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Random variables
Christopher Croke
University of Pennsylvania
Math 115
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Sample Space = {(H, T , H, H, T ...), (H, H, T , ...)...}
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Sample Space = {(H, T , H, H, T ...), (H, H, T , ...)...}
X ( (H, H, T , T , T , H, H, H, T , H) ) = 6.
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Sample Space = {(H, T , H, H, T ...), (H, H, T , ...)...}
X ( (H, H, T , T , T , H, H, H, T , H) ) = 6.
Example: Roll two dice. Let X = sum of the pips.
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Sample Space = {(H, T , H, H, T ...), (H, H, T , ...)...}
X ( (H, H, T , T , T , H, H, H, T , H) ) = 6.
Example: Roll two dice. Let X = sum of the pips.
Sample space = {(1, 1), (1, 2), (1, 3), ..., (6, 6)}.
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Sample Space = {(H, T , H, H, T ...), (H, H, T , ...)...}
X ( (H, H, T , T , T , H, H, H, T , H) ) = 6.
Example: Roll two dice. Let X = sum of the pips.
Sample space = {(1, 1), (1, 2), (1, 3), ..., (6, 6)}.
X ( (3, 5) ) = 8.
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Sample Space = {(H, T , H, H, T ...), (H, H, T , ...)...}
X ( (H, H, T , T , T , H, H, H, T , H) ) = 6.
Example: Roll two dice. Let X = sum of the pips.
Sample space = {(1, 1), (1, 2), (1, 3), ..., (6, 6)}.
X ( (3, 5) ) = 8.
For an example of a continuous random variable:
Example: Measure the height of people and record the result.
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Sample Space = {(H, T , H, H, T ...), (H, H, T , ...)...}
X ( (H, H, T , T , T , H, H, H, T , H) ) = 6.
Example: Roll two dice. Let X = sum of the pips.
Sample space = {(1, 1), (1, 2), (1, 3), ..., (6, 6)}.
X ( (3, 5) ) = 8.
For an example of a continuous random variable:
Example: Measure the height of people and record the result.
X = height in feet.
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Sample Space = {(H, T , H, H, T ...), (H, H, T , ...)...}
X ( (H, H, T , T , T , H, H, H, T , H) ) = 6.
Example: Roll two dice. Let X = sum of the pips.
Sample space = {(1, 1), (1, 2), (1, 3), ..., (6, 6)}.
X ( (3, 5) ) = 8.
For an example of a continuous random variable:
Example: Measure the height of people and record the result.
X = height in feet.
Y = height in inches.
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Sample Space = {(H, T , H, H, T ...), (H, H, T , ...)...}
X ( (H, H, T , T , T , H, H, H, T , H) ) = 6.
Example: Roll two dice. Let X = sum of the pips.
Sample space = {(1, 1), (1, 2), (1, 3), ..., (6, 6)}.
X ( (3, 5) ) = 8.
For an example of a continuous random variable:
Example: Measure the height of people and record the result.
X = height in feet.
Y = height in inches.
These are different random variables.
Christopher Croke
Calculus 115
Random Variables
A random variable X is a function from the sample space to the
real numbers.
Example: Flip a coin 10 times. Let X be the number of heads.
Sample Space = {(H, T , H, H, T ...), (H, H, T , ...)...}
X ( (H, H, T , T , T , H, H, H, T , H) ) = 6.
Example: Roll two dice. Let X = sum of the pips.
Sample space = {(1, 1), (1, 2), (1, 3), ..., (6, 6)}.
X ( (3, 5) ) = 8.
For an example of a continuous random variable:
Example: Measure the height of people and record the result.
X = height in feet.
Y = height in inches.
These are different random variables. We can say things like
Y = 12X .
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3).
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
getting 3 heads.
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
(10)
getting 3 heads. The answer is 2310 .
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
(10)
getting 3 heads. The answer is 2310 .
Pr (X ≥ 2) is Pr (E ) where E is the event of at least two heads.
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
(10)
getting 3 heads. The answer is 2310 .
Pr (X ≥ 2) is Pr (E ) where E is the event of at least two heads.
Pr (X ≥ 2) = 1 − Pr (X = 0) − Pr (X = 1)
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
(10)
getting 3 heads. The answer is 2310 .
Pr (X ≥ 2) is Pr (E ) where E is the event of at least two heads.
Pr (X ≥ 2) = 1 − Pr (X = 0) − Pr (X = 1) = 1 − 2110 − 210
10
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
(10)
getting 3 heads. The answer is 2310 .
Pr (X ≥ 2) is Pr (E ) where E is the event of at least two heads.
Pr (X ≥ 2) = 1 − Pr (X = 0) − Pr (X = 1) = 1 − 2110 − 210
10 =
1013
1 − 211
=
.
10
1024
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
(10)
getting 3 heads. The answer is 2310 .
Pr (X ≥ 2) is Pr (E ) where E is the event of at least two heads.
Pr (X ≥ 2) = 1 − Pr (X = 0) − Pr (X = 1) = 1 − 2110 − 210
10 =
1013
1 − 211
=
.
10
1024
Let X be a discrete random variable, i.e. at most countably many
values say {x1 , x2 , ...}.
