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X1, X2, X3. This network is shown in Fig. 1, which shows that three of the variables (X1, X2, X3) have observed values, with the other three (Y₁, Y₂, Ys) being unobserved. Your goal in this question is use variable elimination to compute p(Y3 = END| X₁ = a, X2 = b, X3 = #). All of these values are described below. D(Y, Y) DIY, Y₂) Y₁ Y, P(X, Y) P(X,-bY₂) X-b P(X,Y) Figure 1: Bayesian network for question 1 Assume that the Y; variables are all defined over three different values (Z1, Z2, and END), where the same conditional distribution p₁(YY-1) is used to define the distribution of Yi, given its parent Yi-1 (where there's a prior distribution over Y₁). Further assume that the Xi variables are all defined over three different values (a, b, and #), where the same conditional distribution pe(XiY;) is used to define the distribution of X; given its parent Yi. These distributions are labeled (for ease) on the arcs of the figure, and are provided below in Table 1 (in both sub-tables, each row is a different distribution): a Z₁ X₁ = b # 7.30 Y₁ = Z₁ Z.15 Z₂ END .8 .05 Z₁ Y₁-1= Y₁₁ = Z2 2.8 0 Z₂ .6 END 0 0 35 .05 .7 Y₁ = Z₂ END 2.1 1 END 0 0 1 (a) The definition of pe, for 1 (b) The definition of p₂, for i = i≤3. 2.3. (c) The prior distribution for Y₁. Table 1: The distributions needed for question 1. In order to compute p(Y3 = END X₁ = a, X2 = b, X3 = #), you'll need to eliminate both Y₁ and Y2. To make the computations a bit easier, we'll start with Y, and then eliminiate to Y2. Assume that the factors have been defined as given in Table 2: Factor e₁(Y₁) Definition Pe(X₁ = a❘Y₁) €2(Y2) Pe(X2=b|Y₂) es(Ys) Pe(X3 = #|Y3) ti (Yi) t₂(Y₁, Y₂) t3(Y2,Y3) P(Y₁) Pt(Y2Y1) Pi (Ya Y2) Table 2: Initial factor definitions.

X1, X2, X3. This network is shown in Fig. 1, which shows that three of the variables (X1, X2, X3) have observed values, with the other three (Y₁, Y₂, Ys) being unobserved. Your goal in this question is use variable elimination to compute p(Y3 = END| X₁ = a, X2 = b, X3 = #). All of these values are described below. D(Y, Y) DIY, Y₂) Y₁ Y, P(X, Y) P(X,-bY₂) X-b P(X,Y) Figure 1: Bayesian network for question 1 Assume that the Y; variables are all defined over three different values (Z1, Z2, and END), where the same conditional distribution p₁(YY-1) is used to define the distribution of Yi, given its parent Yi-1 (where there's a prior distribution over Y₁). Further assume that the Xi variables are all defined over three different values (a, b, and #), where the same conditional distribution pe(XiY;) is used to define the distribution of X; given its parent Yi. These distributions are labeled (for ease) on the arcs of the figure, and are provided below in Table 1 (in both sub-tables, each row is a different distribution): a Z₁ X₁ = b # 7.30 Y₁ = Z₁ Z.15 Z₂ END .8 .05 Z₁ Y₁-1= Y₁₁ = Z2 2.8 0 Z₂ .6 END 0 0 35 .05 .7 Y₁ = Z₂ END 2.1 1 END 0 0 1 (a) The definition of pe, for 1 (b) The definition of p₂, for i = i≤3. 2.3. (c) The prior distribution for Y₁. Table 1: The distributions needed for question 1. In order to compute p(Y3 = END X₁ = a, X2 = b, X3 = #), you'll need to eliminate both Y₁ and Y2. To make the computations a bit easier, we'll start with Y, and then eliminiate to Y2. Assume that the factors have been defined as given in Table 2: Factor e₁(Y₁) Definition Pe(X₁ = a❘Y₁) €2(Y2) Pe(X2=b|Y₂) es(Ys) Pe(X3 = #|Y3) ti (Yi) t₂(Y₁, Y₂) t3(Y2,Y3) P(Y₁) Pt(Y2Y1) Pi (Ya Y2) Table 2: Initial factor definitions.

