# Dy it follows that for a monotonic function g we have

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Unformatted text preview: (y )} = fX (x)dx, −∞ yielding dFY (y ) dg −1 (y ) = fX (g −1 (y )) . dy dy 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 1 fX = a |a| y−b 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 ) = k=1 − fX (gk 1 (y )) · − dgk 1 (y ) dy Example 2.4 : Let Y = g (X ), where g (x) = ax2 with a &gt; 0, as illustrated in ﬁgure 2.5. 0 Figure 2.5: Y = aX 2 with a &gt; 0. Each value of y &gt; 0 corresponds to two values of x, i.e. x= Note that − g1 1 (y ) = − y /a −1 g2 (y ) = y /a − − dg1 1 (y ) dg2 1 (y ) 1y = = dy dy 2a a −1/2 1 =√ . 2 ay It follows that 1 fX − fY (y ) = √ 2 ay y a + fX 3 y 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.

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