MATH 20: PROBABILITY Midterm 2 Xingru Chen - PowerPoint PPT Presentation

MATH 20: PROBABILITY Midterm 2 Xingru Chen xingru.chen.gr@dartmouth.edu XC 2020

Ex Exam How many hours you spend preparing for the exam? Wrapper Wr How many hours you spend on the exam? Content on the last two weeks. for midterm 2 Arrangements for the final. … XC 2020

Problem 1: True or False XC 2020

Can 𝑌 follow a continuous uniform ? distribution? XC 2020

Density Functions of Continuous Random Variable § Let 𝑌 be a continuous real-valued random variable. A density function for 𝑌 is a real-valued function 𝑔 that satis fi es 𝒛 " 𝑔 𝑦 𝑒𝑦 , for 𝑏, 𝑐 ∈ ℝ . 𝑄 𝑏 ≤ 𝑌 ≤ 𝑐 = ∫ ! § If 𝐹 is a subset of ℝ , then 𝑄 𝑦 ∈ 𝐹 = # 𝑔 𝑦 𝑒𝑦 . ∫ § In particular, if 𝐹 is an interval [𝑏, 𝑐] , the probability that the outcome of the experiment falls in 𝐹 is given by 𝑏 𝑐 𝒚 " 𝑔 𝑦 𝑒𝑦 . 𝑄 [𝑏, 𝑐] = ∫ ! XC 2020

Can 𝑌 follow a continuous uniform ? No distribution? 𝑛 𝑦 ? Can 𝑌 follow a discrete uniform distribution? = ⋯ XC 2020

𝑛 𝑦 = 0 ? Can 𝑌 follow a discrete uniform distribution? 𝑛 𝑦 > 0 XC 2020

$% ? Can 𝑌 follow a discrete uniform distribution? 𝑛 𝑦 = 0 , 0 = 0 !"# $% No 𝑛 𝑦 > 0 , 𝑑 = +∞ !"# XC 2020

Problem 1: True or False Discr crete variance ce 𝑊 𝑌 Co Continuous variance 𝑊 𝑌 𝐹 𝑌 & − 𝜈 & = 0 = $% 𝑦 − 𝜈 & 𝑔 𝑦 𝑒𝑦 𝑌 − 𝜈 & = , 0 = 6 (𝑦 − 𝜈) & 𝑛(𝑦) . = 𝐹 *% '∈) 𝑌 = 𝜈 XC 2020

Problem 2: Computation XC 2020

𝐹 𝑌 &#&# = 𝐹 &#&# (𝑌) ? XC 2020

The Product of Two Random Variables § Let 𝑌 and 𝑍 be independent real-valued continuous random variables with fi nite expected values. Then we have 𝐹(𝑌𝑍) = 𝐹(𝑌)𝐹(𝑍) . § More generally, for 𝑜 mutually independent random variables 𝑌 ! , we have 𝐹 𝑌 + 𝑌 & ⋯ 𝑌 , = 𝐹 𝑌 + 𝐹 𝑌 & ⋯ 𝐹(𝑌 , ) . 𝐹 𝑌 &#&# = 𝐹 &#&# (𝑌) ? XC 2020

? Are 𝑌 and 𝑌 independent? If 𝑌 is any random variable and 𝑑 is any constant, then § 𝑊 𝑑𝑌 = 𝑑 & 𝑊(𝑌) , 𝑊 𝑌 + 𝑑 = 𝑊(𝑌) . Let 𝑌 and 𝑍 be two independent random variables. Then § 𝑊(𝑌 + 𝑍) = 𝑊(𝑌) + 𝑊(𝑍) . 𝑊 2𝑌 = 4𝑊 𝑌 ? 𝑊 𝑌 + 𝑌 = 2𝑊 𝑌 ? XC 2020

Problem 2: Computation XC 2020

Moment: ! 𝐹 𝑌 , , where 𝑜 = 1, 2, 3, ⋯ XC 2020

Problem 3: Proof 𝑦 ! − ̅ 𝑦 = (𝑦 ! − 𝜈) + (𝜈 − ̅ 𝑦) XC 2020

𝐹 𝑡 & = 𝑜 − 1 𝜏 & 𝑜 How to redefine 𝑡 & , so 𝐹 𝑡 & = 𝜏 & ? that XC 2020

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Extreme: ! 𝑔 - 𝑦 = 0 𝑔 -- 𝑦 > 0 𝑔 -- 𝑦 < 0 ! ! Minimum: Maximum: XC 2020