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
(10)
getting 3 heads. The answer is 2310 .
Pr (X ≥ 2) is Pr (E ) where E is the event of at least two heads.
Pr (X ≥ 2) = 1 − Pr (X = 0) − Pr (X = 1) = 1 − 2110 − 210
10 =
1013
1 − 211
=
.
10
1024
Let X be a discrete random variable, i.e. at most countably many
values say {x1 , x2 , ...}.
The probability distribution for X is the function f (x) given by:
f (x) = Pr (X = x).
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
(10)
getting 3 heads. The answer is 2310 .
Pr (X ≥ 2) is Pr (E ) where E is the event of at least two heads.
Pr (X ≥ 2) = 1 − Pr (X = 0) − Pr (X = 1) = 1 − 2110 − 210
10 =
1013
1 − 211
=
.
10
1024
Let X be a discrete random variable, i.e. at most countably many
values say {x1 , x2 , ...}.
The probability distribution for X is the function f (x) given by:
f (x) = Pr (X = x). So
f is defined on the range of X .
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
(10)
getting 3 heads. The answer is 2310 .
Pr (X ≥ 2) is Pr (E ) where E is the event of at least two heads.
Pr (X ≥ 2) = 1 − Pr (X = 0) − Pr (X = 1) = 1 − 2110 − 210
10 =
1013
1 − 211
=
.
10
1024
Let X be a discrete random variable, i.e. at most countably many
values say {x1 , x2 , ...}.
The probability distribution for X is the function f (x) given by:
f (x) = Pr (X = x). So
f is defined on the range of X .
0 ≤ f (xi ) ≤ 1.
Christopher Croke
Calculus 115
Going back to the X of the first example. We can talk about
Pr (X = 3). This is the same as Pr (E ) where E is the event of
(10)
getting 3 heads. The answer is 2310 .
Pr (X ≥ 2) is Pr (E ) where E is the event of at least two heads.
Pr (X ≥ 2) = 1 − Pr (X = 0) − Pr (X = 1) = 1 − 2110 − 210
10 =
1013
1 − 211
=
.
10
1024
Let X be a discrete random variable, i.e. at most countably many
values say {x1 , x2 , ...}.
The probability distribution for X is the function f (x) given by:
f (x) = Pr (X = x). So
f is defined on the range of X .
0 ≤ f (xi ) ≤ 1.
Σi f (xi ) = 1.
Christopher Croke
Calculus 115
Binomial trials
Problem: Let X be the number of heads that occur when a coin is
flipped 6 times.
a) Compute the probability distribution of X .
Christopher Croke
Calculus 115
Binomial trials
Problem: Let X be the number of heads that occur when a coin is
flipped 6 times.
a) Compute the probability distribution of X .
b) Compute the probability distribution of (X − 3)2 .
Christopher Croke
Calculus 115
Binomial trials
Problem: Let X be the number of heads that occur when a coin is
flipped 6 times.
a) Compute the probability distribution of X .
b) Compute the probability distribution of (X − 3)2 .
Note that two different random variables can have the same
probability distribution. For example in this problem consider
Y = # tails.
Christopher Croke
Calculus 115
Binomial trials
Problem: Let X be the number of heads that occur when a coin is
flipped 6 times.
a) Compute the probability distribution of X .
b) Compute the probability distribution of (X − 3)2 .
Note that two different random variables can have the same
probability distribution. For example in this problem consider
Y = # tails.
Binomial trials refer to trials that have two possible outcomes and
are repeated.
Christopher Croke
Calculus 115
Binomial trials
Problem: Let X be the number of heads that occur when a coin is
flipped 6 times.
a) Compute the probability distribution of X .
b) Compute the probability distribution of (X − 3)2 .
Note that two different random variables can have the same
probability distribution. For example in this problem consider
Y = # tails.
Binomial trials refer to trials that have two possible outcomes and
are repeated.
Head or Tails.
Christopher Croke
Calculus 115
Binomial trials
Problem: Let X be the number of heads that occur when a coin is
flipped 6 times.
a) Compute the probability distribution of X .
b) Compute the probability distribution of (X − 3)2 .
Note that two different random variables can have the same
probability distribution. For example in this problem consider
Y = # tails.
Binomial trials refer to trials that have two possible outcomes and
are repeated.
Head or Tails.
Red or Black card.
Christopher Croke
Calculus 115
Binomial trials
Problem: Let X be the number of heads that occur when a coin is
flipped 6 times.
a) Compute the probability distribution of X .
b) Compute the probability distribution of (X − 3)2 .
Note that two different random variables can have the same
probability distribution. For example in this problem consider
Y = # tails.
Binomial trials refer to trials that have two possible outcomes and
are repeated.
Head or Tails.
Red or Black card.
Positive or negative in TB test.
Christopher Croke
Calculus 115
Binomial trials
Problem: Let X be the number of heads that occur when a coin is
flipped 6 times.
a) Compute the probability distribution of X .
b) Compute the probability distribution of (X − 3)2 .
Note that two different random variables can have the same
probability distribution. For example in this problem consider
Y = # tails.
Binomial trials refer to trials that have two possible outcomes and
are repeated.
