normal random variables and probability - millersville...
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Normal Random Variables and ProbabilityAn Undergraduate Introduction to Financial Mathematics
J. Robert Buchanan
2010
J. Robert Buchanan Normal Random Variables and Probability
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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
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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
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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
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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
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Continuous Random Variables
Definition
A random variable X has a continuous distribution (orprobability distribution function or probability densityfunction) if there exists a non-negative function f : R → R suchthat for an interval [a, b] the
P (a ≤ X ≤ b) =
∫ b
af (x) dx .
The function f must, in addition to satisfying f (x) ≥ 0, have thefollowing property,
∫∞
−∞
f (x) dx = 1.
J. Robert Buchanan Normal Random Variables and Probability
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Area Under the PDF
a bx
fHxL
J. Robert Buchanan Normal Random Variables and Probability
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Uniformly Distributed Continuous Random Variables
Definition
A continuous random variable X is uniformly distributed inthe interval [a, b] (with b > a) if the probability that X belongs toany subinterval of [a, b] is equal to the length of the subintervaldivided by b − a.
J. Robert Buchanan Normal Random Variables and Probability
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Uniformly Distributed Continuous Random Variables
Definition
A continuous random variable X is uniformly distributed inthe interval [a, b] (with b > a) if the probability that X belongs toany subinterval of [a, b] is equal to the length of the subintervaldivided by b − a.
Question: Assuming the PDF vanishes outside of [a, b] and isconstant on [a, b], what is the PDF?
J. Robert Buchanan Normal Random Variables and Probability
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Example
Example
Random variable X is continuously uniformly randomlydistributed in the interval [−5, 5]. Find the probability that−1 ≤ X ≤ 2.
J. Robert Buchanan Normal Random Variables and Probability
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Example
Example
Random variable X is continuously uniformly randomlydistributed in the interval [−5, 5]. Find the probability that−1 ≤ X ≤ 2.
P (−1 ≤ X ≤ 2) =2 − (−1)
5 − (−5)=
310
J. Robert Buchanan Normal Random Variables and Probability
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Expected Value
Definition
The expected value or mean of a continuous random variableX with probability density function f (x) is
E [X ] =
∫∞
−∞
x f (x) dx .
J. Robert Buchanan Normal Random Variables and Probability
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Example
Example
Find the expected value of X if X is a continuously uniformlydistributed random variable on the interval [−10, 80].
J. Robert Buchanan Normal Random Variables and Probability
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Example
Example
Find the expected value of X if X is a continuously uniformlydistributed random variable on the interval [−10, 80].
E [X ] =
∫∞
−∞
x f (x) dx
=
∫ 80
−10
x90
dx
=x2
180
∣∣∣∣
80
−10
=6400180
− 100180
= 35
J. Robert Buchanan Normal Random Variables and Probability
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Expected Value of a Function
Definition
The expected value of a function g of a continuously distributedrandom variable X which has probability distribution function fis defined as
E [g(X )] =
∫∞
−∞
g(x)f (x) dx ,
provided the improper integral converges absolutely, i.e.,E [g(X )] is defined if and only if
∫∞
−∞
|g(x)|f (x) dx < ∞.
J. Robert Buchanan Normal Random Variables and Probability
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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
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Example
Example
Find the expected value of X 2 if X is continuously distributed on[0,∞) with probability density function f (x) = e−x .
E[
X 2]
=
∫∞
0x2e−x dx
= limM→∞
∫ M
0x2e−x dx
= limM→∞
[
−(x2 + 2x + 2)e−x]∣∣∣
M
0
= limM→∞
[
2 − (M2 + 2M + 2)e−M]
= 2
J. Robert Buchanan Normal Random Variables and Probability
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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
∫∞
−∞
∫∞
−∞
f (x , y) dx dy = 1.
J. Robert Buchanan Normal Random Variables and Probability
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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
∫∞
−∞
∫∞
−∞
f (x , y) dx dy = 1.
Definition
If X and Y are continuous random variables with jointdistribution f (x , y) then the marginal distribution for X isdefined as the function
fX (x) =
∫∞
−∞
f (x , y) dy .
J. Robert Buchanan Normal Random Variables and Probability
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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
∫∞
−∞
∫∞
−∞
f (x , y) dx dy = 1.
Definition
If X and Y are continuous random variables with jointdistribution f (x , y) then the marginal distribution for X isdefined as the function
fX (x) =
∫∞
−∞
f (x , y) dy .
