Transformations of Random Variables - Advanced Probability Theory for Biomedical Engineers - page 50

Example 6.1.4. Random variable x has the PDF

(1

2 )

+ α / 6 , − 1 <α< 2

f x (

α ) =

0 , otherwise .

Find the PDF of random variable z defined by

⎧
⎨

x − 1 , x ≤ 0

=

g ( x )

=

z 0 , 0 < x ≤ 0 . 5

⎩

1 , 0 . 5 < x .

Solution. To find f z , we evaluate F z first, then differentiate this result. The CDF for random
variable x is

⎧
⎨

0 , α < − 1

F x (

α ) =

3

( α + 3 α + 4) / 18 , − 1 ≤ α< 2

⎩

1 , 2 ≤ α.

Figure 6.2 shows a plot of g ( x ). With the aid of Figure 6.2, we find

⎧
⎨
⎩

( −∞ ,γ + 1] , γ ≤− 1

( −∞ , 0] , − 1 ≤ γ< 0

A ( γ ) ={ x : g ( x ) ≤ γ }=

( −∞ , 0 . 5] , 0 ≤ γ< 1

( −∞ , ∞ ) , 1 ≤ γ.

Consequently,

⎨

F x ( γ + 1) , γ ≤− 1

F x (0) , − 1 ≤ γ< 0

F z ( γ ) =

⎩

F x (0 . 5) , 0 ≤ γ< 1

1 , 1 ≤ γ.

g ( x )

1

x

− 1 1 2

− 1

− 2

FIGURE 6.2: Transformation for Example 6.1.4.