Head or Tails.
Red or Black card.
Positive or negative in TB test.
Good or Bad refrigerator.
Christopher Croke
Calculus 115
Binomial trials
Usually use “success” or “failure” (you can pick either label for
either case).
p = Pr (success).
Christopher Croke
Calculus 115
Binomial trials
Usually use “success” or “failure” (you can pick either label for
either case).
p = Pr (success).
q = Pr (failure) = 1 − p.
Christopher Croke
Calculus 115
Binomial trials
Usually use “success” or “failure” (you can pick either label for
either case).
p = Pr (success).
q = Pr (failure) = 1 − p.
The experiment is to repeat the trial n times and our random
variable X is the number of successes.
Christopher Croke
Calculus 115
Binomial trials
Usually use “success” or “failure” (you can pick either label for
either case).
p = Pr (success).
q = Pr (failure) = 1 − p.
The experiment is to repeat the trial n times and our random
variable X is the number of successes. We have solved this before:
n k n−k
Pr (X = k) =
p q
k
for k = 0, 1, ..., n.
Christopher Croke
Calculus 115
Binomial trials
Usually use “success” or “failure” (you can pick either label for
either case).
p = Pr (success).
q = Pr (failure) = 1 − p.
The experiment is to repeat the trial n times and our random
variable X is the number of successes. We have solved this before:
n k n−k
Pr (X = k) =
p q
k
for k = 0, 1, ..., n.
In this case we say X is a binomial random variable with
parameters n and p.
Christopher Croke
Calculus 115
Binomial trials
Usually use “success” or “failure” (you can pick either label for
either case).
p = Pr (success).
q = Pr (failure) = 1 − p.
The experiment is to repeat the trial n times and our random
variable X is the number of successes. We have solved this before:
n k n−k
Pr (X = k) =
p q
k
for k = 0, 1, ..., n.
In this case we say X is a binomial random variable with
parameters n and p. the terminology comes from the fact that;
n k n−k
n
n
(p + q) = Σi=0
p q
.
k
Christopher Croke
Calculus 115
Binomial trials problems
Problem: A baseball player with a .300 batting average comes to
bat four times in a game. Find:
a) The probability that he gets 2 hits.
Christopher Croke
Calculus 115
Binomial trials problems
Problem: A baseball player with a .300 batting average comes to
bat four times in a game. Find:
a) The probability that he gets 2 hits.
b) The probability that he gets at least 3 hits.
Christopher Croke
Calculus 115
Binomial trials problems
Problem: A baseball player with a .300 batting average comes to
bat four times in a game. Find:
a) The probability that he gets 2 hits.
b) The probability that he gets at least 3 hits.
Problem: Storage.com makes writeable DVDs. One percent of the
disks are defective.
a) What is the probability that a box of 100 disks has exactly 2
defective disks?
Christopher Croke
Calculus 115
Binomial trials problems
Problem: A baseball player with a .300 batting average comes to
bat four times in a game. Find:
a) The probability that he gets 2 hits.
b) The probability that he gets at least 3 hits.
Problem: Storage.com makes writeable DVDs. One percent of the
disks are defective.
a) What is the probability that a box of 100 disks has exactly 2
defective disks?
b) What is the probability that there are at least 2 defective disks
in a box of 100?
Christopher Croke
Calculus 115
Binomial trials problems
Problem: A baseball player with a .300 batting average comes to
bat four times in a game. Find:
a) The probability that he gets 2 hits.
b) The probability that he gets at least 3 hits.
Problem: Storage.com makes writeable DVDs. One percent of the
disks are defective.
a) What is the probability that a box of 100 disks has exactly 2
defective disks?
b) What is the probability that there are at least 2 defective disks
in a box of 100?
Problem:: Approximately 20% of a mall’s customers are over 65
years old (we assume that they spread out evenly over the day). A
new store wants to encourage seniors to become customers. It
decides to give a gift certificate to the first 4 people over 65 to
enter the mall. What is the probability that the 4th person over 65
is the 10th customer?
Christopher Croke
Calculus 115
Continuous random variables
X is a continuous random variable if its values range over real
numbers.
Christopher Croke
Calculus 115
Continuous random variables
X is a continuous random variable if its values range over real
numbers.
Example: Pick a cell at random from a culture and let X be its
age (a real number).
Christopher Croke
Calculus 115
Continuous random variables
X is a continuous random variable if its values range over real
numbers.
Example: Pick a cell at random from a culture and let X be its
age (a real number).
Given a cell culture where cells divide in two every T days. Then
the proportion of cells of age between a and b where
0 < a < b < T remains constant over time (after things have been
going for a while).
Christopher Croke
Calculus 115
Continuous random variables
X is a continuous random variable if its values range over real
numbers.
Example: Pick a cell at random from a culture and let X be its
age (a real number).
Given a cell culture where cells divide in two every T days. Then
the proportion of cells of age between a and b where
0 < a < b < T remains constant over time (after things have been
going for a while). It turns out (under ideal circumstances) that
the proportion is just the area under f (x) = 2ke −kx from a to b
where k = ln(2)
T .
Christopher Croke
Calculus 115
Continuous random variables
X is a continuous random variable if its values range over real
numbers.
