Therefore, the upper bound on the derivative is given by:

$${\frac{\partial P r(y\mid\mathbf{e})}{\partial\tau_{z\mid\mathbf{u}}}}\leq{\frac{P r(y,\mathbf{u}\mid\mathbf{e})P r(\overline{{y}}\mid\mathbf{e})-P r(x\mid\mathbf{u})(P r(y,\mathbf{u}\mid\mathbf{e})-P r(y\mid\mathbf{e})P r(\mathbf{u}\mid\mathbf{e}))}{P r(x\mid\mathbf{u})(1-P r(x\mid\mathbf{u}))}},$$

which is equal to the following term:

Pr (y | e)Pr (y, u | e) Pr (x | u)+ Pr (y | e)Pr (y, u | e) 1 − Pr (x | u) =(1 − Pr (x | u))Pr (y | e)Pr (y, u | e) + Pr (x | u)Pr (y | e)Pr (y, u | e) Pr (x | u)(1 − Pr (x | u)) =Pr (y, u | e)Pr (y | e) − Pr (x | u)(Pr(y, u | e) − Pr (y | e)Pr (u | e)) Pr (x | u)(1 − Pr (x | u)) .
Since Pr (y, u | e) ≤ Pr (y | e) and Pr (y, u | e) ≤ Pr (y | e**), the upper bound on the**
derivative is given by:

∂Pr (y | e) ∂τx|u ≤Pr (y | e)Pr (y, u | e) Pr (x | u)+ Pr (y | e)Pr (y, u | e) 1 − Pr (x | u) ≤Pr (y | e)Pr (y | e) Pr (x | u)+ Pr (y | e)Pr (y | e) 1 − Pr (x | u) =Pr (y | e)(1 − Pr (y | e)) Pr (x | u)(1 − Pr (x | u)).
In order to find a lower bound on the derivative, we note that Pr (y | e**) = 1** − Pr (y | e),
and thus ∂Pr (y | e)/∂τx|u = −∂Pr (y | e)/∂τx|u**. Therefore, we can get our lower bound by** finding the upper bound on the derivative ∂Pr (y | e)/∂τx|u **and multiplying by** −1:

$$\frac{\partial P r(y\mid\mathbf{e})}{\partial\tau_{x|\mathbf{u}}}$$
$$\begin{array}{r l}{\geq}&{{}-{\frac{P r({\overline{{y}}}\mid\mathbf{e})(1-P r({\overline{{y}}}\mid\mathbf{e}))}{P r(x\mid\mathbf{u})(1-P r(x\mid\mathbf{u}))}}}\\ {=}&{{}-{\frac{P r(y\mid\mathbf{e})(1-P r(y\mid\mathbf{e}))}{P r(x\mid\mathbf{u})(1-P r(x\mid\mathbf{u}))}}.}\end{array}$$

Combining the upper bound and the lower bound, we have:

$$\left|{\frac{\partial P r(y\mid\mathbf{e})}{\partial\tau_{x\mid\mathbf{u}}}}\right|\leq{\frac{P r(y\mid\mathbf{e})(1-P r(y\mid\mathbf{e}))}{P r(x\mid\mathbf{u})(1-P r(x\mid\mathbf{u}))}}.\qed$$

Theorem 3.2 Assume that τx|u ≤ .5 without loss of generality. Suppose that ∆τx|u **is an**
infinitesimal change applied to the meta parameter τx|u**, leading to a change of** ∆Pr (y | e)
to the query Pr (y | e)**. We then have:**

$$\left|{\frac{\Delta P r(y\mid\mathbf{e})}{P r(y\mid\mathbf{e})}}\right|\leq2\left|{\frac{\Delta\tau_{x|\mathbf{u}}}{\tau_{x|\mathbf{u}}}}\right|.$$
