Random Variables and
Stochastic Processes – 0903720
Dr. Ghazi Al Sukkar
Email: [email protected]
Office Hours: Refer to the website
Course Website:
http://www2.ju.edu.jo/sites/academic/ghazi.alsukkar
1
Chapter 5
Functions of one Random Variable




The Random Variable 𝒈(𝑿)
Distribution Function of 𝒈(𝑿)
The Density Function of 𝒈(𝑿)
The Inverse Problem
2
The Random Variable 𝑔(𝑋)
𝜁1
S
𝜁2
𝜁4
𝑋(𝜁2 )
𝑋(𝜁4 )
𝜁3
𝑋(𝜁3 )
𝑋(𝜁1 )
ℝ
𝑌 = 𝑔(𝑋)
ℝ
• A random variable is a mapping from the sample
space to the set of real numbers.
• Let 𝑋 be a R.V.
• Suppose 𝑔(𝑥) is a function of the real variable 𝑥.
• Then 𝑌 = 𝑔(𝑋) is a new random variable.
3
The R.V. 𝑌 = 𝑔(𝑋)
• Since 𝑋 is a R.V. then X(𝜁) is a number.
• 𝑔 X(𝜁) is another number specified in
terms of X(𝜁) and 𝑔(𝑥).
• This number is the value 𝑌 𝜁 = 𝑔 X(𝜁)
assigned to the R.V. 𝑌.
⟹ A function of a R.V. 𝑋 is a composite
function 𝑌 = 𝑔 𝑋 = 𝑔 X(𝜁) with domain
the set 𝑆.
4
Distribution Function of 𝒈(𝑿)
•
•
•
•
Let 𝑋 be a R.V. with CDF 𝐹𝑋 𝑥 = 𝑃 𝑋 𝜁 ≤ 𝑥 .
𝑌 = 𝑔(𝑋), 𝐹𝑌 𝑦 =?
𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 𝑔(𝑋) ≤ 𝑦
The event 𝑌 ≤ 𝑦 consists of all outcomes 𝜁 such
that 𝑌 𝜁 = 𝑔 X(𝜁) ≤ 𝑦.
• For a specific 𝑦, the values of 𝑥 such that 𝑔(𝑥) ≤
𝑦 form a set on the 𝑥 axis denoted by 𝑅𝑦 .
• 𝑔 X(𝜁) ≤ 𝑦 if X(𝜁) is a number in the set 𝑅𝑦 ⟹
𝐹𝑌 𝑦 = 𝑃 𝑋 ∈ 𝑅𝑦 .
5
𝑔(𝑥)
𝑏
 Case1: General Idea
• Let 𝑎 < 𝑔(𝑥) < 𝑏.
• If 𝑦 ≥ 𝑏, 𝑔 𝑥 ≤ 𝑦, ∀𝑥
⇒𝑃 𝑌≤𝑦 =1
• If 𝑦 < 𝑎, no 𝑥 such that 𝑔 𝑥 ≤ 𝑦
⇒ 𝑃 𝑌 ≤ 𝑦 =0
𝑦1
𝑦2
𝑥21 𝑥22 𝑥23
𝑥1
𝑎
1, 𝑦 ≥ 𝑏
• 𝐹𝑌 𝑦 =
0, 𝑦 < 𝑎
• For 𝑦1 = 𝑔(𝑥1 ), 𝑔 𝑥 ≤ 𝑦1 for 𝑥 ≤ 𝑥1
⇒ 𝐹𝑌 𝑦1 = 𝑃 𝑌 ≤ 𝑦1 = 𝑃 𝑋 ≤ 𝑥1 = 𝐹𝑋 𝑥1
• For 𝑦2 = 𝑔 𝑥21 = 𝑔 𝑥22 = 𝑔(𝑥23 ), 𝑔 𝑥 ≤ 𝑦2 , if 𝑥 ≤
𝑥21 or 𝑥22 ≤ 𝑥 ≤ 𝑥23 .
