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4. Random variables part two. Review. A discrete random variable X assigns a discrete value to every outcome in the sample space. E [ N ]. Probability mass function of X : p ( x ) = P ( X = x ). Expected value of X : E [ X ] = ∑ x x p ( x ). N: number of heads in two coin flips. - PowerPoint PPT Presentation
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ENGG 2040C: Probability Models and Applications
Andrej Bogdanov
Spring 2013
4. Random variablespart 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]
N: number of headsin two coin flips
One die
Example from last time
F = face value of fair 6-sided dieE[F] = 1 + 2 + 3 + 4 + 5 + 6 = 3.5
1616 16 16 16 16
Two dice
S = sum of face values of two fair 6-sided diceSolution 1
2 3 4 5 6 7 89 10 11 12
spS(s) 1
36236
336
436
536
636
536
436
336
236
136
E[S] = 2 + 3 + … + 12
136236 136 =
7
We calculate the p.m.f. of S:
Two dice again
S = sum of face values of two fair 6-sided dice
F1 F2
S = F1 + F2
F1 = outcome of first dieF2 = 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(w) + Y(w) to outcome w.
X assigns value X(w) to outcome wY assigns value Y(w) to outcome w
Sum of random variables
F1 F2 S = F1 + F2
11 1 1 212 1 2 3
21 2 1 3
… …
66 6 6 12
… …
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 diceSolution 2
E[S] = E[F1] + E[F2]
= 3.5 + 3.5 = 7
F1 F2
S = F1 + F2
Balls
We draw 3 balls without replacement from this urn:
-1
1
11
-1
-1
0
What is the expected sum of values on the 3 balls?
0
-1
Balls
-1
1
11
-1
-1
00
-1S = B1 + B2
+ B3where Bi is the value of i-th ball.E[S] = E[B1] + E[B2] + E[B3]
p.m.f of B1:
-1 01
xp(x) 4
9 29
39
E[B1] = -1 (4/9) + 0 (2/9) + 1 (3/9) = -1/9same for B2, B3E[S] = 3 (-1/9) = -1/3.
Three dice
E[N] = E[I1] + E[I2] + E[I3]
Solution I1 I2 I3
N = I1 + I2 + I3
E[I1] = 1 (1/6) + 0(5/6) = 1/6E[I2], E[I3] = 1/6
= 3 (1/6)
= 1/2
Ik =1 if face value of kth 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 variablePerform a trial that succeeds with probability p and fails with probability 1 – p.
01p(x) 1 – p p
x
p = 0.5
p(x)
p = 0.4
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
ExamplesToss 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).
w C(w)
HTHHHHT 3THHHHHTHHHHHHH
20
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 variablesIf X is Binomial(n, p), its p.m.f. is
p(k) = P(X = k) = C(n, k) pk (1 - p)n-k
We can write X = I1 + … + In, where Ii is an indicator random variable for the success of the i-th trial E[X] = E[I1] + … + E[In]
= p + … + p = np.
E[X] = np
Probability mass functionBinomial(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
NA = amount on choice A
NB = amount on choice B 50 × Bernoulli(½)
A: put $50 in one stock B: put $½ in each of 50 stocks
Binomial(50, ½)
E[NA] E[NB]
Variance and standard deviation
Let m = E[X] be the expected value of X.
The variance of X is the quantityVar[X] = E[(X –
m)2]The standard deviation of X is s = √Var[X]
They measure how close X and m are typically.
Calculating variancem = E[NA]
y 252q(y) 1
p.m.f of (NA – m)2
Var[NA] = E[(NA – m)2]
x 050p(x) ½½ p.m.f of NA
= 252
s = std. dev. of NA = 25
m – s m + s
Another formula for variance
Var[X] = E[(X – m)2]
= E[X2 – 2mX + m2]= E[X2] + E[–2mX] + E[m2]= E[X2] – 2m E[X] + m2
= E[X2] – 2m m + m2
= E[X2] – m2
for constant c, E[cX] = cE[X]
for constant c, E[c] = c
Var[X] = E[X2] – E[X]2
Variance of binomial random variableSuppose X is Binomial(n, p).Then X = I1 + … + In, where Ii =
1, if trial i succeeds0, if trial i fails
Var[X] = E[X2] – m2 = E[X2] – (np)2
m = E[X] = np
E[X2]= E[(I1 + … + In)2]= E[I1
2 + … + In2 + I1I2 + … + In-1In]
= E[I12] + … + E[In
2] + E[I1I2] + … + E[In-
1In]
E[Ii2] = E[Ii] = p E[Ii Ij] = P(Ii = 1 and Ij = 1)
= P(Ii = 1) P(Ij = 1) = p2
= n p = n(n-1) p2
= np + n(n-1) p2 – (np)2
Variance of binomial random variable
Suppose X is Binomial(n, p).
Var[X] = np + n(n-1) p2 – (np)2
m = E[X] = np
= np – p2 = np(1-p)
Var[X] = np(1-p)
s = √np(1-p).
The standard deviation of X is
Investments
NA = amount on choice A
NB = amount on choice B 50 × Bernoulli(½)
A: put $50 in one stock B: put $½ in each of 50 stocks
Binomial(50, ½)m m
s = 25 s = √50 ½ ½ = 3.536…
m – s m + s
Apples
About 10% of the apples on your farm are rotten.You sell 10 apples. How many are rotten?
Probability modelNumber of rotten apples you sold isBinomial(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).349
.387
.194
.001
10-
10
Binomial(20, 1/20).354
.377
.189
.002
10-8 10-
26
.367
.367
.183
.003
10-7 10-
19
The Poisson random variable
A Poisson(m) random variable has this p.m.f.:
p(k) = e-m mk/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 m = np is fixed and n is large (so p is small)
pPoisson(m)(k) = limn → ∞ pBinomial(n, m/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 Ei be the event “raindrop hits during interval i.”Assuming E1, …, En 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] = np Var[X] = np(1-
p)When p = m/n, we get
E[X] = m Var[X] = m(1-m/n)
As n → ∞, E[X] → m and Var[X] → m. This suggests
When X is Poisson(m), E[X] = m and Var[X] = m.
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
SolutionOn 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 get
P(N > 100) = 1 - ∑i = 0 P(N = i) ≈ 13.49%99