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Normal Random Variables and Probability
An Undergraduate Introduction to Financial Mathematics
J. Robert Buchanan
2010
J. Robert Buchanan
Normal Random Variables and Probability
Discrete vs. Continuous Random Variables
Think about the probability of selecting X from the interval [0, 1]
when
X ∈ {0, 1}
J. Robert Buchanan
Normal Random Variables and Probability
Discrete vs. Continuous Random Variables
Think about the probability of selecting X from the interval [0, 1]
when
X ∈ {0, 1}
X ∈ {k/10 : k = 0, 1, . . . , 10}
J. Robert Buchanan
Normal Random Variables and Probability
Discrete vs. Continuous Random Variables
Think about the probability of selecting X from the interval [0, 1]
when
X ∈ {0, 1}
X ∈ {k/10 : k = 0, 1, . . . , 10}
X ∈ {k/n : k = 0, 1, . . . , n}
J. Robert Buchanan
Normal Random Variables and Probability
Discrete vs. Continuous Random Variables
Think about the probability of selecting X from the interval [0, 1]
when
X ∈ {0, 1}
X ∈ {k/10 : k = 0, 1, . . . , 10}
X ∈ {k/n : k = 0, 1, . . . , n}
Question: what happens to the last probability as n → ∞?
J. Robert Buchanan
Normal Random Variables and Probability
Continuous Random Variables
Definition
A random variable X has a continuous distribution (or
probability distribution function or probability density
function) if there exists a non-negative function f : R → R such
that for an interval [a, b] the
P (a ≤ X ≤ b) =
Z
b
f (x) dx.
a
The function f must, in addition to satisfying f (x) ≥ 0, have the
following property,
Z
∞
f (x) dx = 1.
−∞
J. Robert Buchanan
Normal Random Variables and Probability
Area Under the PDF
fHxL
x
a
J. Robert Buchanan
b
Normal Random Variables and Probability
Uniformly Distributed Continuous Random Variables
Definition
A continuous random variable X is uniformly distributed in
the interval [a, b] (with b > a) if the probability that X belongs to
any subinterval of [a, b] is equal to the length of the subinterval
divided by b − a.
J. Robert Buchanan
Normal Random Variables and Probability
Uniformly Distributed Continuous Random Variables
Definition
A continuous random variable X is uniformly distributed in
the interval [a, b] (with b > a) if the probability that X belongs to
any subinterval of [a, b] is equal to the length of the subinterval
divided by b − a.
Question: Assuming the PDF vanishes outside of [a, b] and is
constant on [a, b], what is the PDF?
J. Robert Buchanan
Normal Random Variables and Probability
Example
Example
Random variable X is continuously uniformly randomly
distributed in the interval [−5, 5]. Find the probability that
−1 ≤ X ≤ 2.
J. Robert Buchanan
Normal Random Variables and Probability
Example
Example
Random variable X is continuously uniformly randomly
distributed in the interval [−5, 5]. Find the probability that
−1 ≤ X ≤ 2.
P (−1 ≤ X ≤ 2) =
J. Robert Buchanan
3
2 − (−1)
=
5 − (−5)
10
Normal Random Variables and Probability
Expected Value
Definition
The expected value or mean of a continuous random variable
X with probability density function f (x) is
Z ∞
x f (x) dx.
E [X ] =
−∞
J. Robert Buchanan
Normal Random Variables and Probability
Example
Example
Find the expected value of X if X is a continuously uniformly
distributed random variable on the interval [−10, 80].
J. Robert Buchanan
Normal Random Variables and Probability
Example
Example
Find the expected value of X if X is a continuously uniformly
distributed random variable on the interval [−10, 80].
E [X ] =
Z
∞
x f (x) dx
−∞
Z 80
x
dx
−10 90
80
x 2 =
180 −10
6400 100
=
−
180
180
= 35
=
J. Robert Buchanan
Normal Random Variables and Probability
Expected Value of a Function
Definition
The expected value of a function g of a continuously distributed
random variable X which has probability distribution function f
is defined as
Z ∞
g(x)f (x) dx,
E [g(X )] =
−∞
provided the improper integral converges absolutely, i.e.,
E [g(X )] is defined if and only if
Z ∞
|g(x)|f (x) dx < ∞.