⇒ 𝐹𝑌 𝑦2 = 𝑃 𝑌 ≤ 𝑦2 = 𝑃 𝑋 ≤ 𝑥21 + 𝑃 𝑥22 ≤ 𝑥 ≤ 𝑥23
6
= 𝐹𝑋 𝑥21 + 𝐹𝑋 𝑥23 − 𝐹𝑋 𝑥22
𝑥
Example
• 𝑌 = 𝑎𝑋 + 𝑏 ⇒ 𝑔 𝑥 = 𝑎𝑥 + 𝑏
Sol.:
• if 𝑎 > 0, then 𝑔(𝑥) ≤ 𝑦 for 𝑥 ≤ 𝑦 − 𝑏 𝑎
⇒ 𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 𝑋 ≤
𝑔(𝑥)
𝑦
(𝑦 − 𝑏)/𝑎
7
𝑥
• if 𝑎 < 0, then 𝑔(𝑥) ≤ 𝑦 for 𝑥 > 𝑦 − 𝑏 𝑎
⇒ 𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 𝑋 >
𝑔(𝑥)
𝑦
(𝑦 − 𝑏)/𝑎
8
𝑥
Example
• 𝑌 = 𝑋2, ⟹ 𝑔 𝑥 = 𝑥 2
• For 𝑦 < 0, no 𝑥 such that 𝑔(𝑥) ≤ 𝑦
⟹ 𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 0, 𝑦 < 0
• For 𝑦 ≥ 0, 𝑔(𝑥) ≤ 𝑦 for − 𝑦 ≤ 𝑥 ≤ + 𝑦
⟹ 𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 − 𝑦 ≤ 𝑥 ≤
𝑔 𝑥 = 𝑥2
𝑦
𝑥2 = − 𝑦
𝑥1 = + 𝑦
𝑥
9
• 𝑓𝑌 𝑦 =
1
2 𝑦
𝑓𝑋
𝑑𝐹𝑌 𝑦
𝑑𝑦
=
𝑦 + 𝑓𝑋 − 𝑦
,𝑦 ≥ 0
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
• If 𝑓𝑋 𝑥 is an event function (e.g. Normal
distribution), i.e. 𝑓𝑋 𝑥 = 𝑓𝑋 −𝑥
• ⟹ 𝑓𝑌 𝑦 =
1
𝑓𝑋
𝑦
𝑦 𝑈(𝑦), where
1, 𝑦 > 0
𝑈 𝑦 =
is the unit step function.
0, 𝑦 < 0
10
In particular if 𝑋~𝑁(0,1)
⟹ 𝑓𝑋 𝑥 =
1
−𝑥 2 2
𝑒
2𝜋
2
• Then for 𝑌 = 𝑋
𝑓𝑌 𝑦 =
1
𝑓𝑋
𝑦
𝑦 𝑈 𝑦 =
1
2𝜋𝑦
𝑒 −𝑦 2 𝑈(𝑦)
Which is the distribution of Chi-square R.V.
with 𝑛 = 1.
• ⟹ if 𝑋~𝑁(0,1) is Normal R.V. then 𝑋 2 is
Chi-square R.V. with one degree of
freedom.
11
In particular if 𝑋~𝑈(−1,1)
𝐹𝑋 𝑥 =
1
𝑥
+ ,
2
2
⟹ 𝐹𝑌 𝑦 =
𝑥 <1
1, 𝑦 > 1
𝑦, 0 ≤ 𝑦 ≤ 1
0, 𝑦 < 0
12
 Case2: 𝑔(𝑥) is constant in an interval
𝑥0 , 𝑥1 .