Example: Pick a cell at random from a culture and let X be its
age (a real number).
Given a cell culture where cells divide in two every T days. Then
the proportion of cells of age between a and b where
0 < a < b < T remains constant over time (after things have been
going for a while). It turns out (under ideal circumstances) that
the proportion is just the area under f (x) = 2ke −kx from a to b
where k = ln(2)
T .
This means
Z b
Pr (a ≤ X ≤ b) =
f (x)dx
a
Christopher Croke
Calculus 115
Continuous random variables
X is a continuous random variable if its values range over real
numbers.
Example: Pick a cell at random from a culture and let X be its
age (a real number).
Given a cell culture where cells divide in two every T days. Then
the proportion of cells of age between a and b where
0 < a < b < T remains constant over time (after things have been
going for a while). It turns out (under ideal circumstances) that
the proportion is just the area under f (x) = 2ke −kx from a to b
where k = ln(2)
T .
This means
Z b
Z b
2ke −kx = −2e −kb + 2e −ka .
Pr (a ≤ X ≤ b) =
f (x)dx =
a
a
Christopher Croke
Calculus 115
Continuous random variables
In general if X is a continuous random variable where there is a
function on the range of X so that
Z
Pr (a ≤ X ≤ b) =
b
f (x)dx
a
then f is called a probability density function (sometimes
denoted p.d.f.).
Christopher Croke
Calculus 115
Continuous random variables
In general if X is a continuous random variable where there is a
function on the range of X so that
Z
Pr (a ≤ X ≤ b) =
b
f (x)dx
a
then f is called a probability density function (sometimes
denoted p.d.f.).
Christopher Croke
Calculus 115
Probability density functions
If a random variable takes values from A to B (B could be ∞ and
A could be −∞) then
f (x) ≥ 0 for all A ≤ x ≤ B.
Christopher Croke
Calculus 115
Probability density functions
If a random variable takes values from A to B (B could be ∞ and
A could be −∞) then
f (x) ≥ 0 for all A ≤ x ≤ B.
RB
A f (x)dx = 1.
Christopher Croke
Calculus 115
Probability density functions
If a random variable takes values from A to B (B could be ∞ and
A could be −∞) then
f (x) ≥ 0 for all A ≤ x ≤ B.
RB
A f (x)dx = 1.
The integral could be improper.
Christopher Croke
Calculus 115
Probability density functions
If a random variable takes values from A to B (B could be ∞ and
A could be −∞) then
f (x) ≥ 0 for all A ≤ x ≤ B.
RB
A f (x)dx = 1.
The integral could be improper. .
In fact any function satisfying the above two conditions is a
probability density function
R bfor a continuous random variable X .
Define Pr (a ≤ X ≤ b) = a f (x)dx.
Christopher Croke
Calculus 115
Probability density functions
If a random variable takes values from A to B (B could be ∞ and
A could be −∞) then
f (x) ≥ 0 for all A ≤ x ≤ B.
RB
A f (x)dx = 1.
The integral could be improper. .
In fact any function satisfying the above two conditions is a
probability density function
R bfor a continuous random variable X .
Define Pr (a ≤ X ≤ b) = a f (x)dx. Note Pr (X = x0 ) = 0 for all
x0 .
Christopher Croke
Calculus 115
Probability density functions
If a random variable takes values from A to B (B could be ∞ and
A could be −∞) then
f (x) ≥ 0 for all A ≤ x ≤ B.
RB
A f (x)dx = 1.
The integral could be improper. .
In fact any function satisfying the above two conditions is a
probability density function
R bfor a continuous random variable X .
Define Pr (a ≤ X ≤ b) = a f (x)dx. Note Pr (X = x0 ) = 0 for all
x0 . On the other hand we can ask: “what the probability that x is
near x0 ?”.
Christopher Croke
Calculus 115
Probability density functions
If a random variable takes values from A to B (B could be ∞ and
A could be −∞) then
f (x) ≥ 0 for all A ≤ x ≤ B.
RB
A f (x)dx = 1.
The integral could be improper. .
In fact any function satisfying the above two conditions is a
probability density function
R bfor a continuous random variable X .
Define Pr (a ≤ X ≤ b) = a f (x)dx. Note Pr (X = x0 ) = 0 for all
x0 . On the other hand we can ask: “what the probability that x is
near x0 ?”. How near?
Christopher Croke
Calculus 115
Probability density functions
If a random variable takes values from A to B (B could be ∞ and
A could be −∞) then
f (x) ≥ 0 for all A ≤ x ≤ B.
RB
A f (x)dx = 1.
The integral could be improper. .
In fact any function satisfying the above two conditions is a
probability density function
R bfor a continuous random variable X .
Define Pr (a ≤ X ≤ b) = a f (x)dx. Note Pr (X = x0 ) = 0 for all
x0 . On the other hand we can ask: “what the probability that x is
near x0 ?”. How near? ∆x near.
Christopher Croke
Calculus 115
Probability density functions
1
1
Pr (x0 − ∆x ≤ X ≤ x0 + ∆x) ≈ f (x0 ) · ∆x.