Remark: a similar definition may be stated for the marginaldistribution for Y .
J. Robert Buchanan Normal Random Variables and Probability
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Independence of Jointly Distributed RVs
Definition
Two continuous random variables are independent and only ifthe joint probability distribution function factors into the productof the marginal distributions of X and Y . In other words X andY 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
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Example (1 of 2)
Example
Consider the jointly distributed random variables(X , Y ) ∈ [0,∞) × [−2, 2] whose distribution is the functionf (x , y) = 1
4ex . Find the mean of X + Y .
J. Robert Buchanan Normal Random Variables and Probability
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Example (2 of 2)
E [X + Y ] =
∫∞
0
∫ 2
−2(x + y)
(1
4ex
)
dy dx
=
∫∞
0
14
e−x∫ 2
−2(x + y) dy dx
=
∫∞
0
14
e−x(4x) dx
=
∫∞
0xe−x dx
= limM→∞
∫ M
0xe−x dx
= limM→∞
(1 − Me−M − e−M)
= 1
J. Robert Buchanan Normal Random Variables and Probability
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Properties of the Expected Values
Theorem
If X1, X2, . . . , Xk are continuous random variables with jointprobability distribution f (x1, x2, . . . , xk ) thenE [X1 + X2 + · · · + Xk ] = E [X1] + E [X2] + · · · + E [Xk ].
J. Robert Buchanan Normal Random Variables and Probability
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Properties of the Expected Values
Theorem
If X1, X2, . . . , Xk are continuous random variables with jointprobability distribution f (x1, x2, . . . , xk ) thenE [X1 + X2 + · · · + Xk ] = E [X1] + E [X2] + · · · + E [Xk ].
TheoremLet X1, X2, . . . , Xk be pairwise independent random variableswith joint distribution f (x1, x2, . . . , xk ), then
E [X1X2 · · ·Xk ] = E [X1] E [X2] · · ·E [Xk ] .
J. Robert Buchanan Normal Random Variables and Probability
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Variance and Standard Deviation
Definition
If X is a continuously distributed random variable withprobability density function f (x), the variance of X is defined as
Var (X ) = E[
(X − µ)2]
=
∫∞
−∞
(x − µ)2f (x) dx ,
where µ = E [X ]. The standard deviation of X isσ(X ) =
√
Var (X ).
J. Robert Buchanan Normal Random Variables and Probability
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Variance and Standard Deviation
Definition
If X is a continuously distributed random variable withprobability density function f (x), the variance of X is defined as
Var (X ) = E[
(X − µ)2]
=
∫∞
−∞
(x − µ)2f (x) dx ,
where µ = E [X ]. The standard deviation of X isσ(X ) =
√
Var (X ).
Theorem
Let X be a random variable with probability distribution f andmean µ, then Var (X ) = E
[X 2]− µ2.
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Example
Example
Suppose X is continuously distributed on [0,∞) with probabilitydensity function f (x) = e−x . Find Var (X ).
J. Robert Buchanan Normal Random Variables and Probability
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Example
Example
Suppose X is continuously distributed on [0,∞) with probabilitydensity function f (x) = e−x . Find Var (X ).
Var (X ) = E[
X 2]
− (E [X ])2
=
∫∞
0x2e−x dx −
(∫∞
0xe−x dx
)2
= 2 −(∫
∞
0xe−x dx
)2
= 2 − (1)2
= 1
J. Robert Buchanan Normal Random Variables and Probability
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Properties of Variance
TheoremLet X be a continuous random variable with probabilitydistribution f (x) and let a, b ∈ R, then
Var (aX + b) = a2Var (X ) .
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Properties of Variance
TheoremLet X be a continuous random variable with probabilitydistribution f (x) and let a, b ∈ R, then
Var (aX + b) = a2Var (X ) .
TheoremLet X1, X2, . . . , Xk be pairwise independent continuous randomvariables 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
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Normal Random Variable
Assumption: any characteristic of an object subject to a largenumber of independently acting forces typically takes on anormal distribution.
J. Robert Buchanan Normal Random Variables and Probability
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Normal Random Variable
Assumption: any characteristic of an object subject to a largenumber of independently acting forces typically takes on anormal distribution.
We will develop the normal probability density function from theprobability function for the binomial random variable.
P (X = x) =n!
x!(n − x)!px(1 − p)n−x
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Overview of Derivation
Thought Experiment: Imagine standing at the origin of thenumber line and at each tick of a clock taking a step to the leftor the right. In the long run where will you stand?
