本次加拿大留学生统计作业代写主要是完成贝叶斯网络的推理

CMPT 310 Artificial Intelligence Survey

Simon Fraser University Spring 2021

Assignment 4: Probabilistic Reasoning and Learning

贝叶斯网络的概率推理总共75分。

请访问www.aispace.org并启动“信仰和决策网络”工具。加载示例文件“文件/加载示例问题/简单诊断示例”。我们将使用它来测试一些基本的概率定律。 AIspace工具可以为您完成许多这样的计算,但是练习的目的是学习这些计算背后的原理。您可以使用该工具检查答案,但是您应该自己使用概率演算以及贝叶斯网络中的条件概率来自己计算答案。如果您难以在浏览器中运行该工具,请尝试下载jar文件。

联合概率。 25分
计算以下联合概率(最多6个有效数字)。

使用贝叶斯网络的乘积公式和AIspace指定的条件概率参数来计算以下概率:所有节点均为假。

使用贝叶斯网络的乘积公式和Aispace指定的条件概率参数来计算以下概率:除Sore Throat以外,所有节点均为false,Sore Throat为true。

展示如何使用这两个联合概率来计算以下概率:除“喉咙痛”以外的所有其他节点均为假。 (如果未指定“喉咙痛”的值。)

验证产品规则:
P(所有节点均为假)=
P(喉咙痛=假|所有其他节点均为假)×P(所有​​其他节点均为假)。您可以通过使用该工具执行查询来获得第一个条件概率。

计算喉咙痛为真和发烧为真的概率。 (提示:如果使用正确的公式,则只需要4个条件概率。)

您可以在下表中输入计算出的概率。

计算概率
P(所有节点为假)
P(喉咙痛=真,所有其他节点均为假)
P(喉咙痛以外的所有其他节点为false)
P(所有节点均为假)= P(喉咙痛=假|其他所有节点均为假)x P(所有其他节点均为假)。
P(喉咙痛=真,发烧=真)

您的结果

独立。 20分

“如果没有观察到证据,流感和吸烟是概率独立的。”

(10分)从数字语义上证明这一说法,即表明流感和烟的值在所有可能性上都是独立的。您可以在工具中使用查询来执行此操作。

(10分)从拓扑语义学上证明这一点,即使用马尔可夫条件,即给定其父代的值,该变量在概率上独立于其所有后代。

Probabilistic Reasoning With Bayesian Networks 75 points total.

Go to www.aispace.org and start the “belief and decision network” tool. Load the sample file “File/Load Sample Problem/Simple Diagnostic Example”. We will use this to test some of the basic probability laws. The AIspace tool can do many of these calculations for you, but the purpose of the exercise is to learn the principles behind the calculations. You can use the tool to check your answers, but you should compute them yourself using the probability calculus together with the conditional probabilities from the Bayesian network. If you have difficult running the tool in the browser, try downloading the jar file.

Joint Probabilities. 25 points
Compute the following joint probabilities up to 6 significant digits.

  1. Use the product formula of Bayes nets and the conditional probability parameters specified by AIspace to compute the probability that: all nodes are false.
  2. Use the product formula of Bayes nets and the conditional probability parameters specified by Aispace to compute the probability that: all nodes are false except for Sore Throat, and that Sore Throat is true.
  3. Show how can you use these two joint probabilities to compute the probability that: all nodes other than Sore Throat are false. (Where the value of Sore Throat is unspecified.)
  4. Verify the product rule:
    P(all nodes are false) =
    P(Sore Throat = false| all other nodes are false) x P(all other nodes are false). You may get the first conditional probability by executing a query with the tool.
  5. Compute the probability that Sore Throat is true and that Fever is true. (Hint: If you use the right formula, you only need 4 conditional probabilities.)

You can enter the computed probabilities in the table below.

Probability to be Computed
P(all nodes false)
P(Sore Throat = True, all other nodes false)
P(all nodes other than Sore Throat false)
P(all nodes are false) = P(Sore Throat = false| all other nodes are false) x P(all other nodes are false).
P(Sore Throat = true, Fever = True)

Your Result

Independence. 20 points

“If no evidence is observed, Influenza and Smoking are probabilistically independent.”

  1. (10 points) Prove this statement from the numerical semantics, i.e. show that the values of Influenza and Smokes are independent for all possibilities. You may use queries in the tool to do this.
  2. (10 points) Prove this from the topological semantics, i.e. using the Markov condition that given values for its parents, a variable is probabilistically independent of all its nondescendants.

EasyDue™ 支持PayPal, AliPay, WechatPay, Taobao等各种付款方式!

E-mail: easydue@outlook.com  微信:easydue


EasyDue™是一个服务全球中国留学生的专业代写公司
专注提供稳定可靠的北美、澳洲、英国代写服务
专注提供CS、统计、金融、经济、数学等覆盖100+专业的作业代写服务