5. Functions of a Random Variable Let X be a r.v defined on the model ( , F , P ), and suppose g(x) is a function of the variable x. Define Y g ( X ). (5-1) Is Y necessarily a r.v? If so what is its PDF FY ( y ), pdf fY ( y ) ? Clearly if Y is a r.v, then for every Borel set B, the set of for Y ( ) must B which belong to F. Given that X is a r.v, this is gis (also B ) a Borel set, i.e., if g(x) is a Borel assured if function. In that case if X is a r.v, so is Y, and for every Borel set B 1 1 P (Y B ) P ( X g ( B )). (5-2) 1 PILLAI In particular FY ( y ) P (Y ( ) y ) P g ( X ( )) y P X ( ) g ( , y ] . 1 (5-3) Thus the distribution function as well of the density function of Y can be determined in terms of that of X. To obtain the distribution function of Y, we must determine the Borel set on the x-axis such that X ( ) g ( y ) for every given y, and the probability of that set. At this point, we shall consider some of the following functions to illustrate the technical details. 1 aX b X nis 1 X X 2 Y g(X ) |X | X log X e X | X |U ( x) 2 PILLAI Example 5.1: Y Solution: Suppose aX b (5-4) a 0. yb yb FY ( y ) P Y ( ) y P aX ( ) b y P X ( ) F . X a a and yb fY ( y ) fX . a a 1 On the other hand if a 0, (5-5) (5-6) then yb FY ( y ) P Y ( ) y P aX ( ) b y P X ( ) a yb 1 FX , a (5-7) and hence fY ( y ) yb fX . a a 1 (5-8) 3 PILLAI From (5-6) and (5-8), we obtain (for all a) fY ( y ) Example 5.2: yb fX . |a | a 1 (5-9) (5-10) Y X . 2 FY ( y ) P Y ( ) y P X ( ) y . (5-11) 2 If y 0, then the event X FY ( y ) 0 , 2 ( ) y , (5-12) y 0. For y 0 , from Fig. 5.1, the event is equivalent to { x X ( ) x }. 1 and hence {Y ( ) y } { X ( ) y } 2 2 Y X 2 y x1 Fig. 5.1 x2 X 4 PILLAI Hence FY ( y ) P x 1 X ( ) x 2 F X ( x 2 ) F X ( x 1 ) FX ( y ) FX ( y 0. y ), (5-13) By direct differentiation, we get 1 fY ( y ) 2 y If fX (x) f X y ) f X ( ( y) , 0, y 0, otherwise . (5-14) represents an even function, then (5-14) reduces to fY ( y ) In particular if X 1 y fX U ( y ). y (5-15) N ( 0 ,1), so that f X ( x) 1 2 e x 2 /2 , (5-16) 5 PILLAI and substituting this into (5-14) or (5-15), we obtain the p.d.f of Y X 2 to be fY ( y ) 1 2 y e (5-17) y /2 U ( y ). On comparing this with (3-36), we notice that (5-17) represents a Chi-square r.v with n = 1, since (1 / 2 ) . Thus, if X is a Gaussian r.v with 0 , then Y X represents a Chi-square r.v with one degree of freedom (n = 1). 2 Example 5.3: Let X c, Y g ( X ) 0, X c, X c, c X c, X c. 6 PILLAI In this case P ( Y 0 ) P ( c X ( ) c ) F X ( c ) F X ( c ). For we have y 0, x c, and Y ( ) X ( ) c (5-18) so that FY ( y ) P Y ( ) y P ( X ( ) c y ) P X ( ) y c F X ( y c ), Similarly y 0, if x c, and y 0. Y ( ) X ( ) c (5-19) so that FY ( y ) P Y ( ) y P ( X ( ) c y ) Thus P X ( ) y c F X ( y c ), y 0. y 0, f X ( y c ), f Y ( y ) [ F X ( c ) F X ( c )] ( y ), f ( y c ), y 0. X g(X ) FX ( x ) c c (5-21) FY ( y ) X x (a) (5-20) (b) Fig. 5.2 y (c) 7 PILLAI Example 5.4: Half-wave rectifier Y g ( X ); x, g(x) 0, x 0, x 