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Overview of Derivation
Thought Experiment: Imagine standing at the origin of thenumber line and at each tick of a clock taking a step to the leftor 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
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Take a Few Steps
Suppose r out of n steps (0 ≤ r ≤ n) have been to the right.
Question: Where are you?
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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
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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)
=
(nr
)(12
)r (12
)n−r
=n!
r !(n − r)!
(12
)n
=n!(1
2
)n
(12(n + m)
)!(1
2(n − m))!
J. Robert Buchanan Normal Random Variables and Probability
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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?
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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
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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
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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?
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The Sum of Bernoulli Steps
Questions: after n steps,
What is the expected value of where you stand?
E
[n∑
i=1
X
]
= nE [X ] = 0
What is the variance in final position?
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The Sum of Bernoulli Steps
Questions: after n steps,
What is the expected value of where you stand?
E
[n∑
i=1
X
]
= nE [X ] = 0
What is the variance in final position?
Var
(n∑
i=1
X
)
= nVar (X ) = n(∆x)2
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Stirling’s Formula
n! ≈√
2πe−nnn+1/2
Replace all the factorials with Stirling’s Formula.
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Stirling’s Formula
n! ≈√
2πe−nnn+1/2
Replace all the factorials with Stirling’s Formula.
√2πe−nnn+1/2
(12
)n
√2πe−(n+m)/2
(12(n + m)
)(n+m+1)/2 √2πe−(n−m)/2
(12(n − m)
)(n−m+
=2√2nπ
(
1 +mn
)−m/2 (
1 − mn
)m/2(
1 − m2
n2
)−(n+1)/2
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Further Simplification
Since m = x/∆x and n = t/∆t ,
2√2nπ
(
1 +mn
)−m/2 (
1 − mn
)m/2(
1 − m2
n2
)−(n+1)/2
=2√
∆t√2πt
(
1 +x∆tt∆x
)−
x2∆x(
1 − x∆tt∆x
) x2∆x
(
1 −[
x∆tt∆x
]2)
−1+t/∆t
2
=∆x√kπt
[
1 +x ∆x2kt
]−
x2∆x[
1 − x ∆x2kt
] x2∆x
[
1 −(
x ∆x2kt
)2]−
kt(∆x)2
−12
,
since (∆x)2 = 2k∆t .
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Passing to the Limit
As we take the limit with ∆x → 0, the probability of standing atexactly 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).
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Passing to the Limit
As we take the limit with ∆x → 0, the probability of standing atexactly 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)
=1
2√
kπtlim
∆x→0
[
1 +x ∆x2kt
] −x2∆x[
1 − x ∆x2kt
] x2∆x
[
1 −(
x ∆x2kt
)2]−
kt(∆x
=1
2√
kπt
(
ex
2kt
)−
x2(
e−x
2kt
) x2
(
e−x2
4k2 t2
)−kt
=1
2√
kπte−
x2
4kt
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Is f (x , t) a PDF?
Suppose∫
∞
−∞
1
2√
kπte−
x2
4kt dx = S, then
S2 =
∫∞
−∞
1
2√
kπte−
x2
4kt dx∫
∞
−∞
1
2√
kπte−
y2
4kt dy
=1
4kπt
∫∞
−∞
∫∞
−∞
e−(x2+y2)/4kt dx dy
=1
4kπt
∫ 2π
0
∫∞
0re−r2/4kt dr dθ
= 1
J. Robert Buchanan Normal Random Variables and Probability
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Surface Plot
The graph of the PDF resembles:
x
t
z
x
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The Bell Curve
For a fixed value of t , the graph of the PDF resembles:
x
y
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Expected Value and Variance
If X is a continuously distributed random variable with PDF:
f (x , t) =1
2√
kπte−
x2
4kt
then
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Expected Value and Variance
If X is a continuously distributed random variable with PDF:
f (x , t) =1
2√
kπte−
x2
4kt
then
E [X ] =
∫∞
−∞
x
2√
kπte−
x2
4kt dx = 0
and
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Expected Value and Variance
If X is a continuously distributed random variable with PDF:
f (x , t) =1
2√
kπte−
x2
4kt
then
E [X ] =
∫∞
−∞
x
2√
kπte−
x2
4kt dx = 0
and
Var (X ) =
∫∞
−∞
x2
2√
kπte−
x2
4kt dx − (E [X ])2 = 2kt
and thus 2kt = σ2 and we express the PDF for a normallydistributed random variable with mean µ and variance σ2 as
f (x) =1
σ√
2πe−
(x−µ)2
2σ2 .
