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Unit 5: Functions of Random Variables



                                                                                                  Notes
                   Example 2: Suppose X and Y are independent exponential random variables with  the
            density function
                                                           y
                                          -x
                                  f (x) = e   , x > 0 and f(y) = e   , y > 0
                                      = 0 ,        x  0.             = 0, y  0
            Define Z = X + Y and let us find the distribution function of Z.

            From the definition of Z,
                                       Fz(z) = P[Z  z] = 0 if z < 0
            and, for z > 0

                                              Figure  5.2
















                 Fz (z) = P [Z  z]

                            =     f x, y  (x,y) dx dy

                          { (x,y) : x + y  z}
            where f  (x, y) is the joint density of (X,Y). Since X and Y are independent random variables, the
                  x, y
            joint density of (X,Y) is given by
                                         f  (x, y) = f  (x) f  (Y)
                                          x, y     x   y
            where fx (x)and fy (y) are the marginal density functions
            i.e.                  f  (x, y) = e  × e , x > 0, y > 0
                                             –x
                                                  y
                                   x, y
                                           = 0 , otherwise
            Now for z > 0, the set { (x, y) : x + y  z, x > 0, y > 0} is the region shaded in Fig. 2.
            Hence, for z > 0,

                                 z   z x  
                                   
                                              
                         Fz(z)  =       e  y  dy le – x  dx
                                           
                                 0   0    
                                 z     z x
                                        
                                            
                                  
                              =   e   x   0    e – x dx
                                 0
                                 z
                                        
                                             
                                   
                              =      1 e  (z x)   e – x dx
                                          
                                 0
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