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Minimum of Two Exponential
Suppose we want to simulate pseudo random numbers from the following distribution, which I came across in
r - Finding a way to simulate random numbers for this distribution - Cross Validated
,
F
(
x
)
=
1
−
exp
(
−
a
x
−
b
p
+
1
x
p
+
1
)
x
≥
0
F(x) = 1-\exp\left(-ax - \frac{b}{p+1}x^{p+1}\right)\,\quad x\ge 0
F
(
x
)
=
1
−
exp
(
−
a
x
−
p
+
1
b
x
p
+
1
)
x
≥
0
where
a
,
b
>
0
a, b>0
a
,
b
>
0
and
p
∈
(
0
,
1
)
p\in (0,1)
p
∈
(
0
,
1
)
.
Xi'an
provided a very elegant solution.
Note that
(
1
−
F
(
x
)
)
=
exp
{
−
a
x
−
b
p
+
1
x
p
+
1
}
=
exp
{
−
a
x
}
⏟
1
−
F
1
(
x
)
exp
{
−
b
p
+
1
x
p
+
1
}
⏟
1
−
F
2
(
x
)
(1-F(x))=\exp\left\{-ax-\frac{b}{p+1}x^{p+1}\right\}=\underbrace{\exp\left\{-ax\right\}}_{1-F_1(x)}\underbrace{\exp\left\{-\frac{b}{p+1}x^{p+1}\right\}}_{1-F_2(x)}
(
1
−
F
(
x
))
=
exp
{
−
a
x
−
p
+
1
b
x
p
+
1
}
=
1
−
F
1
(
x
)
exp
{
−
a
x
}
1
−
F
2
(
x
)
exp
{
−
p
+
1
b
x
p
+
1
}
the distribution
F
F
F
is the distribution of
X
=
min
{
X
1
,
X
2
}
X
1
∼
F
1
,
X
2
∼
F
2
X=\min\{X_1,X_2\}\qquad X_1\sim F_1\,,X_2\sim F_2
X
=
min
{
X
1
,
X
2
}
X
1
∼
F
1
,
X
2
∼
F
2
since
F
(
x
)
=
P
(
X
≤
x
)
=
1
−
P
(
X
>
x
)
=
1
−
P
(
X
1
>
x
)
P
(
X
2
>
x
)
=
1
−
(
1
−
F
1
(
X
)
)
(
1
−
F
2
(
X
)
)
.
F(x) = P(X\le x) = 1- P(X > x) = 1-P(X_1 > x)P(X_2 > x) = 1-(1-F_1(X))(1-F_2(X))\,.
F
(
x
)
=
P
(
X
≤
x
)
=
1
−
P
(
X
>
x
)
=
1
−
P
(
X
1
>
x
)
P
(
X
2
>
x
)
=
1
−
(
1
−
F
1
(
X
))
(
1
−
F
2
(
X
))
.
Thus, the R code to simulate is simple to be
x
=
pmin
(
rexp
(
n
,
a
),
rexp
(
n
,
b
/
(
p
+
1
))
^
(
1
/
(
p
+
1
)))
References
Finding a way to simulate random numbers for this distribution
simulation fodder for future exams
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Last modified
2yr ago
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