# 统计代写|贝叶斯分析代写Bayesian Analysis代考|Solving the Resolution Problem

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|Solving the Resolution Problem

The resolution problem occurs when the variance of a child node, $Y$, is many orders of magnitude smaller compared with the variance of the parent nodes. As a result, the mass generated by each sample taken on the parent node, $X$, is so small that the conditional probability distribution $f(Y \mid X)$ is sparse and contains areas without mass that should actually contain it.

Consider, for example, a model containing $X$ and $Y$ with ranges $\Omega_X$ and $\Omega_Y$, respectively, and joint density function $f_{X, Y}(x, y)=f_Y(y \mid X=x) f_X(x)$, with distributions:
$$\begin{gathered} f_X(x)=N(0,1 E 8) \ f_Y(y \mid X=x)=N(0,1 E-8) \end{gathered}$$
Here, $f_X(x)$ has extremely high variance, at $1 E 8$, and the conditional distribution $f_Y(y \mid X=x)$ has extremely low variance at $1 E-8$. If we take two samples per parent region, we achieve a marginal posterior density for $Y$ which is jagged, containing areas that are clearly under-sampled, as shown in Figure D.4. We should achieve a smooth function resembling $f(X)$ for $Y$ given it is nearly an identity function. Likewise, the fact that the result on $Y$ is jagged means that the DD algorithm treats the marginal distribution as multi-modal and therefore spends more computation time splitting intervals that are in fact smooth and locally linear.
What is happening here? Samples taken from $X$ generate an extremely small part of the mass of the conditional probability distribution of $Y$ and it would require a very large number of samples taken from $X$ to generate mass such that it would adequately approximate the posterior range of $Y$. In this case, we roughly calculate how many samples would be needed to accurately cover the likelihood: Here, $X$ has prior range $\Omega_X=\left{-2 \times 10^{-5}, 2 \times 10^5\right}$ and each sample would generate a probability mass with width $\left|\Omega_\gamma\right|=\left|10^{-3}\right|$, thus requiring $4 \times 10^5 / 10^{-3}$ samples, which equals 400 million samples in total. This is clearly too many.

We can solve this resolution problem by firstly tuning the sampling process to generate a high number of samples in the target region of the child node’s conditional probability distribution, given the range of the parents and, additionally, by smoothing any under-sampled regions with probability mass, using an additional optimization procedure called uniform smoothing.

Sample tuning is done by computing the resolution, $r_i$, for each sub-region, $w_i$, in the child node $Y$, as a function of the conditional probability mass generated in that sub-region by the states of the parent nodes, $p a{Y}$. This mass on the child node $Y$ is computed from the expectations and standard deviations computed on $Y$ from the parent states sampled on, $p a{Y}$ (to make the notation easier we will simply assume $p a{Y}$ means the sample values from the parent nodes of child $Y)$. Thus, we compute two bounds using $E(Y$ । $p a{Y}) \pm$ s.d. $(Y \mid p a{Y})$ and from this determine: $r_i$ :
$$r_i=\frac{\left|w_i\right|}{\mid E(Y \mid p a{Y})-\text { s.d. }(Y \mid p a{Y}), E(Y \mid p a{Y})+\text { s.d. }(Y \mid p a{Y}) \mid}$$

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|Approximating Mixture Distributions

The preceding cases deal with deterministic and statistical functions but we can also deal with more interesting and more challenging cases, such as mixture distributions where continuous nodes are conditionally dependent on labeled or Boolean nodes, as may be the case in hybrid BNs. Unlike other algorithms, dynamic discretization does not enforce any restriction on whether particular continuous or discrete nodes can be parents or children of others. Neither does the algorithm make any assumptions about whether some nodes in a HBN can or cannot receive evidence.

Mixture distributions are easily declared as hybrid models containing at least one labeled node that specifies the mixtures we wish to model along with the prior probabilities of each mixture indexing a continuous node containing the distributions we wish to mix. A mixture distribution is usually specified as a marginal distribution, indexed by discrete states in parent nodes $A$ and $B$ :
$$P\left(C \in \omega_k\right)=\sum_{j=1}^n\left[\int_{\omega_k} f\left(C \mid A=a_j, B=b_j\right) d C\right] P\left(A=a_j\right) P\left(B=b_j\right)$$
In the $\mathrm{BN}$ we therefore simply specify a set of conditional distribution functions partitioned by the labeled nodes we wish to use to define the mixture (in AgenaRisk we use partitioned expressions to declare these). Thus, for a mixture function where continuous variable, $A$, is conditioned on a discrete variable, $B$, with discrete states, $\left{b_1, b_2, \ldots, b_n\right}$, we could generate a different statistical or deterministic function for each state in the parent $B$ :
$$f\left(A \mid B=\left{b_1, b_2, \ldots, b_n\right}\right)=\left{\begin{array}{c} f\left(A \mid B=b_1\right)=N(0,10) \ f\left(A \mid B=b_2\right)=\operatorname{Gamma}(5,4) \ \cdot \quad \cdot \ \cdot \ f\left(A \mid B=b_n\right)=\text { TNormal }(10,100,0,10) \end{array}\right.$$
Since dynamic discretization is flexible and agnostic about the underlying distributions chosen it will simply substitute the corresponding conditional distribution function into the NPT and execute, producing the mixture as if it was any other function.

# 贝叶斯分析代考

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|Solving the Resolution Problem

$$f_X(x)=N(0,1 E 8) f_Y(y \mid X=x)=N(0,1 E-8)$$

$$r_i=\frac{\left|w_i\right|}{\mid E(Y \mid p a Y)-\text { s.d. }(Y \mid p a Y), E(Y \mid p a Y)+\text { s.d. }(Y \mid p a Y) \mid}$$

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|Approximating Mixture Distributions

$$P\left(C \in \omega_k\right)=\sum_{j=1}^n\left[\int_{\omega_k} f\left(C \mid A=a_j, B=b_j\right) d C\right] P\left(A=a_j\right) P\left(B=b_j\right)$$

$\$ \$$f\left(A \mid B=b_1\right)=N(0,10) f\left(A \mid B=b_2\right)=\operatorname{Gamma}(5,4) \cdot \quad \cdots\left(A \mid B=b_n\right) 正确的。 \ \$$

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