⟹ 𝑔 𝑥 = 𝑚, 𝑥0 ≤ 𝑥 ≤ 𝑥1
Then
𝑃 𝑌 = 𝑚 = 𝑃 𝑥0 ≤ 𝑥 ≤ 𝑥1 = 𝐹𝑋 𝑥1 − 𝐹𝑋 𝑥0
• ⟹ 𝐹𝑌 𝑦 is discontinuous at 𝑦 = 𝑚 and its
discontinuity equal 𝐹𝑋 𝑥1 − 𝐹𝑋 𝑥0
𝑔(𝑥)
𝑚
𝑥0
𝑥1
𝑥
13
Example
• 𝑌 = 𝑔(𝑋)
g (x)
c
𝑥 − 𝑐, 𝑥 > 𝑐
• 𝑔 𝑥 = 0, −𝑐 ≤ 𝑥 ≤ 𝑐
𝑥 + 𝑐, 𝑥 < −𝑐
x
c
⟹ 𝐹𝑌 𝑦 is discontinuous at 𝑦 =0, 𝐹𝑌 0 = 𝐹𝑋 𝑐 − 𝐹𝑋 −𝑐
• For 𝑦 > 0, 𝑥 > 𝑐, then 𝑔(𝑥) ≤ 𝑦, 𝑥 ≤ 𝑦 + 𝑐
⟹ 𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 𝑋 ≤ 𝑦 + 𝑐 = 𝐹𝑋 (𝑦 + 𝑐)
• For 𝑦 < 0, 𝑥 < −𝑐, then 𝑔(𝑥) ≤ 𝑦, 𝑥 ≤ 𝑦 − 𝑐
⟹ 𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 𝑋 ≤ 𝑦 − 𝑐 = 𝐹𝑋 (𝑦 − 𝑐)
• 𝑓𝑌 𝑦 =
𝑓𝑋 𝑦 + 𝑐 , 𝑦 > 0
𝐹𝑋 𝑐 − 𝐹𝑋 −𝑐 𝛿 𝑦 , 𝑦 = 0
𝑓𝑋 𝑦 −1𝑐 F, (x
𝑦) < 0
X
-c
c
x
FY ( y )
1
y
14
Example : Half-wave rectifier
Y  g ( X );
 x,
g ( x)  
0,
x  0,
x  0.
Y
In this case
X
P (Y  0)  P ( X ( )  0)  FX (0).
and for y  0, since Y  X ,
FY ( y )  PY ( )  y   P X ( )  y   FX ( y ).
Thus
y  0,
 f X ( y ),

fY ( y )   FX (0) ( y ) y  0,

0,
y  0,

 f X ( y )U ( y )  FX (0) ( y ).
15
 Case3: 𝑔(𝑥) is a staircase function.
• 𝑔 𝑥 = 𝑔 𝑥𝑖 = 𝑦𝑖 , 𝑥𝑖−1 < 𝑥 ≤ 𝑥𝑖
⟹ 𝑌 = 𝑔(𝑋) is discrete type R.V. with
𝑃 𝑌 = 𝑦𝑖 = 𝑃 𝑥𝑖−1 < 𝑋 ≤ 𝑥𝑖
= 𝐹𝑋 𝑥𝑖 − 𝐹𝑋 𝑥𝑖−1
𝑔(𝑥)
𝑦𝑖
𝑥𝑖−1 𝑥𝑖
16
Example: Quantizer
• 𝑔 𝑥 = 𝑛𝑠, 𝑛 − 1 𝑠 < 𝑥 ≤ 𝑛𝑠, 𝑛 ∈ ℤ
• 𝑌 = 𝑔(𝑋), is discrete type that takes the
values 𝑦𝑛 = 𝑛𝑠 with:
• 𝑃 𝑌 = 𝑛𝑠 = 𝑃 𝑛 − 1 𝑠 < 𝑥 ≤ 𝑛𝑠
= 𝐹𝑋 𝑛𝑠 − 𝐹𝑋 𝑛 − 1 𝑠
17
 Case4: 𝑔(𝑥) is discontinuous at x = x0
and
𝑔 𝑥 < 𝑔 𝑥0− , 𝑓𝑜𝑟 𝑥 < 𝑥0 ,
𝑔 𝑥 > 𝑔 𝑥0+ , 𝑓𝑜𝑟 𝑥 > 𝑥0
+
−
• For 𝑔 𝑥0 ≤ 𝑦 ≤ 𝑔 𝑥0 , 𝑔 𝑥 ≤ 𝑦, for
𝑥 ≤ 𝑥0 .