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Based on the attach please help sovle a , b and c. I will give you a thumb up if you can help with this and please provided the answer as handwritten on paper. Thank you

(a) Eliminate Y1. Name the resulting factor as f.
(i) Identify the factors that must be considered to eliminate Y1.
(ii) Identify the variable(s) that f will be defined over.
(iii) Write the formula for f (in terms of the factors you’ve identified).
(b) Eliminate Y2. Name the resulting factor as g.
(i) Identify the factors that must be considered to eliminate Y2.
(ii) Identify the variable(s) that g will be defined over.
(iii) Write the formula for g (in terms of the factors you’ve identified).
(c) Answer the original question: what is the posterior distribution p(Y3 = END|X1 =
a, X2 = b, X3 = #) in terms of factors, after the variables Y1 and Y2 are eliminated.
For example, in the first example in the class lecture, the final posterior distribution
was:
P(Tampering|Smoke = true ∧ Report = true) = f0(Tampering) × f8(Tampering)
P
Tampering=v
f0(v) × f8(v)
or
P(Tampering|Smoke = true ∧ Report = true) = α × f0(Tampering) × f8(Tampering)
In this assignment, you have been asked for a specific value of the query variable. So
if I wanted the posterior probability of ‘done’ for tempering, the equation would be:
P(Tampering = done|Smoke = true ∧ Report = true) =
α × f0(Tampering = done) × f8(Tampering = done)

X1, X2, X3. This network is shown in Fig. 1, which shows that three of the variables (X1, X2, X3) have observed values, with the other three (Y₁, Y₂, Ys) being unobserved. Your goal in this question is use variable elimination to compute p(Y3 = END| X₁ = a, X2 = b, X3 = #). All of these values are described below. D(Y, Y) DIY, Y₂) Y₁ Y, P(X, Y) P(X,-bY₂) X-b P(X,Y) Figure 1: Bayesian network for question 1 Assume that the Y; variables are all defined over three different values (Z1, Z2, and END), where the same conditional distribution p₁(YY-1) is used to define the distribution of Yi, given its parent Yi-1 (where there's a prior distribution over Y₁). Further assume that the Xi variables are all defined over three different values (a, b, and #), where the same conditional distribution pe(XiY;) is used to define the distribution of X; given its parent Yi. These distributions are labeled (for ease) on the arcs of the figure, and are provided below in Table 1 (in both sub-tables, each row is a different distribution): a Z₁ X₁ = b # 7.30 Y₁ = Z₁ Z.15 Z₂ END .8 .05 Z₁ Y₁-1= Y₁₁ = Z2 2.8 0 Z₂ .6 END 0 0 35 .05 .7 Y₁ = Z₂ END 2.1 1 END 0 0 1 (a) The definition of pe, for 1 (b) The definition of p₂, for i = i≤3. 2.3. (c) The prior distribution for Y₁. Table 1: The distributions needed for question 1. In order to compute p(Y3 = END X₁ = a, X2 = b, X3 = #), you'll need to eliminate both Y₁ and Y2. To make the computations a bit easier, we'll start with Y, and then eliminiate to Y2. Assume that the factors have been defined as given in Table 2: Factor e₁(Y₁) Definition Pe(X₁ = a❘Y₁) €2(Y2) Pe(X2=b|Y₂) es(Ys) Pe(X3 = #|Y3) ti (Yi) t₂(Y₁, Y₂) t3(Y2,Y3) P(Y₁) Pt(Y2Y1) Pi (Ya Y2) Table 2: Initial factor definitions.
Transcribed Image Text:X1, X2, X3. This network is shown in Fig. 1, which shows that three of the variables (X1, X2, X3) have observed values, with the other three (Y₁, Y₂, Ys) being unobserved. Your goal in this question is use variable elimination to compute p(Y3 = END| X₁ = a, X2 = b, X3 = #). All of these values are described below. D(Y, Y) DIY, Y₂) Y₁ Y, P(X, Y) P(X,-bY₂) X-b P(X,Y) Figure 1: Bayesian network for question 1 Assume that the Y; variables are all defined over three different values (Z1, Z2, and END), where the same conditional distribution p₁(YY-1) is used to define the distribution of Yi, given its parent Yi-1 (where there's a prior distribution over Y₁). Further assume that the Xi variables are all defined over three different values (a, b, and #), where the same conditional distribution pe(XiY;) is used to define the distribution of X; given its parent Yi. These distributions are labeled (for ease) on the arcs of the figure, and are provided below in Table 1 (in both sub-tables, each row is a different distribution): a Z₁ X₁ = b # 7.30 Y₁ = Z₁ Z.15 Z₂ END .8 .05 Z₁ Y₁-1= Y₁₁ = Z2 2.8 0 Z₂ .6 END 0 0 35 .05 .7 Y₁ = Z₂ END 2.1 1 END 0 0 1 (a) The definition of pe, for 1 (b) The definition of p₂, for i = i≤3. 2.3. (c) The prior distribution for Y₁. Table 1: The distributions needed for question 1. In order to compute p(Y3 = END X₁ = a, X2 = b, X3 = #), you'll need to eliminate both Y₁ and Y2. To make the computations a bit easier, we'll start with Y, and then eliminiate to Y2. Assume that the factors have been defined as given in Table 2: Factor e₁(Y₁) Definition Pe(X₁ = a❘Y₁) €2(Y2) Pe(X2=b|Y₂) es(Ys) Pe(X3 = #|Y3) ti (Yi) t₂(Y₁, Y₂) t3(Y2,Y3) P(Y₁) Pt(Y2Y1) Pi (Ya Y2) Table 2: Initial factor definitions.
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