0. (5-22) Y In this case P ( Y 0 ) P ( X ( ) 0 ) F X ( 0 ). and for y 0, since X (5-23) Fig. 5.3 Y X , FY ( y ) P Y ( ) y P X ( ) y F X ( y ). (5-24) Thus y 0, f X ( y ), f Y ( y ) F X (0 ) ( y ) y 0, 0, y 0, f X ( y )U ( y ) F X (0 ) ( y ). (5-25) 8 PILLAI Note: As a general approach, given Y g ( X ), first sketch the graph y g ( x ), and determine the range space of y. Suppose a y b is the range space of y g ( x ). Then clearly for y a , FY ( y ) 0 , and for y b , FY ( y ) 1, so that F ( y ) can be nonzero only in a y b. Next, determine whether there are discontinuities in the range space of y. If so evaluate P Y ( ) y at these discontinuities. In the continuous region of y, use the basic approach Y i FY ( y ) P g ( X ( )) y and determine appropriate events in terms of the r.v X for every y. Finally, we must have F ( y ) for y , and obtain Y fY ( y ) dFY ( y ) dy in a y b. 9 PILLAI However, if Y g ( X ) is a continuous function, it is easy to establish a direct procedure to obtain f ( y ). A continuos function g(x) with g ( x ) nonzero at all but a finite number of points, has only a finite number of maxima and minima, and it eventually becomes monotonic as | x | . Consider a specific y on the y-axis, and a positive increment y as shown in Fig. 5.4 Y g(x) y y y x x 3 x3 x3 x 1 x1 x1 x2 x2 x2 Fig. 5.4 fY ( y ) for Y g ( X ), where g () is of continuous type. 10 PILLAI Using (3-28) we can write P y Y ( ) y y y y y f Y ( u ) du f Y ( y ) y . (5-26) But the event y Y ( ) y y can be expressed in terms of X ( ) as well. To see this, referring back to Fig. 5.4, we notice that the equation y g ( x ) has three solutions x1 , x 2 , x 3 (for the specific y chosen there). As a result when y Y ( ) y y , the r.v X could be in any one of the three mutually exclusive intervals { x1 X ( ) x1 x1 }, { x 2 x 2 X ( ) x 2 } or { x 3 X ( ) x 3 x 3 } . Hence the probability of the event in (5-26) is the sum of the probability of the above three events, i.e., P y Y ( ) y y P { x 1 X ( ) x 1 x 1 } P { x 2 x 2 X ( ) x 2 } P { x 3 X ( ) x 3 x 3 } . (5-27) 11 PILLAI For small we get y , xi , making use of the approximation in (5-26), f Y ( y ) y f X ( x1 ) x1 f X ( x 2 )( x 2 ) f X ( x 3 ) x 3 . In this case, rewritten as x1 0 , x 2 0 fY ( y ) f X ( xi ) and | xi | y i and as y 0, so that (5-28) can be x 3 0 , i (5-28) 1 y / xi f X ( xi ) (5-29) (5-29) can be expressed as fY ( y ) i 1 dy / dx f X ( xi ) xi i 1 g ( x i ) f X ( x i ). (5-30) The summation index i in (5-30) depends on y, and for every y the equation y g ( x i ) must be solved to obtain the total number of solutions at every y, and the actual solutions x1 , x 2 , 12 all in terms of y. PILLAI For example, if Y X , then for all y 0 , x1 y and x 2 y represent the two solutions for each y. Notice that the solutions x i are all in terms of y so that the right side of (5-30) is only a function of y. Referring back to the example Y X (Example 5.2) here for each y 0 , there are two solutions given