Download 1 Introduction 2 Discrete Random Variables

Lecture 3: Random Variables
Readings: Sections 1.3, 1.6, 2.1, 2.2
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Introduction
Random Variables
• Random variable (rv): a real-valued function whose domain is a sample space.
• Notation:
– Capital letters X, Y, Z for random variables
– Lower case letters x, y, z for specific values that a random variable might take
Example 1: Toss a fair coin twice. Let X be the number of heads
Discrete vs. Continuous Random Variable
• Discrete random variable: a random variable that can take on only finite or countable
number of possible values
• Continuous random variable: a random variable that can take on any value in an
interval
Example 2: Identify which of the following random variables are discrete and which are
continuous
– X = the number of tosses if you toss a coin until getting a head
– X = lifetime of a light bulb
– X = number of phone calls a receptionist receives in an hour
– X = number of hearts if you select 10 cards at random from a deck of 52 cards
– X = height of a Purdue female student
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Discrete Random Variables
Probability Mass Function (PMF) p(x)
• The probability mass function p(x) of a discrete random variable X is a function that
is defined for every possible value x, i.e., p(x) = P (X = x)
– 0 ≤ p(x) ≤ 1
P
p(x) = 1
–
x
1
– P (a ≤ X ≤ b) =
P
p(x)
a≤x≤b
Example 3: I have an unbalanced die with probabilities:
x
p(x)
1
0.01
2
0.02
3
0.1
4
0.2
5
0.3
6
0.37
Roll the die once and let X be the number that was recorded. Find the following probabilities:
– P (2 ≤ X < 5)
– P (X is even)
Example 4: Let X be a discrete random variable with the following PMF:
 1
 14 (x + 1)2 x = 0, 1, 2
p(x) =

0
otherwise
– P (X = 1) =
– P (X < 2) =
– P (X = 4) =
Expected Value (Mean)
• The expected value of a discrete random variable X with PMF p(x) is given by
X
E(X) =
xp(x),
x
where summation is over all possible values x
• Note: The mean is often denoted as µ, µX , E(X), E[X], E{X}
Example 5: X is a discrete random variable with the following PMF:
x
p(x)
0
0.1
Find the mean of X.
2
1
0.5
2
0.4
Variance and Standard Deviation
• The variance of a discrete random variable X with PMF p(x) is given by
X
V ar(X) =
(x − E(X))2 p(x),
x
where the summation is over all possible values x.
p
• The standard deviation SD(X) = V ar(X)
• Note:
2 , V ar(X), V (X)
– The variance is often denoted as σ 2 , σX
– The standard deviation is often denoted as σ, σX , SD(X)
Example 5 (cont’d): Find the variance and standard deviation of X.
Properties of Mean
Let c be a constant,
a. E(X + c) = E(X) + c
b. E(cX) = cE(X)
c. The expected values of a sum is the sum of the expected values
E(X + Y ) = E(X) + E(Y )
E(X1 + X2 + . . . + Xn ) = E(X1 ) + E(X2 ) + . . . + E(Xn )
d. E(c) = c
Example 6: Suppose that the mean of X is 10 and the mean of Y is 2. Find the mean of
W = 3X − 2Y + 2
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Properties of Variance and Standard Deviation
Let c be a constant,
a. V ar(X + c) = V ar(X) and SD(X + c) = SD(X)
b. V ar(cX) = c2 V ar(X) and SD(cX) = |c|SD(X)
c. The variance of a sum of independent random variables is the sum of the variances
∗ If X and Y are independent, then
V ar(X + Y ) =p
V ar(X) + V ar(Y )
SD(X + Y ) = V ar(X) + V ar(Y )
∗ If X1 , X2 , . . . , Xn are independent, then
V ar(X1 + X2 + . . . + Xn ) =p
V ar(X1 ) + V ar(X2 ) + . . . + V ar(Xn )
SD(X1 + X2 + . . . + Xn ) = V ar(X1 ) + V ar(X2 ) + . . . + V ar(Xn )
d. V ar(c) = 0 and SD(c) = 0
Example 7: Suppose that the variance of X is 5, the variance of Y is 10, and X and Y
are independent. Find the variance and standard deviation of W = 3X − 2Y + 2
Binomial Distribution
• Binomial experiment
– A fixed number n of identical trials
– The trials are independent
– Two possible outcomes for each trial, success and failure
– Probability of success p is constant for all trials
• Binomial random variable: If X is the number of successes in a binomial experiment,
then X is a binomial random variable with parameters n and p, i.e., X ∼ binomial(n, p).
• The PMF for a binomial rv: for X ∼ binomial(n, p),
 n

