) = 1 - ∑ i = 0 P(N = i) ≈ 13.49% 99