2
2
Christopher Croke
Calculus 115
Probability density functions
1
1
Pr (x0 − ∆x ≤ X ≤ x0 + ∆x) ≈ f (x0 ) · ∆x.
2
2
Problem: Show that our first example: f (x) = 2ke −kx for
0 ≤ x ≤ T (where k = ln(2)
T ) is a probability density.
Christopher Croke
Calculus 115
Probability density functions
1
1
Pr (x0 − ∆x ≤ X ≤ x0 + ∆x) ≈ f (x0 ) · ∆x.
2
2
Problem: Show that our first example: f (x) = 2ke −kx for
0 ≤ x ≤ T (where k = ln(2)
T ) is a probability density.
Problem: Let f (x) = k − x 2 for 0 ≤ x ≤ 1.
a) What values of k make f (x) a probability density function?
Christopher Croke
Calculus 115
Probability density functions
1
1
Pr (x0 − ∆x ≤ X ≤ x0 + ∆x) ≈ f (x0 ) · ∆x.
2
2
Problem: Show that our first example: f (x) = 2ke −kx for
0 ≤ x ≤ T (where k = ln(2)
T ) is a probability density.
Problem: Let f (x) = k − x 2 for 0 ≤ x ≤ 1.
a) What values of k make f (x) a probability density function?
b) Find Pr (0 ≤ X ≤ 21 ) for the associated X .
Christopher Croke
Calculus 115
Probability density functions
1
1
Pr (x0 − ∆x ≤ X ≤ x0 + ∆x) ≈ f (x0 ) · ∆x.
2
2
Problem: Show that our first example: f (x) = 2ke −kx for
0 ≤ x ≤ T (where k = ln(2)
T ) is a probability density.
Problem: Let f (x) = k − x 2 for 0 ≤ x ≤ 1.
a) What values of k make f (x) a probability density function?
b) Find Pr (0 ≤ X ≤ 21 ) for the associated X .
Comments
the p.d.f. is not unique!
Christopher Croke
Calculus 115
Probability density functions
1
1
Pr (x0 − ∆x ≤ X ≤ x0 + ∆x) ≈ f (x0 ) · ∆x.
2
2
Problem: Show that our first example: f (x) = 2ke −kx for
0 ≤ x ≤ T (where k = ln(2)
T ) is a probability density.
Problem: Let f (x) = k − x 2 for 0 ≤ x ≤ 1.
a) What values of k make f (x) a probability density function?
b) Find Pr (0 ≤ X ≤ 21 ) for the associated X .
Comments
the p.d.f. is not unique! (e.g. Could change the value of f on
a finite number of points and it doesn’t change integrals.)
Christopher Croke
Calculus 115
Probability density functions
1
1
Pr (x0 − ∆x ≤ X ≤ x0 + ∆x) ≈ f (x0 ) · ∆x.
2
2
Problem: Show that our first example: f (x) = 2ke −kx for
0 ≤ x ≤ T (where k = ln(2)
T ) is a probability density.
Problem: Let f (x) = k − x 2 for 0 ≤ x ≤ 1.
a) What values of k make f (x) a probability density function?
b) Find Pr (0 ≤ X ≤ 21 ) for the associated X .
Comments
the p.d.f. is not unique! (e.g. Could change the value of f on
a finite number of points and it doesn’t change integrals.)
There is usually a continuous one that works and we use that
one.
Christopher Croke
Calculus 115
Probability density functions
1
1
Pr (x0 − ∆x ≤ X ≤ x0 + ∆x) ≈ f (x0 ) · ∆x.
2
2
Problem: Show that our first example: f (x) = 2ke −kx for
0 ≤ x ≤ T (where k = ln(2)
T ) is a probability density.
Problem: Let f (x) = k − x 2 for 0 ≤ x ≤ 1.
a) What values of k make f (x) a probability density function?
b) Find Pr (0 ≤ X ≤ 21 ) for the associated X .
Comments
the p.d.f. is not unique! (e.g. Could change the value of f on
a finite number of points and it doesn’t change integrals.)
There is usually a continuous one that works and we use that
one.
f (x) could be greater than 1 (in fact could be unbounded).
Christopher Croke
Calculus 115
Probability density functions
1
1
Pr (x0 − ∆x ≤ X ≤ x0 + ∆x) ≈ f (x0 ) · ∆x.
2
2
Problem: Show that our first example: f (x) = 2ke −kx for
0 ≤ x ≤ T (where k = ln(2)
T ) is a probability density.
Problem: Let f (x) = k − x 2 for 0 ≤ x ≤ 1.
a) What values of k make f (x) a probability density function?
b) Find Pr (0 ≤ X ≤ 21 ) for the associated X .
Comments
the p.d.f. is not unique! (e.g. Could change the value of f on
a finite number of points and it doesn’t change integrals.)
There is usually a continuous one that works and we use that
one.
f (x) could be greater than 1 (in fact could be unbounded).
Only need the integrals to be less than 1.
