Fundamentals of Probability and Statistics for Engineers

Functions of Random Variables 147where the integration limits are determined from the requirements y x 2 >0,and x 2 > 0. The result isf Y …y† ˆ a2 ye ay ; for y 0;…5:59†0; elsewhere:Let us note that this problem has also been solved in Example 4.16, by means ofcharacteristic functions. It is to be stressed again that the method of characteristicfunctions is another powerful technique for dealing with sums of independentrandom variables. In fact, when the number of random variables involvedin a sum is large, the method of characteristic function is preferred since there isno need to consider only two at a time as required by Equation (5.56).Ex ample 5. 17. Problem: the random variables X 1 and X 2 are independentand uniformly distributed in intervals 0 x 1 1, and 0 x 2 2. Determinethe pdf of Y ˆ X 1 ‡ X 2 .Answer: the convolution of f X 1(x 1 ) ˆ1, 0 x 1 1, and f X 2(x 2 ) ˆ 1/2,0 x 2 2, results inf Y …y† ˆˆˆˆZ 1f X1…y x 2 †f X2…x 2 †dx 2 ;1Z y…1† 1 dx 2 ˆ y ; for 0 < y 1;2 20Z y y 1…1† 1 dx 2 ˆ 1 ; for 1 < y 2;2 2y 1…1† 1 dx 2 ˆ 3 y ; for 2 < y 3;2 2Z 2ˆ 0;elsewhere:In the above, the limits of the integrals are determined from the requirements0 y x 2 1, and 0 x 2 2. The shape of f Y (y) is that of a trapezoid, asshown in Figure 5.21.5.3 m FUNCTIONS OF n RANDOM VARIABLESWe now consider the general transformation given by Equation (5.1), that is,Y j ˆ g j …X 1 ; ...; X n †; j ˆ 1; 2; ...; m; m n: …5:60†TLFeBOOK