Suppose γ ∈ R 1 and β ∈ R k .Let f ( γ , β ) = ( y 2 − γ y 1 ) − ( y 3 − γ y 2 ) exp ⁡ ( x ′ β )Then is f a concave function of ( γ , β ′ )?

Question

Suppose   γ  ∈      R          1       and   β  ∈      R          k      .Let   f  (  γ  ,  β  )  =  (      y          2        −  γ      y          1        )  −  (      y          3        −  γ      y          2        )  exp  ⁡  (      x          ′        β  )Then is f a concave function of   (  γ  ,      β          ′        )?

dyrni0gm · Accepted Answer

Computing the Hessian of  (      y    2    −  γ      y    1    )  −  (      y    3    −  γ      y    2    )      e          x      ⋅      β      we get                    H                            =                              ∂            ∇            f                                ∂            (            γ            ,            β            )                                                            =                              ∂                          (              −                              y                1                            +                              y                2                                            e                                  x                  ⋅                  β                                            ,              (              γ                              y                2                            −                              y                3                            )              x                              e                                  x                  ⋅                  β                                            )                                            ∂            (            γ            ,            β            )                                                            =                  [                                                    0                                                              y                  2                                x                                                                                      y                  2                                                  x                  ′                                                            (                γ                                  y                  2                                −                                  y                  3                                )                                  x                  ′                                x                                              ]                          e                      x            ⋅            β                              Therefore,                              e                      −            x            ⋅            β                                    [                                                    u                                            v                                              ]                H                  [                                                    u                                                                                      v                  ′                                                              ]                                    =                  [                                                                      y                  2                                v                ⋅                x                                                              (                                      y                    2                                    u                  +                  (                  γ                                      y                    2                                    −                                      y                    3                                    )                  v                  ⋅                  x                  )                                x                                              ]                ⋅                  [                                                    u                                            v                                              ]                                                  =                  y          2                u        v        ⋅        x        +                  (                      y            2                    u          +          (          γ                      y            2                    −                      y            3                    )          v          ⋅          x          )                v        ⋅        x                                          =        2                  y          2                u        v        ⋅        x        +        (        γ                  y          2                −                  y          3                )        (        v        ⋅        x                  )          2                    If       y    2    =  0 and       y    3    ≤  0, then the Hessian would be positive.

Suppose gamma in R^1 and beta in R^k. Let f(gamma,beta)=(y2−gamma y1)−(y3−gamma y2)exp(x′beta).Then is f a concave function of (gamma,beta′)?

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New Questions in Calculus 1