ENGG 2040C: Probability Models and Applications
Spring 2013
4. Random variables
part two
Andrej Bogdanov
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
One die
Example from last time
F = face value of fair 6-sided die
E[F] = 1 1 6 + 2 1 6 + 3 1 6 + 4 1 6 + 5 1 6 + 6 1 6 = 3.5
Two dice
S = sum of face values of
two fair 6-sided dice
Solution 1
We calculate the p.m.f. of S:
s
pS(s)
2
1
36
3
2
36
4
3
36
5
4
36
6
5
36
7
6
36
8
5
36
9 10 11 12
4
36
3
36
2
36
1
36
E[S] = 2 1 36 + 3 2 36 + … + 12 1 36 = 7
Two dice again
S = sum of face values of
two fair 6-sided dice
F1
S = F1 + F2
F1 = outcome of first die
F2 = outcome of second die
F2
Sum of random variables
Let X, Y be two random variables.
X assigns value X(w) to outcome w
Y assigns value Y(w) to outcome w
X + Y is the random variable that assigns value
X(w) + Y(w) to outcome w.
Sum of random variables
11
12
1
1
1
2
2
3
2
1
3
6
6
12
…
F2
…
21
…
…
66
S = F1 + F2
F1
Linearity of expectation
For every two random variables X and Y
E[X + Y] = E[X] + E[Y]
Two dice again
S = sum of face values of
two fair 6-sided dice
Solution 2
S = F1 + F2
E[S] = E[F1] + E[F2] = 3.5 + 3.5 = 7
F1
F2
Balls
We draw 3 balls without replacement from this urn:
0
1
-1
0
1
1
-1
-1
-1
What is the expected sum of values on the 3 balls?
Balls
0
1
0
S = B1 + B2 + B3
1
where Bi is the value of i-th ball.
1
E[S] = E[B1] + E[B2] + E[B3]
p.m.f of B1:
-1
x
-1
0
1
p(x)
4
2
3
9
9
9
E[B1] = -1 (4/9) + 0 (2/9) + 1 (3/9) = -1/9
same for B2, B3
E[S] = 3 (-1/9) = -1/3.
-1
-1
-1
Three dice
N = number of
s. Find E[N].
Solution
Ik =
I1
1 if face value of kth die equals
0 if not
N = I1 + I2 + I3
E[N] = E[I1] + E[I2] + E[I3] = 3 (1/6) = 1/2
E[I1] = 1 (1/6) + 0(5/6) = 1/6
E[I2], E[I3] = 1/6
I2
I3
Problem for you to solve
Five balls are chosen without replacement from an
urn with 8 blue balls and 10 red balls.
(a) What is the expected number of blue balls
that are chosen?
(b) What if the balls are chosen with replacement?
The indicator (Bernoulli) random variable
x
p(x)
0
1–p
1
p
p(x)
Perform a trial that succeeds with probability p and
fails with probability 1 – p.
p = 0.5
E[X] = p
p(x)
If X is Bernoulli(p) then
p = 0.4
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).
Examples:
w
C(w)
HHHHHHH
THHHHHT
HTHHHHT
0
2
3
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) 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 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
A: put $50 in one stock
NA = amount on choice A
50 × Bernoulli(½)
E[NA]
B: put $½ in each of 50 stocks
NB = amount on choice B
Binomial(50, ½)
E[NB]
Variance and standard deviation
Let m = E[X] be the expected value of X.
The variance of X is the quantity
Var[X] = E[(X – m)2]
The standard deviation of X is s = √Var[X]
They measure how close X and m are typically.
Calculating variance
m = E[NA]
m–s
m+s
x
0
50
p(x)
½
½
p.m.f of NA
Var[NA] = E[(NA –
m)2]
=
s = std. dev. of NA = 25
252
y
252
q(y)
1
p.m.f of (NA – m)2
Another formula for variance
Var[X] = E[(X –
m)2]
for constant c, E[cX] = cE[X]
= E[X2 – 2mX + m2]
for constant c, E[c] = c
= E[X2] + E[–2mX] + E[m2]
= E[X2] – 2m E[X] + m2
= E[X2] – 2m m + m2
= E[X2] – m2
Var[X] = E[X2] – E[X]2
Variance of binomial random variable
Suppose X is Binomial(n, p).
1, if trial i succeeds
Then X = I1 + … + In, where Ii = 0, if trial i fails
m = E[X] = np
Var[X] = E[X2] – m2 = E[X2] – (np)2 = np + n(n-1) p2 – (np)2
E[X2] = E[(I1 + … + In)2]
= E[I12 + … + In2 + I1I2 + … + In-1In]
= E[I12] + … + E[In2] + E[I1I2] + … + E[In-1In]
=np
E[Ii2] = E[Ii] = p
= n(n-1) p2
E[Ii Ij] = P(Ii = 1 and Ij = 1)
= P(Ii = 1) P(Ij = 1) = p2
Variance of binomial random variable
Suppose X is Binomial(n, p).
m = E[X] = np
Var[X] = np + n(n-1) p2 – (np)2 = np – p2 = np(1-p)
Var[X] = np(1-p)
The standard deviation of X is
s = √np(1-p).
Investments
A: put $50 in one stock
NA = amount on choice A
50 × Bernoulli(½)
m
m–s
B: put $½ in each of 50 stocks
NB = amount on choice B
Binomial(50, ½)
m
m+s
s = 25
s = √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
Number 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).
.387
.349
Binomial(10, 1/10)
.194
.001
10-10
.377
.354
Binomial(20, 1/20)
.189
.002
10-8
10-26
.003
10-7
10-19
.367.367
.183
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)
Raindrops
Rain is falling on your head at an average speed of
2.8 drops/second.
0
1
Divide the second evenly in n intervals of length 1/n.
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.
Raindrops
0
1
Number of drops N is Binomial(n, 2.8/n)
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.
When 100 drops hit you,
your hair gets wet.
You walk for 30 sec from
MTR to bus stop.
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 get
P(N > 100) = 1 - ∑i99= 0 P(N = i) ≈ 13.49%