Christopher Croke
Calculus 115
Cumulative Distribution Functions
If X is a continuous random variable with density function f (x)
defined on [A, B] then the cumulative distribution function (also
known as c.d.f.) is the function
Z x
F (x) =
f (t)dt
A
Christopher Croke
Calculus 115
Cumulative Distribution Functions
If X is a continuous random variable with density function f (x)
defined on [A, B] then the cumulative distribution function (also
known as c.d.f.) is the function
Z x
F (x) =
f (t)dt = Pr (A ≤ X ≤ x) = Pr (X ≤ x).
A
Christopher Croke
Calculus 115
Cumulative Distribution Functions
If X is a continuous random variable with density function f (x)
defined on [A, B] then the cumulative distribution function (also
known as c.d.f.) is the function
Z x
F (x) =
f (t)dt = Pr (A ≤ X ≤ x) = Pr (X ≤ x).
A
Note that the Fundamental Theorem of Calculus says that
F 0 (x) = f (x).
Christopher Croke
Calculus 115
Cumulative Distribution Functions
If X is a continuous random variable with density function f (x)
defined on [A, B] then the cumulative distribution function (also
known as c.d.f.) is the function
Z x
F (x) =
f (t)dt = Pr (A ≤ X ≤ x) = Pr (X ≤ x).
A
Note that the Fundamental Theorem of Calculus says that
F 0 (x) = f (x). Thus
Z b
f (x)dx = F (b) − F (a).
Pr (a ≤ X ≤ b) =
a
Christopher Croke
Calculus 115
Cumulative Distribution Functions
If X is a continuous random variable with density function f (x)
defined on [A, B] then the cumulative distribution function (also
known as c.d.f.) is the function
Z x
F (x) =
f (t)dt = Pr (A ≤ X ≤ x) = Pr (X ≤ x).
A
Note that the Fundamental Theorem of Calculus says that
F 0 (x) = f (x). Thus
Z b
f (x)dx = F (b) − F (a).
Pr (a ≤ X ≤ b) =
a
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Note that F is exactly the antiderivative of f such that F (A) = 0.
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Note that F is exactly the antiderivative of f such that F (A) = 0.
Problem: Find the cumulative distribution function for our first
example (f (x) = 2ke −kx ).
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Note that F is exactly the antiderivative of f such that F (A) = 0.
Problem: Find the cumulative distribution function for our first
example (f (x) = 2ke −kx ).
Problem: Let f (x) = x12 on x > 1.
a) Show that f is a probability density.
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Note that F is exactly the antiderivative of f such that F (A) = 0.
Problem: Find the cumulative distribution function for our first
example (f (x) = 2ke −kx ).
Problem: Let f (x) = x12 on x > 1.
a) Show that f is a probability density.
b) Find the c.d.f for f .
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Note that F is exactly the antiderivative of f such that F (A) = 0.
Problem: Find the cumulative distribution function for our first
example (f (x) = 2ke −kx ).
Problem: Let f (x) = x12 on x > 1.
a) Show that f is a probability density.
b) Find the c.d.f for f .
c) Find Pr (3 ≤ X ≤ 4).
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Note that F is exactly the antiderivative of f such that F (A) = 0.
Problem: Find the cumulative distribution function for our first
example (f (x) = 2ke −kx ).
Problem: Let f (x) = x12 on x > 1.
a) Show that f is a probability density.
b) Find the c.d.f for f .
c) Find Pr (3 ≤ X ≤ 4).
Problem: A point is chosen at random from the unit square
[0, 1] × [0, 1]. Let X be the sum of its two coordinates. Find the
probability density function f (x) and the cumulative distribution
function F (x).
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Note that F is exactly the antiderivative of f such that F (A) = 0.
Problem: Find the cumulative distribution function for our first
example (f (x) = 2ke −kx ).
Problem: Let f (x) = x12 on x > 1.
a) Show that f is a probability density.
b) Find the c.d.f for f .
c) Find Pr (3 ≤ X ≤ 4).
Problem: A point is chosen at random from the unit square
[0, 1] × [0, 1]. Let X be the sum of its two coordinates. Find the
probability density function f (x) and the cumulative distribution
function F (x).
What is Pr (0.5 < X < 1)?
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Comments about a cumulative distribution function F :
0 ≤ F (x) ≤ 1.
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Comments about a cumulative distribution function F :
0 ≤ F (x) ≤ 1.
limx→−∞ F (x) = 0
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Comments about a cumulative distribution function F :
0 ≤ F (x) ≤ 1.
limx→−∞ F (x) = 0
limx→∞ F (x) = 1
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Comments about a cumulative distribution function F :
0 ≤ F (x) ≤ 1.
limx→−∞ F (x) = 0
limx→∞ F (x) = 1
F (x) is nondecreasing.
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Comments about a cumulative distribution function F :
0 ≤ F (x) ≤ 1.
limx→−∞ F (x) = 0
limx→∞ F (x) = 1
F (x) is nondecreasing.
For a discrete probability distribution f (x) then the c.d.f
F (x) = Pr (X ≤ x) = Σxi ≤x f (xi ).
Christopher Croke
Calculus 115
Cumulative Distribution Functions
Comments about a cumulative distribution function F :
0 ≤ F (x) ≤ 1.
limx→−∞ F (x) = 0
limx→∞ F (x) = 1
F (x) is nondecreasing.