px (1 − p)n−x x = 0, 1, 2, . . . , n
x
p(x) =

0
otherwise
n
n
n!
Here
reads “n choose x” and
= x!(n−x)!
x
x
– Mean: E(X) = np
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– Variance: V ar(X) = np(1 − p)
Example 8: Roll a balanced die 10 times and let X be the number of times a “3” is rolled.
– What is the probability that exactly two “3”s are rolled?
– What is the probability that at most two “3”s are rolled?
Poisson Distribution
• Poisson Distribution are often used to model number of times some “event” occurs during
a specified time or in a particular region or space. For example,
– number of times a machine fails in the course of a workday
– number of bacteria colonies in a cubic centimeter of water
• Poisson random variable: If X is the number of events in a particular time or region
and λ > 0 is the average number of events in that time or region, then X is a Poisson
random variable with parameter λ.
• The PMF for a Poisson rv: for X ∼ Poisson(λ),
λx e−λ
x = 0, 1, 2, . . .
x!
p(x) =
0
otherwise
where 0! = 1.
– Mean: E(X) = λ
– Variance: V ar(X) = λ
Example 9: The receptionist in the main office of Statistics Department receives an average
of 5 calls in an hour.
– What is the probability that there will be exactly two calls in the next hour?
– What is the probability that there will be at most two calls in the next hour?
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– What is the probability that there will be exactly five calls in the next two hours?
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Continuous Random Variables
Probability Density Function (PDF) f (x)
The probability density function f (x) of a continuous random variable X is a function
such that for any two numbers a and b (a ≤ b),
Zb
P (a ≤ X ≤ b) =
f (x)dx
a
In other words, the probability that X falls in [a, b] is the area under the function f (x)
above this interval
a
b
Important properties of PDF
• f (x) ≥ 0; a PDF is a non-negative function
•
R∞
f (x)dx = 1; the area under the curve must be 1
−∞
• For any specific value c, P (X = c) = 0
• For any a < b, P (a ≤ X ≤ b) = P (a < X < b) =
Rb
a
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f (x)dx
Expected Value
• The expected value of a continuous random variable X with PDF f (x) is given by
Z∞
E(X) =
xf (x)dx,
−∞
Example 10: X is a continuous random variable with the following PDF:
3x2 0 < x < 1
0
otherwise
f (x) =
Find the mean of X.
Variance and Standard Deviation
• The variance of a continuous random variable X with PDF f (x) is given by
Z∞
V ar(X) =
(x − E(X))2 f (x)dx,
−∞
Example 10 (cont’d): Find the variance and standard deviation of X.
Special Continuous Distributions
• Exponential distribution with parameter λ (λ > 0):
λe−λx x > 0
f (x) =
0
otherwise
– E(X) =
1
λ
– V ar(X) =
1
λ2
• Normal distribution with parameters µ and σ:
f (x) = √
7
1
2πσ 2
e−
(x−µ)2
2σ 2
– E(X) = µ
– V ar(X) = σ 2
Exponential PDF
Normal PDF
Example 11: Suppose that a random variable X has a probability density function given
by
cx(1 − x) 0 ≤ x ≤ 1
f (x) =
0
otherwise
– Find the value of c
– Find P (0.4 ≤ X ≤ 1)
– Find P (X ≤ 0.4|X ≤ 0.8)
– Find P (X > 0.4|X ≤ 0.8)
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Percentile of a Distribution
The pth percentile ηp is the value such that p% of the observations fall at or below it,
i.e.,
P (X ≤ ηp ) = p%
The median is the value such that 50% of the observations fall at or below it. Median is
the 50th percentile.
Example 12: Suppose that a random variable X follows an exponential distribution with
parameter λ = 1.
– Find the median of X
– Find the 25th percentile
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