Two random variables, X and Y, have the joint density function: f ( x , y ) = { 2 0 &lt; x ≤ y &lt; 1 0 i o c Calculate the correlation coefficient between X and Y.

Question

Two random variables, X and Y, have the joint density function:  f  (  x  ,  y  )  =      {                            2                          0          &amp;lt;          x          ≤          y          &amp;lt;          1                                      0                          i          o          c                        Calculate the correlation coefficient between X and Y.

haillarip0c9 · Accepted Answer

Step 1Let X and Y random variables with joint density function given by  f  (  x  ,  y  )  =      {                            2          ,                    if                    0          &amp;lt;          x          ⩽          y          &amp;lt;          1          ,                                      0          ,                    if                    otherwise                          .The coefficient correlation of X and Y is given by,                                          ρ                          X              Y                                =                                                    C                o                v                            (              X              ,              Y              )                                                      σ                                  X                                                            σ                                  Y                                                              =                                    σ                              X                Y                                                                    σ                                  X                                                            σ                                  Y                                                                        where       C    o    v    (  X  ,  Y  )  =      E    [  X  Y  ]  −      E    [  X  ]      E    [  Y  ] is the covariance of X and Y and       σ          X       and       σ          Y       standard deviations.Now,            f              X              (    x    )    =          ∫              x                    1              f    (    x    ,    y    )              d        y    =    2    (    1    −    x    )    .              f              Y              (    y    )    =          ∫              0                    y              f    (    x    ,    y    )              d        x    =    2    y    .              E        [    X    ]    =          ∫              −        ∞                    +        ∞              x          f              X              (    x    )              d        x    =          ∫              0                    1              x    (    2    −    2    x    )              d        x    =          1      3                  E        [          X              2              ]    =          ∫              −        ∞                    +        ∞                    x              2                    f              X              (    x    )              d        x    =          ∫              0                    1                    x              2              (    2    −    2    x    )              d        x    =          1      6                  E        [    Y    ]    =          ∫              −        ∞                    +        ∞              y          f              Y              (    y    )              d        y    =          ∫              0                    1              y    (    2    y    )              d        y    =          2      3                  E        [          Y              2              ]    =          ∫              −        ∞                    +        ∞                    y              2                    f              Y              (    y    )              d        y    =          ∫              0                    1                    y              2              (    2    y    )              d        y    =          1      2                  E        [    X    Y    ]    =          ∫              −        ∞                    +        ∞                    ∫              −        ∞                    +        ∞              x    y    f    (    x    ,    y    )              d        x          d        y    =          ∫              0                    1                    ∫              0                    y              x    y    (    2    )              d        x          d        y    =          ∫              0                    1                    y              3                        d        y    =          1      4                  C      o      v        (    X    ,    Y    )    =          E        [    X    Y    ]    −          E        [    X    ]          E        [    Y    ]    =          1      4        −          1      3        ×          2      3        =          1      36                  σ              X              =                  V        a        r            (      X      )        =                  E            [              X                  2                    ]      −      (              E            [      X      ]              )                  2                      =                  1        6            −                        (                      1            3                    )                          2                      =                  2            6                  σ              Y              =                  V        a        r            (      Y      )        =                  E            [              Y                  2                    ]      −      (              E            [      Y      ]              )                  2                      =                  1        2            −                        (                      2            3                    )                          2                      =                  2            6      Therefore,      ρ          X      Y        =                    C        o        v            (      X      ,      Y      )                      σ                  X                            σ                  Y                      =            1              /            36              (              2                    /            6              )                  2                      =      1    2    &amp;gt;  0.Since       ρ          X      Y        &amp;gt;  0 then X  and Y they are positively, linearly correlated, but not perfectly so.

Norah Small · Accepted Answer

Step 1The correlation coefficient between X and Y is defined as follows:                    ρ                  X          ,          Y                    =                                                                  E                                      ⁡            [            (            X            −                          μ                              X                                      )            (            Y            −                          μ                              Y                                      )            ]                                              σ                              X                                                    σ                              Y                                                        However,   ρ can be expressed in terms of uncentered moments:                    ρ                  X          ,          Y                    =                                                                  E                                      ⁡            [                        X                        Y                        ]            −                                          E                                      ⁡            [                        X                        ]                                          E                                      ⁡            [                        Y                        ]                                                                                                    E                                                  ⁡                [                                                  X                                      2                                                                  ]                −                                                      (                                                                  E                                                              ⁡                    [                                        X                                        ]                    )                                                        2                                                                                                                                                         E                                                  ⁡                [                                                  Y                                      2                                                                  ]                −                                                      (                                                                  E                                                              ⁡                    [                                        Y                                        ]                    )                                                        2                                                                                          .      Step 2It seems that you are struggling with the orders of integration. It helps to recall the Law of Total Expectation, which states that      E    [  X  ]  =      E    [      E    [  X      |    Y  ]  ] and       E    [  Y  ]  =      E    [      E    [  Y      |    X  ]  ]Step 3Then, the integrals you need to compute are:      E    [  X  ]  =      E    [      E    [  X      |    Y  ]  ]  =      ∫    0    1        ∫    0    y    x  f  (  x  ,  y  )    d  x    d  y      E    [      X    2    ]  =      E    [      E    [      X    2        |    Y  ]  ]  =      ∫    0    1        ∫    0    y        x    2    f  (  x  ,  y  )    d  x    d  y      E    [  Y  ]  =      E    [      E    [  Y      |    X  ]  ]  =      ∫    0    1        ∫    x    1    y  f  (  x  ,  y  )    d  y    d  x      E    [      Y    2    ]  =      E    [      E    [      Y    2        |    X  ]  ]  =      ∫    0    1        ∫    x    1        y    2    f  (  x  ,  y  )    d  y    d  x      E    [  X  Y  ]  =      ∫    0    1        ∫    x    1    x  y  f  (  x  ,  y  )    d  y    d  x

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