For a discrete probability distribution f (x) then the c.d.f
F (x) = Pr (X ≤ x) = Σxi ≤x f (xi ).
This is a “step function”.
Christopher Croke
Calculus 115
Bivariate Distributions
This is the joint probability when you are given two random
variables X and Y .
Christopher Croke
Calculus 115
Bivariate Distributions
This is the joint probability when you are given two random
variables X and Y .
Consider the case when both are discrete random variables. Then
the joint probability function f (x, y ) is the function:
f (xi , yj ) = Pr (X = xi , Y = yj ).
Christopher Croke
Calculus 115
Bivariate Distributions
This is the joint probability when you are given two random
variables X and Y .
Consider the case when both are discrete random variables. Then
the joint probability function f (x, y ) is the function:
f (xi , yj ) = Pr (X = xi , Y = yj ).
So of course f (xi , yj ) ≥ 0 and the sum over all pairs (xi , yj ) of
f (xi , yj ) is 1.
Christopher Croke
Calculus 115
Bivariate Distributions
This is the joint probability when you are given two random
variables X and Y .
Consider the case when both are discrete random variables. Then
the joint probability function f (x, y ) is the function:
f (xi , yj ) = Pr (X = xi , Y = yj ).
So of course f (xi , yj ) ≥ 0 and the sum over all pairs (xi , yj ) of
f (xi , yj ) is 1. Often it is given in the form of a table.
X ↓Y →
0
2
4
1
0.1
0.1
0.1
2
0
0.4
0
3
0.2
0
0.1
Christopher Croke
Calculus 115
Bivariate Distributions
This is the joint probability when you are given two random
variables X and Y .
Consider the case when both are discrete random variables. Then
the joint probability function f (x, y ) is the function:
f (xi , yj ) = Pr (X = xi , Y = yj ).
So of course f (xi , yj ) ≥ 0 and the sum over all pairs (xi , yj ) of
f (xi , yj ) is 1. Often it is given in the form of a table.
X ↓Y →
0
2
4
1
0.1
0.1
0.1
2
0
0.4
0
3
0.2
What is Pr(X=0,Y=3)?
0
0.1
Christopher Croke
Calculus 115
Bivariate Distributions
This is the joint probability when you are given two random
variables X and Y .
Consider the case when both are discrete random variables. Then
the joint probability function f (x, y ) is the function:
f (xi , yj ) = Pr (X = xi , Y = yj ).
So of course f (xi , yj ) ≥ 0 and the sum over all pairs (xi , yj ) of
f (xi , yj ) is 1. Often it is given in the form of a table.
X ↓Y →
0
2
4
Pr (X ≥ 3, Y
1
2
0.1
0
0.1 0.4
0.1
0
≥ 2)?
3
0.2
What is Pr(X=0,Y=3)?
0
0.1
Christopher Croke
Calculus 115
Bivariate Distributions
This is the joint probability when you are given two random
variables X and Y .
Consider the case when both are discrete random variables. Then
the joint probability function f (x, y ) is the function:
f (xi , yj ) = Pr (X = xi , Y = yj ).
So of course f (xi , yj ) ≥ 0 and the sum over all pairs (xi , yj ) of
f (xi , yj ) is 1. Often it is given in the form of a table.
X ↓Y →
0
2
4
Pr (X ≥ 3, Y
1
2
3
0.1
0 0.2
What is Pr(X=0,Y=3)?
0.1 0.4
0
0.1
0 0.1
≥ 2)? Pr (X = 2)?
Christopher Croke
Calculus 115
Bivariate Distributions
If X and Y are continuous random variables then the joint
probability density function is a function f (x, y ) of two real
variables such that for any domain A in the plane:
Z Z
Pr ((X , Y ) ∈ A) =
f (x, y )dxdy .
A
Christopher Croke
Calculus 115
Bivariate Distributions
If X and Y are continuous random variables then the joint
probability density function is a function f (x, y ) of two real
variables such that for any domain A in the plane:
Z Z
Pr ((X , Y ) ∈ A) =
f (x, y )dxdy .
A
We want f (x, y ) ≥ 0 and
R∞ R∞
−∞ −∞ f (x, y )dxdy
Christopher Croke
Calculus 115
= 1.
Bivariate Distributions
If X and Y are continuous random variables then the joint
probability density function is a function f (x, y ) of two real
variables such that for any domain A in the plane:
Z Z
Pr ((X , Y ) ∈ A) =
f (x, y )dxdy .
A
R∞ R∞
We want f (x, y ) ≥ 0 and −∞ −∞ f (x, y )dxdy = 1.
Problem: Let f be a joint probability density function (j.p.d.f.) for
X and Y where
c(x + y ) if x ≥ 0, y ≥ 0, y ≤ 1 − x
f (x, y ) =
0
othewise
a) What is c?
Christopher Croke
Calculus 115
Bivariate Distributions
If X and Y are continuous random variables then the joint
probability density function is a function f (x, y ) of two real
variables such that for any domain A in the plane:
Z Z
Pr ((X , Y ) ∈ A) =
f (x, y )dxdy .
A
R∞ R∞
We want f (x, y ) ≥ 0 and −∞ −∞ f (x, y )dxdy = 1.
