The joint density function for random variables X and Y is f(x,y) = {C(x+y) if 0 ≤ x ≤ 3 , 0 ≤ y ≤ 2 , 0 otherwise. Find the value of the constant C.

Question

The joint density function for random variables X and Y is f(x,y) = {C(x+y) if   0  ≤  x  ≤  3  ,  0  ≤  y  ≤  2  ,  0 otherwise. Find the value of the constant C.

Maggie Tanner · Accepted Answer

1) Since from the text of the problem we have that f(x,y) is a joint density function we know the following:  ∫      ∫                  R                    2        f  (  x  ,  y  )  d  A  =  12) Now we can write down our integral. Knowing that limits are given by   0  ≤  x  ≤  3 and   0  ≤  y  ≤  2 and f(x,y) is defined as 0 outside the mentioned region we get:  ∫      ∫                  R                    2        f  (  x  ,  y  )  d  A  =      ∫          0              2            ∫          0              3        C  (  x  +  y  )  d  x  d  y  =  C      ∫          0              2        [      1    2        x          2        +  y  x      ]          0              3        d  y  =  C      ∫          0              2        ∗  [      9    2    +  3  y  ]  d  y3) Now integrate for y.  =  C  ∗  [      9    2    y  +      3    2        y          2            ]          0              2      =C*[9+6-0-0]=15C4) Now remember that our starting integral was equal to 1 because f(x,y) was a joint density function. Therefore we get.  15  C  =  1  ⇒  C  =      1    15  Result:  C  =      1    15

mksfmasterio · Accepted Answer

The Integral of a probability density function over             R              2       must be 1Since f(x,y)=0 outside the Interval   [  0  ,  3  ]  ×  [  0  ,  2  ]We can write      ∫          −      ∞              ∞            ∫          −      ∞              ∞        f  (  x  ,  y  )  d  y  d  x  =      ∫          0              3            ∫          0              2        C  (  x  +  y  )  d  t  d  x  =  1  C      ∫          0              3        [  x  y  +            y              2              2        ]          0              2        d  x  =  1  C      ∫          0              3        2  x  +  2  d  x  =  1  C  [      x          2        +  2  x      ]          0              3        =  1C(9+6)=1  C  =      1    15  Result:  C  =      1    15

The joint density function for random variables X and Y is f(x,y) = {C(x+y) if 0 le x le 3,0 le y le 2,0 otherwise. Find the value of the constant C.

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