⟹ 𝐹𝑌 𝑦 = 𝑃 𝑋 ≤ 𝑥0 = 𝐹𝑋 𝑥0 , 𝑔 𝑥0− ≤ 𝑦 ≤ 𝑔 𝑥0+
18
𝑔(𝑥)
Example
𝑥 + 𝑐, 𝑥 ≥ 0
• 𝑔 𝑥 =
𝑥 − 𝑐, 𝑥 < 0
• 𝑔(𝑥) is discontinuous at 𝑥 = 0
𝑔 0− = −𝑐, 𝑔 0+ = 𝑐
• If 𝑦 ≥ 𝑐, 𝑔(𝑥) ≤ 𝑦 for 𝑥 ≤ 𝑦 − 𝑐
⟹ 𝐹𝑌 𝑦 = 𝑃 𝑋 ≤ 𝑦 − 𝑐 = 𝐹𝑋 (𝑦 − 𝑐)
• If 𝑦 ≤ −𝑐, 𝑔(𝑥) ≤ 𝑦 for 𝑥 ≤ 𝑦 + 𝑐
⟹ 𝐹𝑌 𝑦 = 𝑃 𝑋 ≤ 𝑦 + 𝑐 = 𝐹𝑋 (𝑦 + 𝑐)
• If −𝑐 ≤ 𝑦 ≤ 𝑐, 𝑔(𝑥) ≤ 𝑦 for 𝑥 ≤ 0
⟹ 𝐹𝑌 𝑦 = 𝑃 𝑋 ≤ 0 = 𝐹𝑋 (0)
𝑐
𝑥
−𝑐
𝐹𝑋 (𝑥)
𝑐
𝑥
−𝑐
𝐹𝑌 (𝑦)
𝑐
𝑦
−𝑐
19
 Case5: 𝑋 is discrete type R.V. takes the
values 𝑥𝑘 with probability 𝑝𝑘 ,
then Y = 𝑔(𝑋) is also discrete type R.V.
taking the values 𝑦𝑘 = 𝑔(𝑥𝑘 ).
• If 𝑦𝑘 = 𝑔(𝑥) for only one 𝑥 = 𝑥𝑘 .
⟹ 𝑃 𝑌 = 𝑦𝑘 = 𝑃 𝑋 = 𝑥𝑘 = 𝑝𝑘
• If 𝑦𝑘 = 𝑔(𝑥) for 𝑥 = 𝑥𝑘 and 𝑥 = 𝑥𝑙 .
⟹ 𝑃 𝑌 = 𝑦𝑘 = 𝑃 𝑋 = 𝑥𝑘 + 𝑃 𝑋 = 𝑥𝑙
= 𝑝𝑘 + 𝑝𝑙
20
Density Function of 𝒈(𝑿)
• Let 𝑋 be a R.V. with PDF 𝑓𝑋 𝑥 .
• If 𝑌 = 𝑔(𝑋) is a continuous function, it is easy to establish
a direct procedure to obtain 𝑓𝑌 𝑦 .
• Assume that for all 𝑦 , the equation 𝑔(𝑥) = 𝑦 has a
countable number of solutions and at each solution point,
𝑑𝑔(𝑥)/𝑑𝑥 exists and is nonzero. Then the PDF of 𝑦 =
𝑔(𝑥) is:
𝑓𝑋 (𝑥𝑖 )
𝑓𝑌 𝑦 =
𝑔′ (𝑥𝑖 )
𝑖
Where 𝑔′ (𝑥𝑖 ) is the derivative of 𝑔(𝑥) at 𝑥 = 𝑥𝑖 , and 𝑥𝑖 is
the solution of 𝑔 𝑥 = 𝑦.