−∞
J. Robert Buchanan
Normal Random Variables and Probability
Example
Example
Find the expected value of X 2 if X is continuously distributed on
[0, ∞) with probability density function f (x) = e−x .
J. Robert Buchanan
Normal Random Variables and Probability
Example
Example
Find the expected value of X 2 if X is continuously distributed on
[0, ∞) with probability density function f (x) = e−x .
h
E X
2
i
=
=
=
=
Z
∞
x 2 e−x dx
0
lim
Z
M
M→∞ 0
lim
M→∞
lim
M→∞
= 2
h
h
x 2 e−x dx
iM
−(x 2 + 2x + 2)e−x 0
2
2 − (M + 2M + 2)e
J. Robert Buchanan
−M
i
Normal Random Variables and Probability
Joint and Marginal Distributions
Definition
A joint probability distribution for a pair of random variables,
X and Y , is a non-negative function f (x, y) for which
Z ∞Z ∞
f (x, y) dx dy = 1.
−∞
−∞
J. Robert Buchanan
Normal Random Variables and Probability
Joint and Marginal Distributions
Definition
A joint probability distribution for a pair of random variables,
X and Y , is a non-negative function f (x, y) for which
Z ∞Z ∞
f (x, y) dx dy = 1.
−∞
−∞
Definition
If X and Y are continuous random variables with joint
distribution f (x, y) then the marginal distribution for X is
defined as the function
Z ∞
f (x, y) dy.
fX (x) =
−∞
J. Robert Buchanan
Normal Random Variables and Probability
Joint and Marginal Distributions
Definition
A joint probability distribution for a pair of random variables,
X and Y , is a non-negative function f (x, y) for which
Z ∞Z ∞
f (x, y) dx dy = 1.
−∞
−∞
Definition
If X and Y are continuous random variables with joint
distribution f (x, y) then the marginal distribution for X is
defined as the function
Z ∞
f (x, y) dy.
fX (x) =
−∞
Remark: a similar definition may be stated for the marginal
distribution for Y .
J. Robert Buchanan
Normal Random Variables and Probability
Independence of Jointly Distributed RVs
Definition
Two continuous random variables are independent and only if
the joint probability distribution function factors into the product
of the marginal distributions of X and Y . In other words X and
Y are independent if and only if
f (x, y) = fX (x)fY (y)
for all real numbers x and y.
J. Robert Buchanan
Normal Random Variables and Probability
Example (1 of 2)
Example
Consider the jointly distributed random variables
(X , Y ) ∈ [0, ∞) × [−2, 2] whose distribution is the function
f (x, y) = 4e1x . Find the mean of X + Y .
J. Robert Buchanan
Normal Random Variables and Probability
Example (2 of 2)
E [X + Y ] =
=
=
=
Z
0
Z
∞Z 2
(x + y)
−2
∞
Z0 ∞
Z0 ∞
1
4ex
dy dx
Z
1 −x 2
e
(x + y) dy dx
4
−2
1 −x
e (4x) dx
4
xe−x dx
0
=
=
lim
Z
M→∞ 0
M
xe−x dx
lim (1 − Me−M − e−M )
M→∞
= 1
J. Robert Buchanan
Normal Random Variables and Probability
Properties of the Expected Values
Theorem
If X1 , X2 , . . . , Xk are continuous random variables with joint
probability distribution f (x1 , x2 , . . . , xk ) then
E [X1 + X2 + · · · + Xk ] = E [X1 ] + E [X2 ] + · · · + E [Xk ].
J. Robert Buchanan
Normal Random Variables and Probability
Properties of the Expected Values
Theorem
If X1 , X2 , . . . , Xk are continuous random variables with joint
probability distribution f (x1 , x2 , . . . , xk ) then
E [X1 + X2 + · · · + Xk ] = E [X1 ] + E [X2 ] + · · · + E [Xk ].
Theorem
Let X1 , X2 , . . . , Xk be pairwise independent random variables
with joint distribution f (x1 , x2 , . . . , xk ), then
E [X1 X2 · · · Xk ] = E [X1 ] E [X2 ] · · · E [Xk ] .
