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The question is dependent on the definition of optimality. Mathematically, the optimal choice is the target $f$ itself as maths do not account for the cost of producing a simulation. If the cost of computing is taken into account, only functions requesting the same number of basic operations can be compared. For instance, all exponential families can be compared. In which case, the optimal choice is the one leading to the highest acceptance probability. This means, if $$\frac{f(x)}{g_\lambda(x)}\le M_\lambda$$finding $\lambda$ that minimises $M_\lambda$. And the resolution of $$x^\star(\lambda)=\arg\max_x \frac{f(x)}{g_\lambda(x)} = \arg\max_x \lambda^{-1} x e^{(\lambda-4)x}$$that leads to $$x^\star(\lambda)=(4-\lambda)^{-1}$$ makes the constant $$\lambda^{-1}(4-\lambda)^{-1}e^{-1}$$rather straightforward to minimise! The optimum is thus for $\lambda^\star=2$, with an acceptance probability of $$\frac{1}{M^\star}=8\frac{1}{2}\frac{1}{2}e^{-1}=2e^{-1}$$

The question is dependent on the definition of optimality. Mathematically, the optimal choice is the target $f$ itself as maths do not account for the cost of producing a simulation. If the cost of computing is taken into account, only functions requesting the same number of basic operations can be compared. For instance, all exponential families can be compared. In which case, the optimal choice is the one leading to the highest acceptance probability. This means, if $$\frac{f(x)}{g_\lambda(x)}\le M_\lambda$$finding $\lambda$ that minimises $M_\lambda$. And the resolution of $$x^\star(\lambda)=\arg\max_x \frac{f(x)}{g_\lambda(x)} = \arg\max_x \lambda^{-1} x e^{(\lambda-4)x}$$that leads to $$x^\star(\lambda)=(4-\lambda)^{-1}$$ makes the constant $$\lambda^{-1}(4-\lambda)^{-1}e^{-1}$$rather straightforward to minimise! The optimum is thus for $\lambda^\star=2$, with an acceptance probability of $$\frac{1}{M^\star}=8\frac{1}{2}\frac{1}{2}e^{-1}=2e^{-1}\approx0.73$$

The question is dependent on the definition of optimality. Mathematically, the optimal choice is the target $f$ itself as maths do not account for the cost of producing a simulation. If the cost of computing is taken into account, only functions requesting the same number of basic operations can be compared. For instance, all exponential families can be compared. In which case, the optimal choice is the one leading to the highest acceptance probability. This means, if $$\frac{f(x)}{g_\lambda(x)}\le M_\lambda$$finding $\lambda$ that minimises $M_\lambda$. And the resolution of $$x^*(\lambda)=\arg\max_x \frac{f(x)}{g_\lambda(x)} = \arg\max_x \lambda^{-1} x e^{(\lambda-4)x}$$that leads to $$x^*(\lambda)=(4-\lambda)^{-1}$$ makes the constant $$\lambda(4-\lambda)^{-1}e^{-1}$$rather delicate to minimise!

The question is dependent on the definition of optimality. Mathematically, the optimal choice is the target $f$ itself as maths do not account for the cost of producing a simulation. If the cost of computing is taken into account, only functions requesting the same number of basic operations can be compared. For instance, all exponential families can be compared. In which case, the optimal choice is the one leading to the highest acceptance probability. This means, if $$\frac{f(x)}{g_\lambda(x)}\le M_\lambda$$finding $\lambda$ that minimises $M_\lambda$. And the resolution of $$x^\star(\lambda)=\arg\max_x \frac{f(x)}{g_\lambda(x)} = \arg\max_x \lambda^{-1} x e^{(\lambda-4)x}$$that leads to $$x^\star(\lambda)=(4-\lambda)^{-1}$$ makes the constant $$\lambda^{-1}(4-\lambda)^{-1}e^{-1}$$rather straightforward to minimise! The optimum is thus for $\lambda^\star=2$, with an acceptance probability of $$\frac{1}{M^\star}=8\frac{1}{2}\frac{1}{2}e^{-1}=2e^{-1}$$

The question is dependent on the definition of optimality. Mathematically, the optimal choice is the target $f$ itself as maths do not account for the cost of producing a simulation. If the cost of computing is taken into account, only functions requesting the same number of basic operations can be compared. For instance, all exponential families can be compared. In which case, the optimal choice is the one leading to the highest acceptance probability. This means, if $$\frac{f(x)}{g_\lambda(x)}\le M_\lambda$$finding $\lambda$ that minimises $M_\lambda$.

The question is dependent on the definition of optimality. Mathematically, the optimal choice is the target $f$ itself as maths do not account for the cost of producing a simulation. If the cost of computing is taken into account, only functions requesting the same number of basic operations can be compared. For instance, all exponential families can be compared. In which case, the optimal choice is the one leading to the highest acceptance probability. This means, if $$\frac{f(x)}{g_\lambda(x)}\le M_\lambda$$finding $\lambda$ that minimises $M_\lambda$. And the resolution of $$x^*(\lambda)=\arg\max_x \frac{f(x)}{g_\lambda(x)} = \arg\max_x \lambda^{-1} x e^{(\lambda-4)x}$$that leads to $$x^*(\lambda)=(4-\lambda)^{-1}$$ makes the constant $$\lambda(4-\lambda)^{-1}e^{-1}$$rather delicate to minimise!