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Standard Normal Distribution
When µ = 0 and σ = 1 the PDF f (x) =1√2π
e−x2
2 is called the
standard normal distribution.
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Standard Normal Distribution
When µ = 0 and σ = 1 the PDF f (x) =1√2π
e−x2
2 is called the
standard normal distribution.
The cumulative distribution function φ(x) is defined as
φ(x) = P (X < x) =
∫ x
−∞
1√2π
e−t2
2 dt .
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Change of Variable
Theorem
If X is a normally distributed random variable with expectedvalue µ and variance σ2, then Z = (X − µ)/σ is normallydistributed with an expected value of zero and a variance ofone.
J. Robert Buchanan Normal Random Variables and Probability
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Central Limit Theorem (1 of 2)
Suppose the random variables X1, X2, . . . , Xn
1 are pairwise independent but not necessarily identicallydistributed,
2 have means µ1, µ2, . . . , µn and variances σ21, σ2
2 , . . . , σ2n,
and we define a new random variable Yn as
Yn =
∑ni=1 (Xi − µi)√∑n
i=1 σ2i
.
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Central Limit Theorem (1 of 2)
Suppose the random variables X1, X2, . . . , Xn
1 are pairwise independent but not necessarily identicallydistributed,
2 have means µ1, µ2, . . . , µn and variances σ21, σ2
2 , . . . , σ2n,
and we define a new random variable Yn as
Yn =
∑ni=1 (Xi − µi)√∑n
i=1 σ2i
.
A Central Limit Theorem due to Liapounov implies that Yn hasthe standard normal distribution.
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Central Limit Theorem (2 of 2)
Theorem
Suppose that the infinite collection {Xi}∞i=1 of random variablesare pairwise independent and that for each i ∈ N we haveE[|Xi − µi |3
]< ∞. If in addition,
limn→∞
∑ni=1 E
[|Xi − µi |3
]
(∑ni=1 σ2
i
)3/2= 0
then for any x ∈ R
limn→∞
P (Yn ≤ x) = φ(x)
where random variable Yn is defined as above.
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Example (1 of 2)
Example
Suppose the annual snowfall in Millersville, PA is 14.6 incheswith a standard deviation of 3.2 inches and is normallydistributed. Snowfall amounts in different years areindependent. What is the probability that the sum of thesnowfall amounts in the next two years will exceed 30 inches?
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Example (2 of 2)
Solution: If X represents the random variable standing for thesnowfall received in Millersville, PA for one year then X + X isthe random variable representing the snowfall of two years. Therandom variable X + X has mean µ = 2(14.6) = 29.2 inchesand variance σ2 = (3.2)2 + (3.2)2.
P (X + X > 30) = P
(
Z >30 − 29.2
√
(3.2)2 + (3.2)2
)
= 1 − P (Z ≤ 0.176777)
= 1 − φ(0.176777)
= 0.429842
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Lognormal Random Variables
Definition
A random variable X is a lognormal random variable withparameters µ and σ if ln X is a normally distributed randomvariable with mean µ and variance σ2.
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Lognormal Random Variables
Definition
A random variable X is a lognormal random variable withparameters µ and σ if ln X is a normally distributed randomvariable with mean µ and variance σ2.
Remarks:
The parameters µ is sometimes called the drift.
The parameter σ is sometimes called the volatility.
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Lognormal PDF (1 of 2)
Suppose X is lognormal, then Y = ln X is normal and
P (X < x) = P (Y < ln x)
=1√2π
∫ ln x
−∞
e−(t−µ)2/2σ2dt
If we let u = et and du = et dt , then
P (Y < ln x) =1√2π
∫ x
−∞
1u
e−(ln u−µ)2/2σ2du
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Lognormal PDF (2 of 2)
f (x) =1
(σ√
2π)xe−(ln x−µ)2/2σ2
x
y
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Mean and Variance of a Lognormal RV
Lemma
If X is a lognormal random variable with parameters µ and σthen
E [X ] = eµ+σ2/2
Var (X ) = e2µ+σ2(
eσ2 − 1)
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Derivation of Mean of a Lognormal RV
E [X ] =1
σ√
2π
∫∞
0x(
1x
e−(ln x−µ)2/2σ2)
dx
=1
σ√
2π
∫∞
−∞
ete−(t−µ)2/2σ2dt
= eµ+σ2/2 1
σ√
2π
∫∞
−∞
e−(t−(µ+σ2))2/2σ2dt
= eµ+σ2/2
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Derivation of Variance of a Lognormal RV
Var (X ) = E[
X 2]
− (E [X ])2
=1
σ√
2π
∫∞
0x2(
1x
e−(ln x−µ)2/2σ2)
dx −(
eµ+σ2/2)2
=1
σ√
2π
∫∞
−∞
e2te−(t−µ)2/2 dt − e2µ+σ2
= e2(µ+σ2) 1
σ√
2π
∫∞
−∞
e−(t−(µ+2σ))2/2 dt − e2µ+σ2
= e2µ+σ2(
eσ2 − 1)
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Lognormal RVs and Security Prices
Observation:
Let S(0) denote the price of a security at some startingtime arbitrarily chosen to be t = 0.
