document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1163566>Question 1163566</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Consider the following Bayesian graph: <BR>\n" );
document.write( "B←A→C<BR>\n" );
document.write( "The probabilities and conditional probabilities are given below:<BR>\n" );
document.write( "A    P(A)  ¦    A     B       P(B︳A)    ¦   A     B       P(C︳A)  <BR>\n" );
document.write( "f    0.2   ¦    f     f        0.8       ¦   f     f         0.3<BR>\n" );
document.write( "t    0.8   ¦    f     t        0.2       ¦   f     t         0.7<BR>\n" );
document.write( "           ¦    t     f        0.1       ¦   t     f         0.2<BR>\n" );
document.write( "           ¦    t     t        0.9       ¦   t     t         0.8<BR>\n" );
document.write( "a) Prove that B and C are conditionally independent given A<BR>\n" );
document.write( "b) Prove that B and C are not unconditionally dependent. \r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #854373 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2264)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=854373\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To address these proofs, we first note that the graph $B \leftarrow A \rightarrow C$ is a **diverging connection**. In Bayesian networks, the joint probability distribution for this graph is factorized as:<BR>\n" );
document.write( "$$P(A, B, C) = P(A)P(B|A)P(C|A)$$\r<BR>\n" );
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document.write( "### a) Prove that $B$ and $C$ are conditionally independent given $A$\r<BR>\n" );
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document.write( "By definition, $B$ and $C$ are conditionally independent given $A$ if:<BR>\n" );
document.write( "$$P(B, C | A) = P(B|A)P(C|A)$$\r<BR>\n" );
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document.write( "**Proof:**<BR>\n" );
document.write( "Using the definition of conditional probability:<BR>\n" );
document.write( "$$P(B, C | A) = \frac{P(A, B, C)}{P(A)}$$\r<BR>\n" );
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document.write( "Substitute the factorization of the joint distribution from the graph structure:<BR>\n" );
document.write( "$$P(B, C | A) = \frac{P(A)P(B|A)P(C|A)}{P(A)}$$\r<BR>\n" );
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document.write( "The $P(A)$ terms cancel out, leaving:<BR>\n" );
document.write( "$$P(B, C | A) = P(B|A)P(C|A)$$<BR>\n" );
document.write( "This equality holds for all values of $A, B,$ and $C$. Thus, **$B$ and $C$ are conditionally independent given $A$**.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### b) Prove that $B$ and $C$ are not unconditionally independent\r<BR>\n" );
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document.write( "*Note: The prompt asks to prove they are \"not unconditionally dependent,\" but the standard property of a diverging graph is that they are **not unconditionally independent** (meaning they are dependent). I will prove they are dependent by showing $P(B, C) \neq P(B)P(C)$.*\r<BR>\n" );
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document.write( "**1. Calculate $P(B=t)$:**<BR>\n" );
document.write( "Using the law of total probability:<BR>\n" );
document.write( "$$P(B=t) = P(B=t|A=t)P(A=t) + P(B=t|A=f)P(A=f)$$<BR>\n" );
document.write( "$$P(B=t) = (0.9)(0.8) + (0.2)(0.2) = 0.72 + 0.04 = \mathbf{0.76}$$\r<BR>\n" );
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document.write( "**2. Calculate $P(C=t)$:**<BR>\n" );
document.write( "$$P(C=t) = P(C=t|A=t)P(A=t) + P(C=t|A=f)P(A=f)$$<BR>\n" );
document.write( "$$P(C=t) = (0.8)(0.8) + (0.7)(0.2) = 0.64 + 0.14 = \mathbf{0.78}$$\r<BR>\n" );
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document.write( "**3. Calculate $P(B=t, C=t)$:**<BR>\n" );
document.write( "$$P(B=t, C=t) = \sum_{a \in \{t,f\}} P(B=t|A=a)P(C=t|A=a)P(A=a)$$<BR>\n" );
document.write( "$$P(B=t, C=t) = (0.9 \times 0.8 \times 0.8) + (0.2 \times 0.7 \times 0.2)$$<BR>\n" );
document.write( "$$P(B=t, C=t) = 0.576 + 0.028 = \mathbf{0.604}$$\r<BR>\n" );
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document.write( "**4. Compare $P(B, C)$ to $P(B)P(C)$:**<BR>\n" );
document.write( "$$P(B=t)P(C=t) = 0.76 \times 0.78 = \mathbf{0.5928}$$\r<BR>\n" );
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document.write( "Since $0.604 \neq 0.5928$, the joint probability is not equal to the product of the marginal probabilities. Therefore, **$B$ and $C$ are unconditionally dependent**. Knowing the state of $B$ provides information about the likely state of $A$, which in turn provides information about the likely state of $C$.\r<BR>\n" );
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document.write( "How would the relationship between B and C change if the arrow between A and B was reversed? </SPAN>\n" );
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