by x1 y and x 2 y . ( f Y ( y ) 0 for y 0 ). Y X Moreover 2 2 2 dy 2 x so that dx dy dx 2 y y x xi and using (5-30) we get 1 fY ( y ) 2 y f x1 X ( which agrees with (5-14). y ) f X ( 0, y) , Fig. 5.5 X x2 y 0, (5-31) otherwise , 13 PILLAI Example 5.5: 1 Y X . Find Solution: Here for every y, dy dx 1 x 2 (5-32) f Y ( y ). x1 1 / y dy so that dx is the only solution, and 1 1/ y x x1 y , 2 2 and substituting this into (5-30), we obtain fY ( y ) 1 f X . y 1 y 2 (5-33) In particular, suppose X is a Cauchy r.v as in (3-39) with parameter so that fX (x) / x 2 In that case from (5-33), Y fY ( y ) / 1 y 2 2 (1 / y ) 2 2 , x . 1/ X has the p.d.f (1 / ) / (1 / ) y 2 (5-34) 2 , y . (5-35) 14 PILLAI But (5-35) represents the p.d.f of a Cauchy r.v with parameter 1 / . Thus if X C ( ), then 1 / X C (1 / ). Example 5.6: Suppose f X ( x ) 2 x / 2 , 0 x , and Y sin X . Determine f ( y ). Solution: Since X has zero probability of falling outside the interval ( 0 , ), y sin x has zero probability of falling outside the interval ( 0 ,1). Clearly f ( y ) 0 outside this interval. For any 0 y 1, from Fig.5.6(b), the equation y sin x has an infinite number of solutions , x1 , x 2 , x 3 , , where x1 sin 1 y is the principal solution. Moreover, using the symmetry we also get x 2 x1 etc. Further, Y Y dy cos x 1 sin 2 x 1 y 2 dx so that dy dx x xi 1 y . 2 15 PILLAI fX (x) x3 (a) x y sin x y x1 x 1 x2 x x3 (b) Fig. 5.6 Using this in (5-30), we obtain for fY ( y ) 1 1 y i i0 2 0 y 1, f X ( x i ). (5-36) But from Fig. 5.6(a), in this case f ( x ) f ( x ) f ( x (Except for f ( x ) and f ( x ) the rest are all zeros). X X 1 X 2 1 X 3 X 4 ) 0 16 PILLAI Thus (Fig. 5.7) fY ( y ) 1 1 y 2 f X ( x1 ) 2 ( x1 x1 ) 2 1 y 2 f X ( x2 ) 2 1 y 0, 2 , 1 1 y 2 fY ( y ) 2 x2 2 x1 2 2 2 0 y 1, (5-37) y 1 otherwise. Fig. 5.7 Example 5.7: Let Y tan X where X U / 2 , / 2 . Determine f ( y ). Solution: As x moves from / 2 , / 2 , y moves from , . From Fig.5.8(b), the function Y tan X is one-to-one for / 2 x / 2 . For any y, x1 tan 1 y is the principal solution. Further Y dy dx d tan x sec 2 x 1 tan 2 x 1 y 2 dx 17 PILLAI so that using (5-30) fY ( y ) 1 f X ( x1 ) | dy / dx | x x1 1/ 1 y 2 , y , (5-38) which represents a Cauchy density function with parameter equal to unity (Fig. 5.9). fX (x) /2 /2 x fY ( y ) 1 1 y 2 (a) y tan x y y Fig. 5.9 /2 x1 / 2 (b) Fig. 5.8 x 18 PILLAI Functions of a discrete-type r.v Suppose X is a discrete-type r.v with P ( X x i ) p i , x x1 , x 2 , , x i , (5-39) and Y g ( X ). Clearly Y is also of discrete-type, and when x x i , y i g ( x i ), and for those y i P (Y y i ) P ( X x i ) p i , y y 1 , y 2 , , y i , Example 5.8: Suppose X (5-40) P ( ), so that P( X k ) e k , k 0 ,1, 2 , (5-41) k! Define Y X 2 1 . Find the p.m.f of Y. Solution: X takes the values 0 ,1, 2 , , k , so that Y only takes the value 1, 2 , 5, , k 2 1, and 19 PILLAI P (Y k 1) P ( X k ) 2 j k 1 so that for 2 P (Y j ) P X j 1 e ( j 1 j 1)! , j 1, 2, 5, , k 1, 2 . (5-42) 20 PILLAI