J. Robert Buchanan
Normal Random Variables and Probability
Variance and Standard Deviation
Definition
If X is a continuously distributed random variable with
probability density function f (x), the variance of X is defined as
i Z ∞
h
(x − µ)2 f (x) dx,
Var (X ) = E (X − µ)2 =
−∞
where µp
= E [X ]. The standard deviation of X is
σ(X ) = Var (X ).
J. Robert Buchanan
Normal Random Variables and Probability
Variance and Standard Deviation
Definition
If X is a continuously distributed random variable with
probability density function f (x), the variance of X is defined as
i Z ∞
h
(x − µ)2 f (x) dx,
Var (X ) = E (X − µ)2 =
−∞
where µp
= E [X ]. The standard deviation of X is
σ(X ) = Var (X ).
Theorem
Let X be a random variable
with
probability distribution f and
mean µ, then Var (X ) = E X 2 − µ2 .
J. Robert Buchanan
Normal Random Variables and Probability
Example
Example
Suppose X is continuously distributed on [0, ∞) with probability
density function f (x) = e−x . Find Var (X ).
J. Robert Buchanan
Normal Random Variables and Probability
Example
Example
Suppose X is continuously distributed on [0, ∞) with probability
density function f (x) = e−x . Find Var (X ).
h i
Var (X ) = E X 2 − (E [X ])2
Z
Z ∞
2 −x
x e dx −
=
0
= 2−
Z
∞
0
= 2 − (1)2
xe−x dx
∞
0
2
xe
−x
dx
2
= 1
J. Robert Buchanan
Normal Random Variables and Probability
Properties of Variance
Theorem
Let X be a continuous random variable with probability
distribution f (x) and let a, b ∈ R, then
Var (aX + b) = a2 Var (X ) .
J. Robert Buchanan
Normal Random Variables and Probability
Properties of Variance
Theorem
Let X be a continuous random variable with probability
distribution f (x) and let a, b ∈ R, then
Var (aX + b) = a2 Var (X ) .
Theorem
Let X1 , X2 , . . . , Xk be pairwise independent continuous random
variables with joint probability distribution f (x1 , x2 , . . . , xk ), then
Var (X1 + X2 + · · · + Xk ) = Var (X1 ) + Var (X2 ) + · · · + Var (Xk ) .
J. Robert Buchanan
Normal Random Variables and Probability
Normal Random Variable
Assumption: any characteristic of an object subject to a large
number of independently acting forces typically takes on a
normal distribution.
J. Robert Buchanan
Normal Random Variables and Probability
Normal Random Variable
Assumption: any characteristic of an object subject to a large
number of independently acting forces typically takes on a
normal distribution.
We will develop the normal probability density function from the
probability function for the binomial random variable.
P (X = x) =
n!
p x (1 − p)n−x
x!(n − x)!
J. Robert Buchanan
Normal Random Variables and Probability
Overview of Derivation
Thought Experiment: Imagine standing at the origin of the
number line and at each tick of a clock taking a step to the left
or the right. In the long run where will you stand?
J. Robert Buchanan
Normal Random Variables and Probability
Overview of Derivation
Thought Experiment: Imagine standing at the origin of the
number line and at each tick of a clock taking a step to the left
or the right. In the long run where will you stand?
Assumptions:
1
n steps/ticks,
2
random walk takes place during time interval [0, t], which
implies a “tick” lasts ∆t,
3
on each tick move a distance ∆x > 0,
4
n(∆x)2 = 2kt,
5
probability of moving left/right is 1/2,
6
all steps are independent.
J. Robert Buchanan
Normal Random Variables and Probability
Take a Few Steps
Suppose r out of n steps (0 ≤ r ≤ n) have been to the right.
Question: Where are you?
J. Robert Buchanan
Normal Random Variables and Probability
Take a Few Steps
Suppose r out of n steps (0 ≤ r ≤ n) have been to the right.
Question: Where are you?
(r − (n − r ))∆x = (2r − n)∆x = m∆x
Question: What is the probability of standing there?
J. Robert Buchanan
Normal Random Variables and Probability
Take a Few Steps
Suppose r out of n steps (0 ≤ r ≤ n) have been to the right.