Problem 4: Manipulation XC 2020

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Calvin At Ca Atkeson, , Max ax Te Telemaque v 𝑉 = 𝑏𝑌 + 𝑐 , 𝐹 𝑉 = 𝑏𝐹 𝑌 + 𝑐 , 𝑊 𝑉 = 𝑏 & 𝑊(𝑌) 𝐹 𝑉 = + 𝑊 𝑉 = + 𝑉 : uniform on [0, 1] , & , +& 𝑊 𝑌 = (/*.) ! 𝐹 𝑌 = .$/ 𝑌 : uniform on [𝑑, 𝑒] , & , +& XC 2020

Problem 5: Educational Attainment XC 2020

Which one to use? Bin Binomia ial d dist istrib ibutio ion 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞 ! 𝑟 "#! Po Poisson Distribution tw two parameter eters ! 𝑄 𝑌 = 𝑙 = 𝜇 ! 𝑙! 𝑓 #$ 𝒐 𝒒 on one parameter ! 𝝁 XC 2020

Which one to use? Bin Binomia ial d dist istrib ibutio ion 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞 ! 𝑟 "#! Po Poisson Distribution 𝒐 < +∞ ! 𝑄 𝑌 = 𝑙 = 𝜇 ! 𝑙! 𝑓 #$ 𝒐 → +∞ ! XC 2020

Which one to use? Bin Binomia ial d dist istrib ibutio ion 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞 ! 𝑟 "#! Poisson Po Distribution 𝒐 = 𝟑, 𝟔, ⋯ ! 𝑄 𝑌 = 𝑙 = 𝜇 ! 𝑙! 𝑓 #$ 𝒐 = 𝟔𝟏, 𝟐𝟏𝟏, ⋯ ! 𝒐𝒒 = 𝝁 = XC 2020

MATH 20 BABY PROBABILISTS When you cannot explain something: use Po Poisson di distribu bution (Poisson process)! XC 2020

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Problem 6: Cupid’s Arrow XC 2020

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𝑄(𝑍 > 𝑌) $% $% 6 6 𝑒𝑧𝑒𝑦 '"# 2"' XC 2020

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𝑄(𝑌 > 𝑍) $% ' 6 6 𝑒𝑧𝑒𝑦 '"# 2"# XC 2020

𝑄(𝑍 ≥ 𝑌 + 300) $% $% 6 6 𝑒𝑧𝑒𝑦 '"# 2"'$3## XC 2020

𝑄 𝑌 > 𝑍 = 2 ? Daphne will break free from the enchantment first? 7 1 1 golden arrow: 200 years 𝜇 . = = 2 500 200 200 + 1 1 7 500 1 lead arrow: 500 years 𝜇 / = 500 𝜇 4 𝑄 𝑌 > 𝑍 = 𝜇 5 + 𝜇 4 XC 2020

Apollo will break free from the enchantment first and ? Daphne has to wait another 300 years or more before her arrow wears off? 𝑄 𝑍 > 𝑌 ∩ 𝑍 ≥ 𝑌 + 300 𝑄 𝑍 ≥ 𝑌 + 300 = 5 7 𝑓 *3/7 XC 2020

1 𝜇 4 = 2 𝜇 5 = 5 𝜇 . = 𝑄 𝑌 > 𝑍 = 𝑄 𝑍 > 𝑌 = 200 𝜇 5 + 𝜇 4 7 𝜇 5 + 𝜇 4 7 1 𝜇 / = 500 𝑄 𝑍 ≥ 𝑌 + 300 = 5 7 𝑓 *3/7 𝑄 𝑍 > 300 = 𝑓 *3/7 𝑄 𝑍 > 𝑧 = 𝑓 *8 " 2 XC 2020

