An interesting identity about the expectation E ( X I { X &lt; a } I { Y &lt; b } )Given (X,Y) follows a bivariate normal distribution N ( ( 0 0 ) ; ( 1 r r 1 ) ) with − 1 &lt; r &lt; 1

Question

An interesting identity about the expectation       E    (  X            I              {      X      &amp;lt;      a      }                  I              {      Y      &amp;lt;      b      }        )Given (X,Y) follows a bivariate normal distribution       N        (          (                                    0                                                0                              )        ;          (                                    1                                r                                                r                                1                              )        )   with   −  1  &amp;lt;  r  &amp;lt;  1

Bernard Scott · Accepted Answer

Step 1I denote the correlation by   ρ and the standard normal pdf by   ϕ  (  ⋅  ).The conditional distribution of Y given   X  =  x is known to be   N  (  ρ  x  ,  1  −      ρ    2    ).Then,                               E                          [          X                                    1                                      X              &amp;lt;              a                                                          1                                      Y              &amp;lt;              b                                ]                                    =                  E                                  E                          [          X                                    1                                      X              &amp;lt;              a                                                          1                                      Y              &amp;lt;              b                                ∣          X          ]                                                  =                  E                          [          X                                    1                                      X              &amp;lt;              a                                                      P                    (          Y          &amp;lt;          b          ∣          X          )          ]                                                  =                  E                          [          X                                    1                                      X              &amp;lt;              a                                          Φ                      (                                          b                −                ρ                X                                            1                −                                  ρ                  2                                                      )                    ]                                                  =                  ∫                      −            ∞                    a                x                ϕ        (        x        )                Φ                  (                                    b              −              ρ              x                                      1              −                              ρ                2                                              )                                  d                x            Using       ϕ    ′    (  x  )  =  −  x  ϕ  (  x  ), integrate by parts to get                              ∫                      −            ∞                    a                x                Φ                  (                                    b              −              ρ              x                                      1              −                              ρ                2                                              )                ϕ        (        x        )                          d                x                            =                  lim                      A            →            −            ∞                                    (          −          Φ                      (                                          b                −                ρ                x                                            1                −                                  ρ                  2                                                      )                    ϕ          (          x          )                                                    |                                                    A                        a                    )                −        ρ        I                                          =        −        Φ                  (                                    b              −              ρ              a                                      1              −                              ρ                2                                              )                ϕ        (        a        )        −        ρ        I                ,            where   I  =      ∫          −      ∞        a        1          1      −              ρ        2              ϕ      (                  b        −        ρ        x                    1        −                  ρ          2                      )    ϕ  (  x  )        d    xStep 2To find I, write down the expressions of   ϕ  (  ⋅  ) and complete squares to get a normal pdf within the integral:                    I                            =                  1                      2            π                          1              −                              ρ                2                                                              ∫                      −            ∞                    a                exp        ⁡                  [          −                      1            2                                {                                          (                                                      b                    −                    ρ                    x                                                        1                    −                                          ρ                      2                                                                      )                            2                        +                          x              2                        }                    ]                                  d                x                                          =                  1                      2            π                          1              −                              ρ                2                                                              ∫                      −            ∞                    a                exp        ⁡                  [          −                      1                          2              (              1              −                              ρ                2                            )                                            (                          x              2                        +                          b              2                        −            2            b            ρ            x            )                    ]                                  d                x                                          =                              e                          −                              b                2                                            /                            2                                            2            π                                    ∫                      −            ∞                    a                          1                                    2              π                                      1              −                              ρ                2                                                    exp        ⁡                  [          −                      1                          2              (              1              −                              ρ                2                            )                                (          x          −          b          ρ                      )            2                    ]                                  d                x                                          =        ϕ        (        b        )                Φ                  (                                    a              −              b              ρ                                      1              −                              ρ                2                                              )                    Thus,       E        [    X                  1                    X        &amp;lt;        a                            1                    Y        &amp;lt;        b              ]    =  −  ϕ  (  a  )  Φ      (                  b        −        ρ        a                    1        −                  ρ          2                      )    −  ρ    ϕ  (  b  )    Φ      (                  a        −        b        ρ                    1        −                  ρ          2                      )

An interesting identity about the expectation \bb E(X\mathbb{I}_{\{X<a\}}\mathbb{I}_{\{Y<b\}}).

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New Questions in High school geometry