Question: Where are you?
(r − (n − r ))∆x = (2r − n)∆x = m∆x
Question: What is the probability of standing there?
P (X = m∆x) = P (X = (2r − n)∆x)
r n−r
n
1
1
=
r
2
2
n
1
n!
=
r !(n − r )! 2
n
n! 12
1
=
1
2 (n + m) ! 2 (n − m) !
J. Robert Buchanan
Normal Random Variables and Probability
Bernoulli Steps
Each step is a Bernoulli experiment with outcomes ∆x and
−∆x.
Questions:
What is the expected value of a single step?
What is the variance in the outcomes?
J. Robert Buchanan
Normal Random Variables and Probability
Bernoulli Steps
Each step is a Bernoulli experiment with outcomes ∆x and
−∆x.
Questions:
What is the expected value of a single step?
E [X ] = 0
What is the variance in the outcomes?
J. Robert Buchanan
Normal Random Variables and Probability
Bernoulli Steps
Each step is a Bernoulli experiment with outcomes ∆x and
−∆x.
Questions:
What is the expected value of a single step?
E [X ] = 0
What is the variance in the outcomes?
Var (X ) = (∆x)2
J. Robert Buchanan
Normal Random Variables and Probability
The Sum of Bernoulli Steps
Questions: after n steps,
What is the expected value of where you stand?
What is the variance in final position?
J. Robert Buchanan
Normal Random Variables and Probability
The Sum of Bernoulli Steps
Questions: after n steps,
What is the expected value of where you stand?
" n #
X
X = nE [X ] = 0
E
i=1
What is the variance in final position?
J. Robert Buchanan
Normal Random Variables and Probability
The Sum of Bernoulli Steps
Questions: after n steps,
What is the expected value of where you stand?
" n #
X
X = nE [X ] = 0
E
i=1
What is the variance in final position?
!
n
X
X = nVar (X ) = n(∆x)2
Var
i=1
J. Robert Buchanan
Normal Random Variables and Probability
Stirling’s Formula
n! ≈
√
2πe−n nn+1/2
Replace all the factorials with Stirling’s Formula.
J. Robert Buchanan
Normal Random Variables and Probability
Stirling’s Formula
n! ≈
√
2πe−n nn+1/2
Replace all the factorials with Stirling’s Formula.
√
n
2πe−n nn+1/2 12
√
(n+m+1)/2 √
(n−m+
2πe−(n+m)/2 12 (n + m)
2πe−(n−m)/2 12 (n − m)
−(n+1)/2
m −m/2 m m/2
2 m2
1+
1−
= √
1− 2
n
n
n
2nπ
J. Robert Buchanan
Normal Random Variables and Probability
Further Simplification
Since m = x/∆x and n = t/∆t,
−(n+1)/2
m −m/2 m m/2
m2
√
1+
1−
1− 2
n
n
n
2nπ
√ x x
!− 1+t/∆t
2
x∆t − 2∆x
x∆t 2
x∆t 2∆x
2 ∆t
1+
1−
1−
= √
t∆x
t∆x
t∆x
2πt
x x "
#− kt 2 − 12
x ∆x 2 (∆x )
∆x
x ∆x − 2∆x
x ∆x 2∆x
1−
= √
1+
1−
,
2kt
2kt
2kt
kπt
2
since (∆x)2 = 2k∆t.
J. Robert Buchanan
Normal Random Variables and Probability
Passing to the Limit
As we take the limit with ∆x → 0, the probability of standing at
exactly one, specific location becomes 0.
Instead we must change our thinking and ask for
P ((m − 1)∆x < X < (m + 1)∆x) ≈ 2(∆x)f (x, t).
J. Robert Buchanan
Normal Random Variables and Probability
Passing to the Limit
As we take the limit with ∆x → 0, the probability of standing at
exactly one, specific location becomes 0.
