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Unformatted text preview: (y )} = fX (x)dx,
−∞ yielding dFY (y )
dg −1 (y )
= fX (g −1 (y ))
Similarly, when g is monotonically decreasing,
fY (y ) = fY (y ) = −fX (g −1 (y )) dg −1 (y )
dy It follows that, for a monotonic function g , we have
fY (y ) = fX (g −1 (y )) ·
2 dg −1 (y )
dy Example 2.3 : Let Y = g (X ), where g (x) = ax + b. Then, g −1 (y ) = (y − b)/a, yielding
dg −1 (y )/dy = 1/a. It follows that
fY (y ) = fX y−b
a · 1
a . Nonmonotonic Functions
If g is not monotonic, then several values of x can correspond to a single value of y , as
illustrated in ﬁgure 2.4. Figure 2.4: Nonmonotonic function of random variable X .
We can view g as having multiple monotonic components g1 , . . . , gK , where K is the
number of monotonic components, and sum the PDFs from these components, i.e.
K fY (y ) =
fX (gk 1 (y )) · −
dgk 1 (y )
dy Example 2.4 : Let Y = g (X ), where g (x) = ax2 with a > 0, as illustrated in ﬁgure 2.5. 0
Figure 2.5: Y = aX 2 with a > 0.
Each value of y > 0 corresponds to two values of x, i.e.
Note that −
g1 1 (y ) = − y /a
g2 (y ) = y /a −
dg1 1 (y )
dg2 1 (y )
2a a −1/2 1
2 ay It follows that
fY (y ) = √
2 ay y
a + fX
a for y ≥ 0....
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This note was uploaded on 01/28/2014 for the course ELECTRONIC TC401 taught by Professor Gong during the Winter '14 term at Aarhus Universitet.
- Winter '14