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Giskard
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By definition we have $$ \begin{align*} u(\theta) & = \theta q(\theta)-t(\theta) \\ \\ u(\theta + \delta) & = (\theta + \delta) q(\theta + \delta)-t(\theta + \delta) \end{align*} $$ By incentive compatibility (where $\theta + \delta$ is the true type, $\theta$ is the false type) we have $$ u(\theta + \delta) \geq (\theta + \delta) q(\theta)-t(\theta) $$ Using these (I used square brackets for clearer notation, there is no mathematical function to them) $$ \begin{align*} u(\theta + \delta) - u(\theta) & = \left[(\theta + \delta) q(\theta + \delta)-t(\theta + \delta)\right] - \left[\theta q(\theta)-t(\theta)\right] \\ \\ u(\theta + \delta) - u(\theta) & \geq \left[(\theta + \delta) q(\theta)-t(\theta)\right] - \left[\theta q(\theta)-t(\theta)\right] \end{align*} $$ This also holds if you divide by $\delta$$\delta > 0$.

Similar argument for 2.8., using true type $\theta - \delta$ instead of $\theta + \delta$.

By definition we have $$ \begin{align*} u(\theta) & = \theta q(\theta)-t(\theta) \\ \\ u(\theta + \delta) & = (\theta + \delta) q(\theta + \delta)-t(\theta + \delta) \end{align*} $$ By incentive compatibility (where $\theta + \delta$ is the true type, $\theta$ is the false type) we have $$ u(\theta + \delta) \geq (\theta + \delta) q(\theta)-t(\theta) $$ Using these (I used square brackets for clearer notation, there is no mathematical function to them) $$ \begin{align*} u(\theta + \delta) - u(\theta) & = \left[(\theta + \delta) q(\theta + \delta)-t(\theta + \delta)\right] - \left[\theta q(\theta)-t(\theta)\right] \\ \\ u(\theta + \delta) - u(\theta) & \geq \left[(\theta + \delta) q(\theta)-t(\theta)\right] - \left[\theta q(\theta)-t(\theta)\right] \end{align*} $$ This also holds if you divide by $\delta$.

Similar argument for 2.8., using true type $\theta - \delta$ instead of $\theta + \delta$.

By definition we have $$ \begin{align*} u(\theta) & = \theta q(\theta)-t(\theta) \\ \\ u(\theta + \delta) & = (\theta + \delta) q(\theta + \delta)-t(\theta + \delta) \end{align*} $$ By incentive compatibility (where $\theta + \delta$ is the true type, $\theta$ is the false type) we have $$ u(\theta + \delta) \geq (\theta + \delta) q(\theta)-t(\theta) $$ Using these (I used square brackets for clearer notation, there is no mathematical function to them) $$ \begin{align*} u(\theta + \delta) - u(\theta) & = \left[(\theta + \delta) q(\theta + \delta)-t(\theta + \delta)\right] - \left[\theta q(\theta)-t(\theta)\right] \\ \\ u(\theta + \delta) - u(\theta) & \geq \left[(\theta + \delta) q(\theta)-t(\theta)\right] - \left[\theta q(\theta)-t(\theta)\right] \end{align*} $$ This also holds if you divide by $\delta > 0$.

Similar argument for 2.8., using true type $\theta - \delta$ instead of $\theta + \delta$.

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Giskard
  • 24.4k
  • 6
  • 39
  • 67

By definition we have $$ \begin{align*} u(\theta) & = \theta q(\theta)-t(\theta) \\ \\ u(\theta + \delta) & = (\theta + \delta) q(\theta + \delta)-t(\theta + \delta) \end{align*} $$ By incentive compatibility (where $\theta + \delta$ is the true type, $\theta$ is the false type) we have $$ u(\theta + \delta) \geq (\theta + \delta) q(\theta)-t(\theta) $$ Using these (I used square brackets for clearer notation, there is no mathematical function to them) $$ \begin{align*} u(\theta + \delta) - u(\theta) & = \left[(\theta + \delta) q(\theta + \delta)-t(\theta + \delta)\right] - \left[\theta q(\theta)-t(\theta)\right] \\ \\ u(\theta + \delta) - u(\theta) & \geq \left[(\theta + \delta) q(\theta)-t(\theta)\right] - \left[\theta q(\theta)-t(\theta)\right] \end{align*} $$ This also holds if you divide by $\delta$.

Similar argument for 2.8., using true type $\theta - \delta$ instead of $\theta + \delta$.