Instead we must change our thinking and ask for
P ((m − 1)∆x < X < (m + 1)∆x) ≈ 2(∆x)f (x, t).
f (x, t)
=
=
=
−x x "
#− k
1
x ∆x 2 (∆
x ∆x 2∆x
x ∆x 2∆x
√
1−
1−
lim 1 +
2kt
2kt
2kt
2 kπt ∆x→0
−kt
2
1 x − x2 − x x2
− x
√
e 2kt
e 4k 2 t 2
e 2kt
2 kπt
x2
1
√
e− 4kt
2 kπt
J. Robert Buchanan
Normal Random Variables and Probability
Is f (x, t) a PDF?
Suppose
Z
∞
−∞
S2 =
=
=
=
x2
1
√
e− 4kt dx = S, then
2 kπt
Z ∞
Z ∞
y2
x2
1
1
√
√
e− 4kt dx
e− 4kt dy
−∞ 2 kπt
−∞ 2 kπt
Z ∞Z ∞
1
2
2
e−(x +y )/4kt dx dy
4kπt −∞ −∞
Z 2π Z ∞
2
1
re−r /4kt dr d θ
4kπt 0
0
1
J. Robert Buchanan
Normal Random Variables and Probability
Surface Plot
The graph of the PDF resembles:
z
t
x
J. Robert Buchanan
Normal Random Variables and Probability
The Bell Curve
For a fixed value of t, the graph of the PDF resembles:
y
x
J. Robert Buchanan
Normal Random Variables and Probability
Expected Value and Variance
If X is a continuously distributed random variable with PDF:
x2
1
f (x, t) = √
e− 4kt
2 kπt
then
J. Robert Buchanan
Normal Random Variables and Probability
Expected Value and Variance
If X is a continuously distributed random variable with PDF:
x2
1
f (x, t) = √
e− 4kt
2 kπt
then
E [X ] =
Z
∞
−∞
x2
x
√
e− 4kt dx = 0
2 kπt
and
J. Robert Buchanan
Normal Random Variables and Probability
Expected Value and Variance
If X is a continuously distributed random variable with PDF:
x2
1
f (x, t) = √
e− 4kt
2 kπt
then
E [X ] =
Z
∞
−∞
and
Var (X ) =
Z
∞
−∞
x2
x
√
e− 4kt dx = 0
2 kπt
x2
x2
√
e− 4kt dx − (E [X ])2 = 2kt
2 kπt
σ2
and thus 2kt =
and we express the PDF for a normally
distributed random variable with mean µ and variance σ 2 as
2
1
− (x −µ)
f (x) = √ e 2σ2 .
σ 2π
J. Robert Buchanan
Normal Random Variables and Probability
Standard Normal Distribution
x2
1
When µ = 0 and σ = 1 the PDF f (x) = √ e− 2 is called the
2π
standard normal distribution.
J. Robert Buchanan
Normal Random Variables and Probability
Standard Normal Distribution
x2
1
When µ = 0 and σ = 1 the PDF f (x) = √ e− 2 is called the
2π
standard normal distribution.
The cumulative distribution function φ(x) is defined as
Z x
t2
1
√ e− 2 dt.
φ(x) = P (X < x) =
2π
−∞
J. Robert Buchanan
Normal Random Variables and Probability
Change of Variable
Theorem
If X is a normally distributed random variable with expected
value µ and variance σ 2 , then Z = (X − µ)/σ is normally
distributed with an expected value of zero and a variance of
one.
J. Robert Buchanan
Normal Random Variables and Probability
Central Limit Theorem (1 of 2)
Suppose the random variables X1 , X2 , . . . , Xn
1
2
are pairwise independent but not necessarily identically
distributed,
have means µ1 , µ2 , . . . , µn and variances σ12 , σ22 , . . . , σn2 ,
and we define a new random variable Yn as
Pn
i=1 (Xi − µi )
.
Yn = q
Pn
2
σ
i=1 i
J. Robert Buchanan
Normal Random Variables and Probability
Central Limit Theorem (1 of 2)
Suppose the random variables X1 , X2 , . . . , Xn
1
2
are pairwise independent but not necessarily identically
distributed,
have means µ1 , µ2 , . . . , µn and variances σ12 , σ22 , . . . , σn2 ,
and we define a new random variable Yn as
Pn
i=1 (Xi − µi )
.
Yn = q
Pn
2
σ
i=1 i
A Central Limit Theorem due to Liapounov implies that Yn has
the standard normal distribution.
