

![0_image_0.png](0_image_0.png)

be a small change in the absolute value of the given query. Interestingly enough, if y **is a**
disease which is diagnosed by finding e—that is, the probability Pr (y | e**) is quite high—**
then it is not surprising that such queries would be robust against small perturbations to network parameters. This seems to explain some of the results by Pradhan et al. (1996),
where robustness have been confirmed for queries with Pr (y | e) ≥ .9.

Another implication of the above result is that one has to be careful when changing parameters that are extreme. Such parameters are potentially very influential and one must handle them with care.

Therefore, the worst situation from a robustness viewpoint **materializes if one has extreme parameters with non-extreme queries. In such a case, the queries can be very sensitive**
to small variations in the parameters.

Example 3.1 **Consider the network structure in Figure 5. We have two binary nodes,** X
and Y with respective parameters θx, θx and θy, θy. We assume that E **is a deterministic**
binary node where the value of E is e iff X = Y **. This dictates the following CPT for** E:
Pr (e | x, y) = 1, Pr (e | x, y) = 1, Pr (e | x, y) = 0 and Pr (e | x, y) = 0**. The conditional**
probability Pr (y | e) can be expressed using the root parameters θx and θy as:

$$P r(y\mid e)={\frac{\theta_{x}\theta_{y}}{\theta_{x}\theta_{y}+\theta_{\overline{{x}}}\theta_{\overline{{y}}}}}.$$
$\mathbf{a}=\mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a}$ is the sum of $\mathbf{a}$. 
Since ∂θx/∂τx = 1 and ∂θx/∂τx = −1, the derivative of Pr (y | e) **with respect to the meta**
_parameter $\tau_{x}$ is given by:_  $$\frac{\partial Pr(y\mid e)}{\partial\tau_{x}}=\frac{(\theta)}{(\theta)}$$ $$=\frac{(\theta)}{(\theta)}$$  _This is equal to the upper bound given is_  $$\frac{Pr(y\mid e)(1-Pr(y))}{Pr(x)(1-Pr(x))}$$
$$\frac{\frac{(\theta_{x}\theta_{y}+\theta_{\overline{{{x}}}}\theta_{\overline{{{y}}}})\theta_{y}-\theta_{x}\theta_{y}(\theta_{y}-\theta_{\overline{{{y}}}})}{(\theta_{x}\theta_{y}+\theta_{\overline{{{x}}}}\theta_{\overline{{{y}}}})^{2}}}{\frac{\theta_{y}\theta_{\overline{{{y}}}}}{(\theta_{x}\theta_{y}+\theta_{\overline{{{x}}}}\theta_{\overline{{{y}}}})^{2}}}.$$
$$\frac{Pr(y\mid e)(1-Pr(y\mid e))}{Pr(x)(1-Pr(x))}=\frac{(\theta_{x}\theta_{y})(\theta_{\overline{x}}\theta_{\overline{y}})}{\theta_{x}\theta_{\overline{x}}(\theta_{x}\theta_{y}+\theta_{\overline{x}}\theta_{\overline{y}})^{2}}$$ $$=\frac{\theta_{y}\theta_{\overline{y}}}{(\theta_{x}\theta_{y}+\theta_{\overline{x}}\theta_{\overline{y}})^{2}}.$$  _Now, if we set $\theta_{x}=\theta_{\overline{y}}$, the derivative becomes:_  $$\partial Pr(y\mid e)\qquad\qquad1$$
$$\frac{\partial P r(y\mid e)}{\partial\tau_{x}}=\frac{1}{4\theta_{x}\theta_{\overline{{{x}}}}},$$
$$274$$