21
Proof: Consider a specific 𝑦 on the y-axis, and a positive
increment 𝑑𝑦 as shown in the figure
𝑓𝑌 𝑦 𝑑𝑦 = 𝑃 𝑦 < 𝑌 ≤ 𝑦 + 𝑑𝑦
• The set of 𝑥 such that 𝑦 < 𝑔(𝑋) ≤ 𝑦 + 𝑑𝑦 consist of the
following intervals:
𝑥1 < 𝑥 < 𝑥1 + 𝑑𝑥1 , 𝑥2 + 𝑑𝑥2 < 𝑥 < 𝑥2 , 𝑥3 < 𝑥
< 𝑥3 + 𝑑𝑥3
• 𝑑𝑥1 > 0, 𝑑𝑥2 < 0, 𝑑𝑥3 > 0.
g ( x)
y  dy
y
x1 x1  dx1
x2  dx2 x2
x3 x3  dx3
x
22
• 𝑃 𝑦 < 𝑌 ≤ 𝑦 + 𝑑𝑦 = 𝑃 𝑥1 < 𝑋 < 𝑥1 + 𝑑𝑥1 +
𝑃 𝑥2 + 𝑑𝑥2 < 𝑋 < 𝑥2 + 𝑃 𝑥3 < 𝑋 < 𝑥3 + 𝑑𝑥3
• 𝑃 𝑥1 < 𝑋 < 𝑥1 + 𝑑𝑥1 = 𝑓𝑋 𝑥1 𝑑𝑥1
𝑑𝑦
𝑑𝑥1
= 𝑔′ (𝑥1 )
• 𝑃 𝑥2 + 𝑑𝑥2 < 𝑋 < 𝑥2 = 𝑓𝑋 𝑥2 𝑑𝑥2
𝑑𝑦
𝑑𝑥2
= 𝑔′ (𝑥2 )
• 𝑃 𝑥3 < 𝑋 < 𝑥3 + 𝑑𝑥3 = 𝑓𝑋 𝑥3 𝑑𝑥3
𝑑𝑦
𝑑𝑥3
= 𝑔′ (𝑥3 )
𝑓
𝑥
𝑓
𝑥
𝑓
𝑥
• ⟹ 𝑓𝑌 𝑦 𝑑𝑦 = 𝑋′ 1 𝑑𝑦 + 𝑋′ 2 𝑑𝑦 + 𝑋′ 3 𝑑𝑦
𝑔 𝑥1
𝑔 𝑥3
𝑔 𝑥2
1
1
fY ( y )  
f X ( xi )  
f X ( xi ).
i dy / dx x
i g ( xi )
i
23
𝑌 = 𝑎𝑋 + 𝑏
• 𝑔 𝑥 = 𝑦 = 𝑎𝑥 + 𝑏 has a single solution
𝑥1 = 𝑦 − 𝑏 𝑎, 𝑔′ 𝑥 = 𝑎
⟹ 𝑓𝑌 𝑦 =
1
𝑎
𝑦−𝑎
𝑓𝑋 𝑏
• Example: The voltage 𝑉 is a R.V.
𝑉 = 0.01(𝑅 + 1000), 𝑅~𝑈(900,1000)
⇒ 𝑉~𝑈(19,21)
24
1
𝑌=
𝑋
1
𝑥
• 𝑔 𝑥 = 𝑦 = , has a single solution 𝑥1 =
′
𝑔 𝑥 =
⟹ 𝑓𝑌 𝑦
1
− 2 , 𝑔′ 𝑥1
𝑥
1
1
= 𝑦2 𝑓𝑋 𝑦
1
𝑦
= −𝑦 2 .