J. Robert Buchanan
Normal Random Variables and Probability
Central Limit Theorem (2 of 2)
Theorem
Suppose that the infinite collection {Xi }∞
i=1 of random variables
are
pairwiseindependent and that for each i ∈ N we have
E |Xi − µi |3 < ∞. If in addition,
lim
n→∞
Pn
then for any x ∈ R
3
i=1 E |Xi − µi |
Pn
2 3/2
i=1 σi
=0
lim P (Yn ≤ x) = φ(x)
n→∞
where random variable Yn is defined as above.
J. Robert Buchanan
Normal Random Variables and Probability
Example (1 of 2)
Example
Suppose the annual snowfall in Millersville, PA is 14.6 inches
with a standard deviation of 3.2 inches and is normally
distributed. Snowfall amounts in different years are
independent. What is the probability that the sum of the
snowfall amounts in the next two years will exceed 30 inches?
J. Robert Buchanan
Normal Random Variables and Probability
Example (2 of 2)
Solution: If X represents the random variable standing for the
snowfall received in Millersville, PA for one year then X + X is
the random variable representing the snowfall of two years. The
random variable X + X has mean µ = 2(14.6) = 29.2 inches
and variance σ 2 = (3.2)2 + (3.2)2 .
!
30 − 29.2
P (X + X > 30) = P Z > p
(3.2)2 + (3.2)2
= 1 − P (Z ≤ 0.176777)
= 1 − φ(0.176777)
= 0.429842
J. Robert Buchanan
Normal Random Variables and Probability
Lognormal Random Variables
Definition
A random variable X is a lognormal random variable with
parameters µ and σ if ln X is a normally distributed random
variable with mean µ and variance σ 2 .
J. Robert Buchanan
Normal Random Variables and Probability
Lognormal Random Variables
Definition
A random variable X is a lognormal random variable with
parameters µ and σ if ln X is a normally distributed random
variable with mean µ and variance σ 2 .
Remarks:
The parameters µ is sometimes called the drift.
The parameter σ is sometimes called the volatility.
J. Robert Buchanan
Normal Random Variables and Probability
Lognormal PDF (1 of 2)
Suppose X is lognormal, then Y = ln X is normal and
P (X < x) = P (Y < ln x)
Z ln x
1
2
2
= √
e−(t−µ) /2σ dt
2π −∞
If we let u = et and du = et dt, then
Z x
1
1 −(ln u−µ)2 /2σ2
P (Y < ln x) = √
e
du
2π −∞ u
J. Robert Buchanan
Normal Random Variables and Probability
Lognormal PDF (2 of 2)
f (x) =
1
2
2
√
e−(ln x−µ) /2σ
(σ 2π)x
y
x
J. Robert Buchanan
Normal Random Variables and Probability
Mean and Variance of a Lognormal RV
Lemma
If X is a lognormal random variable with parameters µ and σ
then
E [X ] = eµ+σ
2 /2
Var (X ) = e2µ+σ
J. Robert Buchanan
2
2
eσ − 1
Normal Random Variables and Probability
Derivation of Mean of a Lognormal RV
Z ∞ 1
1 −(ln x−µ)2 /2σ2
√
e
dx
E [X ] =
x
x
σ 2π 0
Z ∞
1
2
2
√
=
et e−(t−µ) /2σ dt
σ 2π −∞
Z ∞
1
2 2
2
µ+σ2 /2
√
e−(t−(µ+σ )) /2σ dt
= e
σ 2π −∞
= eµ+σ
2 /2
J. Robert Buchanan
Normal Random Variables and Probability
Derivation of Variance of a Lognormal RV
h i
Var (X ) = E X 2 − (E [X ])2
Z ∞ 2
1
2
2 1 −(ln x−µ)2 /2σ2
√
x
=
e
dx − eµ+σ /2
x
σ 2π 0
Z ∞
1
2
2
√
e2t e−(t−µ) /2 dt − e2µ+σ
=
σ 2π −∞
Z ∞
1
2
2
2(µ+σ2 )
√
= e
e−(t−(µ+2σ)) /2 dt − e2µ+σ
σ 2π −∞
2
2
= e2µ+σ eσ − 1
J. Robert Buchanan
Normal Random Variables and Probability
Lognormal RVs and Security Prices
Observation:
Let S(0) denote the price of a security at some starting
time arbitrarily chosen to be t = 0.