Problem: Let f be a joint probability density function (j.p.d.f.) for
X and Y where
c(x + y ) if x ≥ 0, y ≥ 0, y ≤ 1 − x
f (x, y ) =
0
othewise
a) What is c?
b) Find Pr (X ≤ 21 ).
Christopher Croke
Calculus 115
Bivariate Distributions
If X and Y are continuous random variables then the joint
probability density function is a function f (x, y ) of two real
variables such that for any domain A in the plane:
Z Z
Pr ((X , Y ) ∈ A) =
f (x, y )dxdy .
A
R∞ R∞
We want f (x, y ) ≥ 0 and −∞ −∞ f (x, y )dxdy = 1.
Problem: Let f be a joint probability density function (j.p.d.f.) for
X and Y where
c(x + y ) if x ≥ 0, y ≥ 0, y ≤ 1 − x
f (x, y ) =
0
othewise
a) What is c?
b) Find Pr (X ≤ 21 ).
c) Set up integral for Pr (Y ≤ X ).
Christopher Croke
Calculus 115
Bivariate Distributions
In some cases X might be discrete and Y continuous (or vice
versa).
Christopher Croke
Calculus 115
Bivariate Distributions
In some cases X might be discrete and Y continuous (or vice
versa).
R∞
In this case we would want −∞ Σxi f (xi , y )dy = 1.
Christopher Croke
Calculus 115
Bivariate Distributions
In some cases X might be discrete and Y continuous (or vice
versa).
R∞
In this case we would want −∞ Σxi f (xi , y )dy = 1.
The (cumulative) joint distribution function for continuous
random variables X and Y is
Z x Z y
F (x, y ) = Pr (X ≤ x, Y ≤ y ) =
f (s, t)dtds.
−∞
Christopher Croke
Calculus 115
−∞
Bivariate Distributions
In some cases X might be discrete and Y continuous (or vice
versa).
R∞
In this case we would want −∞ Σxi f (xi , y )dy = 1.
The (cumulative) joint distribution function for continuous
random variables X and Y is
Z x Z y
F (x, y ) = Pr (X ≤ x, Y ≤ y ) =
f (s, t)dtds.
−∞
(Use sums if discrete.)
Christopher Croke
Calculus 115
−∞
Bivariate Distributions
In some cases X might be discrete and Y continuous (or vice
versa).
R∞
In this case we would want −∞ Σxi f (xi , y )dy = 1.
The (cumulative) joint distribution function for continuous
random variables X and Y is
Z x Z y
F (x, y ) = Pr (X ≤ x, Y ≤ y ) =
f (s, t)dtds.
−∞
(Use sums if discrete.)
Thus we see that if F is differentiable
f (x, y ) =
Christopher Croke
∂2F
∂x∂y
Calculus 115
−∞
Bivariate Distributions
What is the probability in a rectangle? I.e.
Pr (a ≤ X ≤ b, c ≤ Y ≤ d)?
Christopher Croke
Calculus 115
Bivariate Distributions
What is the probability in a rectangle? I.e.
Pr (a ≤ X ≤ b, c ≤ Y ≤ d)?
F (b, d) − F (a, d) − F (b, c) + F (a, c).
Christopher Croke
Calculus 115
Bivariate Distributions
What is the probability in a rectangle? I.e.
Pr (a ≤ X ≤ b, c ≤ Y ≤ d)?
F (b, d) − F (a, d) − F (b, c) + F (a, c).
Given F (x, y ) a c.j.d.f. for X and Y we can find the two c.d.f.’s by:
F1 (x) = Pr (X ≤ x)
Christopher Croke
Calculus 115
Bivariate Distributions
What is the probability in a rectangle? I.e.
Pr (a ≤ X ≤ b, c ≤ Y ≤ d)?
F (b, d) − F (a, d) − F (b, c) + F (a, c).
Given F (x, y ) a c.j.d.f. for X and Y we can find the two c.d.f.’s by:
F1 (x) = Pr (X ≤ x) = lim Pr (X ≤ x, Y ≤ y )
y →∞
Christopher Croke
Calculus 115
Bivariate Distributions
What is the probability in a rectangle? I.e.
Pr (a ≤ X ≤ b, c ≤ Y ≤ d)?
F (b, d) − F (a, d) − F (b, c) + F (a, c).
Given F (x, y ) a c.j.d.f. for X and Y we can find the two c.d.f.’s by:
F1 (x) = Pr (X ≤ x) = lim Pr (X ≤ x, Y ≤ y ) = lim F (x, y ).
y →∞
Christopher Croke
y →∞
Calculus 115
Bivariate Distributions
What is the probability in a rectangle? I.e.
Pr (a ≤ X ≤ b, c ≤ Y ≤ d)?
F (b, d) − F (a, d) − F (b, c) + F (a, c).
Given F (x, y ) a c.j.d.f. for X and Y we can find the two c.d.f.’s by:
F1 (x) = Pr (X ≤ x) = lim Pr (X ≤ x, Y ≤ y ) = lim F (x, y ).
y →∞
y →∞
Similarly
F2 (y ) = lim F (x, y ).
x→∞
Christopher Croke
Calculus 115