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=== Question 1: Useful theorems in probability theory === | === Question 1: Useful theorems in probability theory === | ||

- | [50 points; 10 per sub‐problem] | + | [50 points; 10 per sub-problem] |

(Adapted from: Manning & Schuetze, p. 59, exercise 2.1) | (Adapted from: Manning & Schuetze, p. 59, exercise 2.1) | ||

Use the [[Set Theory Identities]] and [[Axioms of Probability Theory]] to prove each of the following five statements. Develop your proof first in terms of sets and then translate into probabilities; use set theoretic operations on sets and arithmetic operators on probabilities. Be sure to apply [[Proofs|good proof technique]]: justify each step in your proofs; set up your proofs in two-column format, with each step showing a statement on the left and a justification on the right. Remember that in order to invoke an axiom as justification, you must first satisfy the conditions / pre-requisites of the axiom. | Use the [[Set Theory Identities]] and [[Axioms of Probability Theory]] to prove each of the following five statements. Develop your proof first in terms of sets and then translate into probabilities; use set theoretic operations on sets and arithmetic operators on probabilities. Be sure to apply [[Proofs|good proof technique]]: justify each step in your proofs; set up your proofs in two-column format, with each step showing a statement on the left and a justification on the right. Remember that in order to invoke an axiom as justification, you must first satisfy the conditions / pre-requisites of the axiom. | ||

- | # <math>P(B - A) = P(B) - P(A \cap B)</math> | + | # $P(B - A) = P(B) - P(A \cap B)$ |

- | # <math>P(A \cup B) = P(A) + P(B) - P(A \cap B)</math> (the addition rule) | + | # $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ (the addition rule) |

#* Hint: refer to part result #1 as a part of your proof of part #2 | #* Hint: refer to part result #1 as a part of your proof of part #2 | ||

- | # <math>P(\varnothing) = 0</math> | + | # $P(\varnothing) = 0$ |

- | # <math>P(\neg A) = 1 - P(A)</math> | + | # $P(\neg A) = 1 - P(A)$ |

#* Hint: refer to part result #1 as a part of your proof of part #4 | #* Hint: refer to part result #1 as a part of your proof of part #4 | ||

- | # <math>A \subseteq B \Rightarrow P(A) \leq P(B)</math> | + | # $A \subseteq B \Rightarrow P(A) \leq P(B)$ |

- | #* Hint: let <math>C = B - A</math> (set difference). | + | #* Hint: let $C = B - A$ (set difference). |

=== Question 2: Joint Probability === | === Question 2: Joint Probability === | ||

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* Let <tt>is-abbreviation</tt> denote the event "this period indicates an abbreviation", and let <tt>three-letter-word</tt> denote "a period occurs after a three letter word" | * Let <tt>is-abbreviation</tt> denote the event "this period indicates an abbreviation", and let <tt>three-letter-word</tt> denote "a period occurs after a three letter word" | ||

* Assume the following probabilities: | * Assume the following probabilities: | ||

- | ** <math>P(</math><tt>is-abbreviation</tt><math> | </math><tt>three-letter-word</tt><math>) = 0.8</math> | + | ** $P($<tt>is-abbreviation</tt>$ | $<tt>three-letter-word</tt>$) = 0.8$ |

- | ** <math>P(</math><tt>three-letter-word</tt><math>) = 0.0003</math> | + | ** $P($<tt>three-letter-word</tt>$) = 0.0003$ |

=== Question 3: Independence === | === Question 3: Independence === | ||

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(Adapted from: Manning & Schuetze, p. 59, exercise 2.4) | (Adapted from: Manning & Schuetze, p. 59, exercise 2.4) | ||

- | Are <math>X</math> and <math>Y</math> as defined in the following table independently distributed? | + | Are $X$ and $Y$ as defined in the following table independently distributed? |

<table border=1 cellspacing=0> | <table border=1 cellspacing=0> | ||

<tr> | <tr> | ||

- | <td><math>x</math></td> | + | <td>$x$</td> |

<td>0</td> | <td>0</td> | ||

<td>0</td> | <td>0</td> | ||

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</tr> | </tr> | ||

<tr> | <tr> | ||

- | <td><math>y</math></td> | + | <td>$y$</td> |

<td>0</td> | <td>0</td> | ||

<td>1</td> | <td>1</td> | ||

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</tr> | </tr> | ||

<tr> | <tr> | ||

- | <td><math>P(X=x, Y=y)</math></td> | + | <td>$P(X=x, Y=y)$</td> |

<td>0.32</td> | <td>0.32</td> | ||

<td>0.08</td> | <td>0.08</td> | ||

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[10 points] | [10 points] | ||

- | # How many possible ways can you completely factor the joint distribution <math>P(X_1, X_2, X_3, X_4, X_5, X_6)</math>? | + | # How many possible ways can you completely factor the joint distribution $P(X_1, X_2, X_3, X_4, X_5, X_6)$? |

- | # For some arbitrary joint probability distribution on six random variables <math>P(X_1, X_2, X_3, X_4, X_5, X_6)</math>, apply the chain rule to completely factor this distribution in one way. | + | # For some arbitrary joint probability distribution on six random variables $P(X_1, X_2, X_3, X_4, X_5, X_6)$, apply the chain rule to completely factor this distribution in one way. |

=== Question 5: Conditional Probability === | === Question 5: Conditional Probability === | ||

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[10 points] | [10 points] | ||

- | Prove the chain rule, namely that <math>P(\cap^n_{i=1} A_i) = P(A_1)\cdot P(A_2|A_1)\cdot P(A_3|A_1\cap A_2)\cdot ...\cdot P(A_n | \cap^{n-1}_{i=1} A_i)</math>. Justify each step. Use the same standard of proof as in problem #1 above. | + | Prove the chain rule, namely that $P(\cap^n_{i=1} A_i) = P(A_1)\cdot P(A_2|A_1)\cdot P(A_3|A_1\cap A_2)\cdot ...\cdot P(A_n | \cap^{n-1}_{i=1} A_i)$. Justify each step. Use the same standard of proof as in problem #1 above. |

* Hint: you could use induction (but that isn't the only way). | * Hint: you could use induction (but that isn't the only way). | ||