𝜇 5 = 5 𝑄 𝑍 > 300 = 𝑓 *3/7 𝑄 𝑍 > 𝑌 = 𝜇 5 + 𝜇 4 7 𝑄 𝑍 ≥ 𝑌 + 300 = 5 7 𝑓 *3/7 𝑄 𝑍 ≥ 𝑌 + 300 = 𝑄 𝑍 > 300 𝑄(𝑍 > 𝑌) 𝑄 𝑍 > 𝑌 + 300|𝑍 > 𝑌 𝑄(𝑍 > 𝑌) Memoryless Property 𝑄 𝑌 > 𝑠 + 𝑡|𝑌 > 𝑠 = 𝑄 𝑌 > 𝑡 XC 2020

Problem 7: Man with No Name: A fi stful of Nuts XC 2020

Conditional Expectation § If 𝐺 is any event and 𝑌 is a random variable with sample space Ω = {𝑦 + , 𝑦 & , ⋯ } , then the conditional expectation given 𝐺 is de fi ned by 𝐹 𝑌 𝐺 = ∑ 9 𝑦 9 𝑄(𝑌 = 𝑦 9 |𝐺) . § Let 𝑌 be a random variable with sample space Ω . If 𝐺 + , 𝐺 & , ⋯ , 𝐺 : are events such that 𝐺 ! ∩ 𝐺 9 = ∅ for 𝑗 ≠ 𝑘 and Ω =∪ 9 𝐺 9 , then 9 ) . 𝐹 𝑌 = ∑ 9 𝐹 𝑌 𝐺 9 𝑄(𝐺 XC 2020

Conditional Expectation § Conditional density joint density 5 $,& (7, 8) 5 & (8) . 𝑔 .|/ 𝑦 𝑧 = marginal density § Conditional expected value .|/ 𝑦 𝑧 𝑒𝑦 . 𝐹 𝑌 𝑍 = 𝑧 = ∫ 𝑦𝑔 § Expected value / 𝑧 𝑒𝑧 . 𝐹 𝑌 = ∫ 𝐹 𝑌 𝑍 = 𝑧 𝑔 XC 2020

EXAMPLE Farming Sim XC 2020

Example § A point 𝑍 is chosen at random from [0, 1] uniformly. A second point 𝑌 is then uniformly and randomly chosen from the interval [0, 𝑍] . Find the expected value for 𝑌 . 1 𝑧 0 5|4 𝑦 𝑧 = 𝑔 5,4 (𝑦, 𝑧) 𝑔 𝑔 4 (𝑧) 𝐹 𝑌 𝑍 = 𝑧 = 6 𝑦𝑔 5|4 𝑦 𝑧 𝑒𝑦 XC 2020

tw two Ge Geometric Di Distribu bution ! 𝑄 𝑈 = 𝑜 = 𝑟 ,*+ 𝑞 same sa me ! tr trial 2 fi first ! tr trial 1 independent in ! XC 2020

Geometric Ge Di Distribu bution 𝑄 𝑈 = 𝑜 = 𝑟 ,*+ 𝑞 Ge Geometric 𝐹 𝑌 = 1 𝑄 𝑈 = 𝑜 = 𝑟 ,*+ 𝑞 𝑞 𝑄 𝑂 = 𝑜 = 𝑦 , (1 − 𝑦) XC 2020

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… v 𝑄 𝑂 = 𝑜 𝑌 = 𝑦 = 𝑦 , (1 − 𝑦) + 𝐹 𝑂 = 𝑜 𝑌 = 𝑦 = 6 𝑦𝑄 𝑂 = 𝑜 𝑌 = 𝑦 𝑒𝑦 # $% 𝐹 𝑂 = 𝑜 𝑌 = 𝑦 = , 𝑜𝑄 𝑂 = 𝑜 𝑌 = 𝑦 ,"# XC 2020

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Problem 8: Man with No Name: Out of San Pecan XC 2020

Expectation of Functions of Random Variables § If 𝑌 is a real-valued random variable and if 𝜚: 𝑆 → 𝑆 is a continuous real-valued function with domain [𝑏, 𝑐] , then $% 𝜚(𝑦)𝑔 𝑦 𝑒𝑦 , 𝐹 𝜚(𝑌) = ∫ *% provided the integral exists. Discr crete expect cted value 𝑭 𝝔(𝒀) Con ontinuou ous expect cted value 𝑭 𝝔(𝒀) $% , 𝜚(𝑦)𝑛(𝑦) 6 𝜚(𝑦)𝑔 𝑦 𝑒𝑦 *% '∈) XC 2020

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