Another function we will encounter often in this book is the power function: \\[ y=x^{\delta} \\] the form \(y=x^{8} / 8\) to ensure that the derivatives have the proper sign). a. Show that this function is concave (and therefore also, by the result of Problem \(2.9,\) quasi-concave). Notice that the \(\delta=1\) is a special case and that the function is "strictly" concave only for \(\delta<1\) b. Show that the multivariate form of the power function \\[ y=f\left(x_{1}, x_{2}\right)=\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8} \\] is also concave (and quasi-concave). Explain why, in this case, the fact that \(f_{12}=f_{21}=0\) makes the determination of concavity especially simple. c. One way to incorporate "scale" effects into the function described in part (b) is to use the monotonic transformation \\[ g\left(x_{1}, x_{2}\right)=y^{y}=\left[\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8}\right]^{y} \\] where \(\gamma\) is a positive constant. Does this transformation preserve the concavity of the function? Is \(g\) quasi-concave?

To demonstrate the concavity of the univariate power function \(y = x^{\delta}\), we need to find the second derivative and check if it is nonpositive. Let's compute the derivatives: 1. First derivative: \\[ \frac{dy}{dx} = \delta x^{\delta - 1} \\] 2. Second derivative: \\[ \frac{d^2 y}{dx^2} = \delta(\delta - 1) x^{\delta - 2} \\] For \(y = x^{\delta}\) to be concave, the second derivative must be nonpositive. Given that \(x\) is positive, we can analyze the sign of the constant term, \(\delta(\delta - 1)\): - For \(\delta < 1\), the second derivative is negative, implying strict concavity. - For \(\delta = 1\), it is a special case where the second derivative is zero, which means the function is linear but not strictly concave. However, it is still concave since nonpositive second derivative still holds. Since the function is concave, it is also quasi-concave by the result of Problem 2.9.

To show that the multivariate power function \(y=f(x_{1}, x_{2})=(x_{1})^{8}+(x_{2})^{8}\) is concave, we will find its Hessian matrix and check if it is negative semidefinite. The Hessian matrix is given by: \\[ H = \begin{bmatrix} \frac{\partial^2 y}{\partial x_1^2} & \frac{\partial^2 y}{\partial x_1 \partial x_2} \\ \frac{\partial^2 y}{\partial x_2 \partial x_1} & \frac{\partial^2 y}{\partial x_2^2} \end{bmatrix} \\] Finding partial derivatives for the matrix elements: 1. \(\frac{\partial^2 y}{\partial x_1^2} = 56x_1^6\) 2. \(\frac{\partial^2 y}{\partial x_2^2} = 56x_2^6\) 3. \(\frac{\partial^2 y}{\partial x_1 \partial x_2} = \frac{\partial^2 y}{\partial x_2 \partial x_1} = 0\) Our Hessian matrix looks like this: \\[ H = \begin{bmatrix}56x_1^6 & 0 \\ 0 & 56x_2^6\end{bmatrix} \\] Since \(H\) is a diagonal matrix with nonnegative elements, it is negative semidefinite. Therefore, the multivariate power function \(y=f(x_{1}, x_{2})=(x_{1})^{8}+(x_{2})^{8}\) is concave and by Problem 2.9, it's also quasi-concave. The fact that \(f_{12} = f_{21} = 0\) makes the determination of concavity simple because there are no mixed partial derivative terms, which simplifies the condition for a function to be concave.

To analyze the effect of the monotonic transformation on concavity, let's find the Hessian matrix of the transformed function \(g(x_{1}, x_{2}) = [\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8}]^{\gamma}\). Applying the chain rule to compute partial derivatives: 1. \(\frac{\partial^2 g}{\partial x_1^2} = \gamma(\gamma - 1)\left[x_1^{8} + x_2^{8}\right]^{\gamma - 2}(56x_1^6)\), 2. \(\frac{\partial^2 g}{\partial x_2^2} = \gamma(\gamma - 1)\left[x_1^{8} + x_2^{8}\right]^{\gamma - 2}(56x_2^6)\), 3. \(\frac{\partial^2 g}{\partial x_1 \partial x_2} = \frac{\partial^2 g}{\partial x_2 \partial x_1} = 0\). The transformed Hessian matrix is: \\[ H_g = \begin{bmatrix} \gamma(\gamma - 1)\left[x_1^8 + x_2^8\right]^{\gamma - 2}(56x_1^6) & 0 \\ 0 & \gamma(\gamma - 1)\left[x_1^8 + x_2^8\right]^{\gamma - 2}(56x_2^6) \end{bmatrix} \\] To preserve concavity, the Hessian matrix of \(g\) must be negative semidefinite. Considering all elements of the Hessian matrix, it is negative semidefinite only if \(\gamma(\gamma - 1) \le 0\), which implies either \(\gamma \le 0\) or \(1\le\gamma\). However, we were given that \(\gamma\) is a positive constant, so the only case preserving concavity is when \(1\le\gamma\). For quasi-concavity, the Hessian matrix's signs are not strictly determined, so the transformed function may still be quasi-concave regardless of the value of \(\gamma\).