For n ≥ 1, let S(n) denote the price of the security on dayn.
The random variable X (n) = S(n)/S(n − 1) for n ≥ 1 islognormally distributed, i.e., ln X (n) = ln S(n) − ln S(n − 1)is normally distributed.
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Closing Prices of Sony (SNE) Stock
Closing prices of Sony Corporation stock:
Oct Jan Apr Jul
40
45
50
55
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Lognormal Behavior of Sony (SNE) Stock
Lognormal behavior of closing prices:
-0.04 -0.02 0 0.02 0.04 0.06
5
10
15
20
25
30
µ = 0.000416686 σ = 0.0160606
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Example (1 of 2)
Example
What is the probability that the closing price of SonyCorporation stock will be higher today than yesterday?
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Example (1 of 2)
Example
What is the probability that the closing price of SonyCorporation stock will be higher today than yesterday?
P
S(n)
S(n − 1)︸ ︷︷ ︸
lognormal
> 1
= P
lnS(n)
S(n − 1)︸ ︷︷ ︸
normal
> ln 1
= P (X > 0)
= P
(
Z >0 − 0.000416686
0.0160606
)
= 1 − P (Z ≤ −0.0259446)
= 1 − φ(−0.0259446)
= 0.510349J. Robert Buchanan Normal Random Variables and Probability
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Example (2 of 2)
Example
What is the probability that tomorrow’s closing price will behigher than yesterday’s closing price?
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Example (2 of 2)
Example
What is the probability that tomorrow’s closing price will behigher than yesterday’s closing price?
P
(S(n + 1)
S(n − 1)> 1
)
= P
(S(n + 1)
S(n)
S(n)
S(n − 1)> 1
)
= P
(
lnS(n + 1)
S(n)+ ln
S(n)
S(n − 1)> 0
)
= P (X + X > 0)
= P
(
Z >0 − 2(0.000416686)√
0.01606062 + 0.01606062
)
= 1 − P (Z ≤ −0.0366912)
= 1 − φ(−0.0366912)
= 0.514634
J. Robert Buchanan Normal Random Variables and Probability
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Properties of Expected Value and Variance
If an item is worth K but can only be sold for X , a rationalinvestor would sell only if X ≥ K .
The payoff of the sale can be expressed as
(X − K )+ =
{X − K if X ≥ K ,0 if X < K .
J. Robert Buchanan Normal Random Variables and Probability
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Payoff When X is Normal
Corollary
If X is normal random variable with mean µ and variance σ2
and K is a constant, then
E[(X − K )+
]=
σ√2π
e−(µ−K )2/2σ2+ (µ − K )φ
(µ − K
σ
)
,
Var((X − K )+
)
=(
(µ − 2K )2 + σ2)
φ
(µ − 2K
σ
)
+(µ − 2K )σ√
2πe−(µ−2K )2/2σ2
−(
σ√2π
e−(µ−K )2/2σ2+ (µ − K )φ
(µ − K
σ
))2
.
J. Robert Buchanan Normal Random Variables and Probability
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Payoff When X is Lognormal
Corollary
If X is a lognormally distributed random variable withparameters µ and σ2 and K > 0 is a constant then
E[(X − K )+
]= eµ+σ2/2φ
(µ − ln K
σ+ σ
)
− Kφ
(µ − ln K
σ
)
,
Var((X − K )+
)= e2(µ+σ2)φ(w + 2σ) + K 2φ(w)
− 2Keµ+σ2/2φ(w + σ)
−(
eµ+σ2/2φ(w + σ) − Kφ(w))2
where w = (µ − ln K )/σ.
J. Robert Buchanan Normal Random Variables and Probability