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# how to open pdf file on button click in mvc : Search pdf files for text programmatically application control utility azure web page html visual studio Simulation%20with%20Financial%20Applications2-part1266

21
D.
WEIBULL DISTRIBUTION
Parameters α and β, the location and scale, where x ≥ 0.
Probability density:
β α
α
α
αβ
)
(
1
( )
x
x e
f x
=
Cumulative distribution function F(x):
β α
)
(
( ) ) 1
x
e
F x
= −
Expected value of x:
= Γ
α
αβ
1
E(x)
Variance of x:
[
]
)
(1
)
(1 2
( )
1
2
1
+
−Γ
= Γ Γ +
a
a
V x
β
To generate a random number from a Weibull distribution:
F(x)
u
s
=
So that:
( )
1
s
u
x F
=
Solve for x:
α
β
1
))
ln(1
(
s
u
x
=
EXCEL:
= \$B\$2 * ( -LN( 1 - RAND() ) ) ^ ( 1 / \$B\$1 )
VBA:
Function Random_Weibull( alpha As Double, beta As Double ) As Double
Random_Weibull = beta * ( -Log(1 - Rnd())) ^ ( 1 / alpha )
End Function
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22
E.
TRIANGULAR DISTRIBUTION
Parameters a, b, and m, the lower and upper bounds and the mode or most likely value, so
that a ≤ m ≤ b.
Probability density:
≤ ≤
≤ ≤
=
x b
if m
b a a b b m
b x
if a a x x m
b a a m m a
x a
f x
)
)(
(
)
2(
)
)(
(
)
2(
( )
Cumulative distribution function F(x):
≤ ≤
≤ ≤
=
x b
if m
b a a b b m
b x
if a a x x m
b a a m m a
x a
F x
)
)(
(
)
(
1
)
)(
(
)
(
( )
2
2
Expected value of x:
3
( )
a b b m
E x
+ +
+
=
Variance of x:
18
( )
2
2
2
bm
am
ab
m
b
a
V x
+
+
=
To generate a random number from a triangular distribution:
u =F(x)
So that:
( )
1
u
x F
=
Solve for x
s
is standard triangular, where a = 0, b = 1, and where:
b a
m a
m
s
=
And, therefore:
>
=
s
s
s
s
s
s
s s
s
m
if u
u
m
m
ifu
mu
x
)
)(1
(1
1
So that x is triangular( a, b, m ):
)
x a x x (b b a
s
= +
=
EXCEL:
VBA:
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23
Function STriangular( m As Double ) As Double
Dim us As Double
us = Rnd()
If us < m Then
…
Else
…
End If
End Function
Function Triangular( a As Double, b As Double, m As Double ) As Double
Dim ms As Double
ms = (m - a) / (b - a)
Triangular = a + STriangular( ms ) * (b - a)
End Function
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24
F.
NORMAL DISTRIBUTION
Parameters µ and σ.
Probability density:
2
2
) 2
(
2
2
1
( )
σ
μ
πσ
− −
=
x
e
f x
Cumulative distribution function F(x):
F(x)=
F
approximation?
The cdf of the standard normal distribution, where µ = 0 and σ = 1, is approximated in
Excel in the NormsDist() function.
EXCEL:
=NORMSDIST( z )
VBA:
Function SNormCDF( z As Double ) As Double
Dim a As Double, b As Double, c As Double, d As Double
Dim e As Double, f As Double, x As Double, y As Double, z As Double
a = 2.506628
b = 0.3193815
c = -0.3565638
d = 1.7814779
e = -1.821256
f = 1.3302744
If z > 0 Or z = 0 Then
x = 1
Else
x = -1
End If
y = 1 / (1 + 0.2316419 * x * z)
SNormCDF = 0.5 + x * (0.5 - (Exp(-z * z / 2) / a) * _
(y * (b + y * (c + y * (d + y * (e + y * f))))))
End Function
Expected value of x:
μ
E(x)=
E
Variance of x:
2
( )
σ
V x x =
To generate a random number from a normal distribution:
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25
u =F(z)
So that:
( )
1
u
z F
=
Solve for x:
z=approximation?
Generating Random Numbers from the Standard Normal Distribution:
To generate a z
s
, a random number drawn from the standard normal distribution, µ = 0
and σ = 1.
EXCEL:
= NORMSINV( RAND() )
VBA:
There are three ways to generate standard normal random numbers.  Here is the first way:
Function Random_SNorm1() As Double
Dim u1 As Double
Dim u2 As Double
u1 = Rnd()
u2 = Rnd()
Random_SNorm1 = Sqr(-2 * Ln(u1)) * Cos(2 * 3.1415927 * u2)
End Function
Here is the second way using an approximation to the Normal Inverse CDF:
Function SNorm_InverseCDF( p As Double ) As Double
‘ Dims are left out for brevity.
a1 = -39.6968303
a2 = 220.9460984
a3 = -275.9285104
a4 = 138.3577519
a5 = -30.6647981
a6 = 2.5066283
b1 = -54.4760988
b2 = 161.5858369
b3 = -155.6989799
b4 = 66.8013119
b5 = -13.2806816
c1 = -0.0077849
c2 = -0.3223965
c3 = -2.4007583
c4 = -2.5497325
c5 = 4.3746641
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26
c6 = 2.938164
d1 = 0.0077847
d2 = 0.3224671
d3 = 2.4451341
d4 = 3.7544087
p_low = 0.02425
p_high = 1 - p_low
q = 0#
r = 0#
Select Case p
Case Is < p_low
q = Sqr(-2 * Log(p))
SNorm_InverseCDF = (((((c1 * q + c2) * q + c3) * q + c4) _
* q + c5) * q + c6) / ((((d1 * q + d2) _
* q + d3) * q + d4) * q + 1)
Case Is < p_high
q = p - 0.5
r = q * q
SNorm_InverseCDF = (((((a1 * r + a2) * r + a3) * r + a4) _
*  r + a5) * r + a6) * q / (((((b1 * r _
+ b2) * r + b3) * r + b4) * r + b5) * _
r + 1)
Case Is < 1
q = Sqr(-2 * Log(1 - p))
SNorm_InverseCDF = -(((((c1 * q + c2) * q + c3) * q + c4) _
* q + c5) * q + c6) / ((((d1 * q + d2))_
* q + d3) * q + d4) * q + 1)
End Select
End Function
Function Random_SNorm2() As Double
Random_SNorm2 = SNorm_InverseCDF( Rnd() )
End Function
Here is the third way which works because of the central limit theorem.  However, the
previous two ways should be preferred.
Function Random_SNorm3() As Double
Random_SNorm3 = Rnd + Rnd + Rnd + Rnd + Rnd + Rnd + Rnd + Rnd + _
Rnd + Rnd + Rnd + Rnd - 6
End Function