• Example: The resistance 𝑅~𝑈(900,1100),
1
𝑅
The Conductance 𝐺 =
1
1
1
𝑓𝐺 𝑔 =
,
<g<
2
200𝑔 1100
900
25
𝑌=
2
𝑎𝑥
• Assume 𝑎 > 0. 𝑔′ 𝑥 = 2𝑎𝑥
• If 𝑦 ≤ 0, then 𝑦 = 𝑎𝑥 2 has no real solutions
⟹ 𝑓𝑌 0 = 0
• If y > 0, then 𝑦 = 𝑎𝑥 2 has two solutions:
𝑦
𝑦
𝑥1 =
, 𝑥2 = −
𝑎
𝑎
⟹ 𝑓𝑌 𝑦 =
1
2𝑎 𝑦 𝑎
𝑓𝑋
𝑦
𝑎
+ 𝑓𝑋 −
𝑦
𝑎
26
𝑌= 𝑋
• 𝑔 𝑥 = 𝑦 = 𝑥,
𝑔′
𝑥 =
1
2 𝑥
• 𝑦 = 𝑥 has a single solution 𝑥1 = 𝑦 2 for 𝑦 > 0,
and no solution for 𝑦 < 0.
⟹ 𝑓𝑌 𝑦 = 2𝑦𝑓𝑋 𝑦 2 𝑈(𝑦)
• If 𝑋~𝜒 2 is Chi-square: 𝑓𝑋 𝑥 =
If 𝑌 = 𝑋 ⇒ 𝑓𝑌 𝑦 =
2
𝑛
2𝑛 2 Γ 2
𝑦
1
2𝑛 2 Γ
𝑛
2
𝑛−1 −𝑦 2 2
𝑒
𝑥𝑛
2−1 −𝑥 2
𝑒
𝑈(𝑥)
𝑈(𝑦)
Called Chi density with 𝑛 degree of freedom.
– For 𝑛 = 3, we get Maxwell density, 𝑓𝑌 𝑦 =
– For 𝑛 = 2, we get Rayleigh density, 𝑓𝑌 𝑦 =
2 2 −𝑦 2 2
𝑦 𝑒
𝑈(𝑦)
𝜋
2 2
−
𝑦
𝑦𝑒
𝑈(𝑦)
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𝑌=𝑒
𝑋
• 𝑔 𝑥 = 𝑦 = 𝑒 𝑥 , 𝑔′ 𝑥 = 𝑒 𝑥
• 𝑦 = 𝑒 𝑥 has a single solution 𝑥1 = ln 𝑦 for
𝑦 > 0, and no solution for 𝑦 < 0.
1
⟹ 𝑓𝑌 𝑦 = 𝑓𝑋 ln 𝑦 𝑈(𝑦)
𝑦
• If 𝑋~𝑁(𝜇, 𝜎 2 ), then 𝑌 = 𝑒 𝑋 is lognormal
⟹ 𝑓𝑌 𝑦 =
1
𝑦
2𝜋𝜎 2
𝑒
− ln 𝑦 −𝜇 2 2𝜎 2
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𝑌 = 𝑎 sin 𝑋 + 𝜃
• Assume 𝑎 > 0
• 𝑔 𝑥 = 𝑦 = 𝑎 sin 𝑥 + 𝜃
• 𝑔′ 𝑥 = 𝑎 cos 𝑥 + 𝜃 = 𝑎2 − 𝑦 2
• For 𝑦 > 𝑎, 𝑔(𝑥) has no solution ⟹ 𝑓𝑌 𝑦 = 0.
• For 𝑦 < 𝑎, 𝑔(𝑥) has infinitely many solutions 𝑥𝑖 =
𝑦
−1
sin
− 𝜃 , 𝑖 = ⋯ , −2. −1,0,1,2, …
𝑎
⟹ 𝑓𝑌 𝑦 =
𝑔(𝑥)
1
∞
𝑎2 − 𝑦 2 𝑖=−∞
𝑓𝑋 (𝑥𝑖 )
𝑦
𝑥0 𝑥1
𝑥2
𝑥
29
• Example: 𝑋~𝑈(−𝜋, 𝜋), 𝑦 = 𝑎 sin(𝑥 + 𝜃)
has exactly two solutions in the interval
−𝜋, 𝜋 for any 𝜃.
