<p>Why are you doing $\frac{\partial^2 V}{\partial w'^2}$ ? </p>

<p>Even if it is said that $w^{'}=w$ at the optimum, it should be taken different when you differentiate it for first order conditions. So, you differentiate it according to $w^{'}$ and $w$.</p>

CC BY-SA 3.0
2015-04-30T08:22:11.780Z
2015-04-30T08:22:11.780Z
optimal control
<p>I think it is the chain rule. Let $w'(w) = w$, since we are looking for revealing mechanisms. The condition
$$
\frac{\partial V}{\partial w'} (w'(w),w) = 0
$$
holds for all $w$ because the mechanism is revealing for all types. As the (not partial) differentiate w.r.t. $w$ of the right hand side is 0, the same goes for the differentiate of the left hand size, hence
$$
\frac{d \frac{\partial V}{\partial w'} (w'(w),w)}{d w} = 0.
$$
According to the chain rule
$$
\frac{d \frac{\partial V}{\partial w'} (w'(w),w)}{d w} = \frac{d w'(w)}{d w}\cdot \frac{\partial^2 V}{\partial^2 w'} (w'(w),w) + \frac{d w}{d w}\cdot \frac{\partial^2 V}{\partial w \partial w'} (w'(w),w),
$$
which in a simplified form is
$$
\frac{\partial^2 V}{\partial^2 w'} (w'(w),w) + \frac{\partial^2 V}{\partial w \partial w'} (w'(w),w).
$$</p>

CC BY-SA 3.0
2015-04-30T09:19:52.593Z
2015-04-30T09:19:52.593Z
Giskard
<p>Starting on p. 98 of Salanie's "The Economic of Taxation" (2nd edition), it explains</p>

<blockquote>
  <p>To probe it more rigorously, let us define the utility of taxpayer $w$ when he claims to have productivity $w'$:
  $$
V(w', w) = u(C(w'), Y(w'), w).
$$
  For the mechanism to be revealing, $V$ must be maximal in $w' = w$. Assume that all functions are differentiable and that income $Y$ is positive. Then we have the first-order necessary condition
  $$
\frac{\partial V}{\partial w'} (w, w) = 0 \tag{NC1}
$$
  and the second-order necessary condition
  $$
\frac{\partial^2 V}{\partial w'^2}(w,w) \leq 0 \tag{NC2}.
$$
  Differentiating (NC1) gives us
  $$
\frac{\partial^2 V}{\partial w'^2} + \frac{\partial^2 V}{\partial w' \partial w} = 0
$$
  ...</p>
</blockquote>

<p>My question is this: How do we arrive at the last equation from (NC1)? If I differentiate 
$$
\frac{\partial V}{\partial w'} (w', w)
$$
with respect to $w$, don't I just end up with $\frac{\partial^2 V}{\partial w'\partial w}$? Likewise, if I differentiate with respect to $w'$, I end up with
$\frac{\partial^2 V}{\partial w'^2}$. What am I missing?</p>

<taxation><mechanism-design><optimal-taxation><asymmetric-information>
Derivation on p.99 of Salanie, The Economics of Taxation (2nd edition)
jmbejara
# Derivation on p.99 of Salanie, The Economics of Taxation (2nd edition)

(Asked by: username_0 on 2015-04-30)

Starting on p. 98 of Salanie's "The Economic of Taxation" (2nd edition), it explains

> To probe it more rigorously, let us define the utility of taxpayer $w$ when he claims to have productivity $w'$:
> $$
> V(w', w) = u(C(w'), Y(w'), w).
> $$
> For the mechanism to be revealing, $V$ must be maximal in $w' = w$. Assume that all functions are differentiable and that income $Y$ is positive. Then we have the first-order necessary condition
> $$
> \frac{\partial V}{\partial w'} (w, w) = 0 \tag{NC1}
> $$
> and the second-order necessary condition
> $$
> \frac{\partial^2 V}{\partial w'^2}(w,w) \leq 0 \tag{NC2}.
> $$
> Differentiating (NC1) gives us
> $$
> \frac{\partial^2 V}{\partial w'^2} + \frac{\partial^2 V}{\partial w' \partial w} = 0
> $$
> ...

My question is this: How do we arrive at the last equation from (NC1)? If I differentiate
$$
\frac{\partial V}{\partial w'} (w', w)
$$
with respect to $w$, don't I just end up with $\frac{\partial^2 V}{\partial w'\partial w}$? Likewise, if I differentiate with respect to $w'$, I end up with
$\frac{\partial^2 V}{\partial w'^2}$. What am I missing?

**Tags:** taxation, mechanism-design, optimal-taxation, asymmetric-information

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## 2 answers

### Answer by username_2 on 2015-04-30. Score: 3 | Accepted answer

I think it is the chain rule. Let $w'(w) = w$, since we are looking for revealing mechanisms. The condition
$$
\frac{\partial V}{\partial w'} (w'(w),w) = 0
$$
holds for all $w$ because the mechanism is revealing for all types. As the (not partial) differentiate w.r.t. $w$ of the right hand side is 0, the same goes for the differentiate of the left hand size, hence
$$
\frac{d \frac{\partial V}{\partial w'} (w'(w),w)}{d w} = 0.
$$
According to the chain rule
$$
\frac{d \frac{\partial V}{\partial w'} (w'(w),w)}{d w} = \frac{d w'(w)}{d w}\cdot \frac{\partial^2 V}{\partial^2 w'} (w'(w),w) + \frac{d w}{d w}\cdot \frac{\partial^2 V}{\partial w \partial w'} (w'(w),w),
$$
which in a simplified form is
$$
\frac{\partial^2 V}{\partial^2 w'} (w'(w),w) + \frac{\partial^2 V}{\partial w \partial w'} (w'(w),w).
$$


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### Answer by username_1 on 2015-04-30. Score: 3

Why are you doing $\frac{\partial^2 V}{\partial w'^2}$ ?

Even if it is said that $w^{'}=w$ at the optimum, it should be taken different when you differentiate it for first order conditions. So, you differentiate it according to $w^{'}$ and $w$.


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