For n ≥ 1, let S(n) denote the price of the security on day
n.
The random variable X (n) = S(n)/S(n − 1) for n ≥ 1 is
lognormally distributed, i.e., ln X (n) = ln S(n) − ln S(n − 1)
is normally distributed.
J. Robert Buchanan
Normal Random Variables and Probability
Closing Prices of Sony (SNE) Stock
Closing prices of Sony Corporation stock:
55
50
45
40
Oct
Jan
J. Robert Buchanan
Apr
Jul
Normal Random Variables and Probability
Lognormal Behavior of Sony (SNE) Stock
Lognormal behavior of closing prices:
30
25
20
15
10
5
-0.04
-0.02
0
0.02
0.04
0.06
µ = 0.000416686 σ = 0.0160606
J. Robert Buchanan
Normal Random Variables and Probability
Example (1 of 2)
Example
What is the probability that the closing price of Sony
Corporation stock will be higher today than yesterday?
J. Robert Buchanan
Normal Random Variables and Probability
Example (1 of 2)
Example
What is the probability that the closing price of Sony
Corporation stock will be higher today than yesterday?








 S(n)

S(n)

 = P
P
>
1
>
ln
1
ln


 S(n − 1)

S(n
−
1)


 | {z }

{z
}
|
lognormal
normal
= P (X > 0)
0 − 0.000416686
= P Z >
0.0160606
= 1 − P (Z ≤ −0.0259446)
= 1 − φ(−0.0259446)
= 0.510349
J. Robert Buchanan
Normal Random Variables and Probability
Example (2 of 2)
Example
What is the probability that tomorrow’s closing price will be
higher than yesterday’s closing price?
J. Robert Buchanan
Normal Random Variables and Probability
Example (2 of 2)
Example
What is the probability that tomorrow’s closing price will be
higher than yesterday’s closing price?
P
S(n + 1)
>1
S(n − 1)
=
=
=
=
=
S(n + 1) S(n)
P
>1
S(n) S(n − 1)
S(n)
S(n + 1)
+ ln
>0
P ln
S(n)
S(n − 1)
P (X + X > 0)
0 − 2(0.000416686)
P Z >√
0.01606062 + 0.01606062
1 − P (Z ≤ −0.0366912)
= 1 − φ(−0.0366912)
= 0.514634
J. Robert Buchanan
Normal Random Variables and Probability
Properties of Expected Value and Variance
If an item is worth K but can only be sold for X , a rational
investor would sell only if X ≥ K .
The payoff of the sale can be expressed as
X − K if X ≥ K ,
(X − K )+ =
0
if X < K .
J. Robert Buchanan
Normal Random Variables and Probability
Payoff When X is Normal
Corollary
If X is normal random variable with mean µ and variance σ 2
and K is a constant, then
µ−K
σ −(µ−K )2 /2σ2
+
+ (µ − K )φ
E (X − K ) = √ e
,
σ
2π
Var (X − K )+
µ − 2K (µ − 2K )σ
2
2
2
2
√
e−(µ−2K ) /2σ
+
=
(µ − 2K ) + σ φ
σ
2π
2
σ −(µ−K )2 /2σ2
µ−K
.
− √ e
+ (µ − K )φ
σ
2π
J. Robert Buchanan
Normal Random Variables and Probability
Payoff When X is Lognormal
Corollary
If X is a lognormally distributed random variable with
parameters µ and σ 2 and K > 0 is a constant then
µ − ln K
µ − ln K
+
µ+σ2 /2
E (X − K ) = e
φ
+ σ − Kφ
,
σ
σ
Var (X − K )+
2
= e2(µ+σ ) φ(w + 2σ) + K 2 φ(w )
where w = (µ − ln K )/σ.
2
− 2Keµ+σ /2 φ(w + σ)
2
2
− eµ+σ /2 φ(w + σ) − K φ(w )
J. Robert Buchanan
Normal Random Variables and Probability