𝑓𝑋 𝑥1 = 𝑓𝑋 𝑥2 =
⟹ 𝑓𝑌 𝑦 =
𝑓𝑋 (𝑥)
1
2𝜋
1
𝜋 𝑎2 − 𝑦 2
, 𝑦 <𝑎
𝑓𝑌 (𝑦)
1
2𝜋
−𝜋
𝜋
𝑥
−𝑎
𝑎
𝑦
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𝑌 = tan 𝑋
• 𝑔 𝑥 = 𝑦 = tan 𝑥
•
𝑔′
𝑥 =
1
𝑐𝑜𝑠2 𝑥
= 𝑠𝑒𝑐 2 𝑥 = 1 + 𝑡𝑎𝑛2 𝑥 = 1 + 𝑦 2
• 𝑔(𝑥) has infinitely many solutions 𝑥𝑖 =
tan−1 𝑦 , 𝑖 = ⋯ , −2. −1,0,1,2, …
∞
1
⟹ 𝑓𝑌 𝑦 =
𝑓𝑋 (𝑥𝑖 )
2
1+𝑦
𝑖=−∞
𝑔(𝑥)
𝑦
𝑥0
𝑥1
𝜋
𝑥2
𝑥
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•
𝜋 𝜋
Example: 𝑋~𝑈(− , ), then 𝑦 = tan 𝑥 has
2 2
one solution 𝑥1 = tan−1 𝑦 in this interval
1
and 𝑓𝑋 𝑥1 =
𝜋
1/𝜋
⟹ 𝑓𝑌 𝑦 =
1 + 𝑦2
So 𝑌 is Cauchy R.V.
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The Inverse Problem
Given the distribution 𝐹𝑋 (𝑥) of the R.V. 𝑋,
find the function 𝑔(𝑥) such that the
distribution of the R.V. 𝑌 = 𝑔(𝑋) equals a
specific function 𝐹𝑌 (𝑦).
• From 𝐹𝑋 (𝑥) to a uniform distribution:
Given 𝐹𝑋 (𝑥) then 𝑈 = 𝑔 𝑋 = 𝐹𝑋 (𝑋) is a
uniformly distributed R.V in the interval (0,1)
⟹ 𝐹𝑈 𝑢 = 𝑢 , 0 ≤ 𝑢 ≤ 1
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• Proof: put 𝑢 = 𝐹𝑋 (𝑥) and knowing that
𝐹𝑋 (𝑥) is monotonic then: 𝑈 ≤ 𝑢 𝑖𝑓𝑓 𝑋 ≤ 𝑥
𝐹𝑈 𝑢 = 𝑃 𝑈 ≤ 𝑢 = 𝑃 𝑋 ≤ 𝑥 = 𝐹𝑋 𝑥
=𝑢
𝑋
𝐹𝑋 (𝑥)
𝑈
𝐹𝑋−1 (𝑢)
𝑈 = 𝐹𝑋 (𝑋)
𝑋 = 𝐹𝑋−1 (𝑈)
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• From Uniform to 𝐹𝑌 (𝑦):
Given 𝑈~𝑈(0,1) then 𝑌 = 𝑔 𝑈 = 𝐹𝑌−1 (𝑈)
is a R.V with the distribution 𝐹𝑌 𝑦
From 𝐹𝑋 (𝑥) to 𝐹𝑌 (𝑦):
Find 𝑈 = 𝐹𝑋 (𝑋) then 𝑌 = 𝐹𝑌−1 (𝑈)
⟹ 𝑌 = 𝐹𝑌−1 𝐹𝑋 (𝑋)
𝑋
𝐹𝑋 (𝑥)
𝑈 = 𝐹𝑋 (𝑋)
𝐹𝑌−1 (𝑢)
𝑌 = 𝐹𝑌−1 (𝑈)
35