MATH 20: PROBABILITY Variance of Discrete Random Variables - PowerPoint PPT Presentation

MATH 20: PROBABILITY Variance of Discrete Random Variables Xingru Chen xingru.chen.gr@dartmouth.edu XC 2020

Important Distributions Hypergeometric Discrete Uniform Distribution Distribution 𝑛 𝜕 = 1 $ &"$ % !"% ℎ 𝑂, 𝑙, 𝑜, 𝑦 = 𝑜 & ! Negative Binomial Binomial Distribution Distribution 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑣 𝑦, 𝑙, 𝑞 𝑙 𝑞 $ 𝑟 !"$ = 𝑦 − 1 𝑙 − 1 𝑞 $ 𝑟 %"$ Geometric Poisson Distribution Distribution 𝑄 𝑌 = 𝑙 = 𝜇 $ 𝑄 𝑈 = 𝑜 = 𝑟 !"# 𝑞 𝑙! 𝑓 "' XC 2020

Binomia Bin ial 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞 ! 𝑟 "#! 𝐹 𝑌 = 𝑜𝑞 Ge Geometric 𝐹 𝑌 = 1 𝑄 𝑈 = 𝑜 = 𝑟 "#$ 𝑞 𝑞 Po Poisson 𝑄 𝑌 = 𝑙 = 𝜇 ! 𝐹 𝑌 = 𝜇 𝑙! 𝑓 #% Negative Ne binomi omial 𝐹 𝑌 = 𝑙 𝑟 𝑣 𝑦, 𝑙, 𝑞 = 𝑦 − 1 𝑙 − 1 𝑞 ! 𝑟 &#! 𝑞 Hy Hyper ergeo eomet etric ic ! '#! 𝐹 𝑌 = 𝑜 𝑙 & "#& ℎ 𝑂, 𝑙, 𝑜, 𝑦 = ' 𝑂 " XC 2020

How Much Does a Hershey Kiss Weight? § A single standard Hershey's Kiss weighs 0.16 ounces. XC 2020

Same 𝜈 , different 𝜏 . Ho How t to m mea easu sure the t q qualit ity o of ? prod oduct cts? XC 2020

Home Made Versus Factory Made XC 2020

Variance of Discrete Random Variables § Let 𝑌 be a numerically-valued random variable with expected value 𝜈 = 𝐹(𝑌) . Then the variance of 𝑌 , denoted by 𝑊(𝑌) , is 𝑊 𝑌 = 𝐹((𝑌 − 𝜈) ( ) . Ex Expected value 𝑭(𝒀) Va Variance 𝑊 𝑌 : 𝑦𝑛(𝑦) %∈) (𝑦 − 𝜈) * 𝑛(𝑦) : %∈) Ex Expected value 𝑭 𝝔(𝒀) : 𝜚(𝑦)𝑛(𝑦) %∈) XC 2020

Variance of Discrete Random Variables § Let 𝑌 be a numerically-valued random variable with expected value 𝜈 = 𝐹(𝑌) . Then the variance of 𝑌 , denoted by 𝑊(𝑌) , is 𝑊 𝑌 = 𝐹((𝑌 − 𝜈) ( ) . § Standard deviation of 𝑌 , denoted by 𝐸(𝑌) , is 𝐸 𝑌 = 𝑊(𝑌) . 𝜏 ( for § We often write 𝜏 for 𝐸(𝑌) and 𝑊 𝑌 . Va Variance 𝑊 𝑌 (𝑦 − 𝜈) * 𝑛(𝑦) : %∈) XC 2020

Variance of Discrete Random Variables § Let 𝑌 be a numerically-valued random variable with expected value 𝜈 = 𝐹(𝑌) . Then the variance of 𝑌 , denoted by 𝑊(𝑌) , is Va Variance 𝑊 𝑌 𝑊 𝑌 = 𝐹((𝑌 − 𝜈) ( ) . (𝑦 − 𝜈) * 𝑛(𝑦) : %∈) § If 𝑌 is any random variable with 𝜈 = 𝐹(𝑌) , then 𝑦 * 𝑛(𝑦) − : : 𝑦𝑛(𝑦) 𝑊 𝑌 = 𝐹 𝑌 ( − 𝜈 ( . %∈) %∈) XC 2020

Proof Let 𝑌 be a numerically-valued random variable with expected value 𝜈 = 𝐹(𝑌) . (𝑦 − 𝜈) * 𝑛(𝑦) 𝑊 𝑌 = : (𝑦 − 𝜈) $ 𝑛(𝑦) 𝑊 𝑌 = $ %∈) !∈# (𝑦 * − 2𝜈𝑦 + 𝜈 * )𝑛(𝑦) = : %∈) 𝑦𝑛 𝑦 + 𝜈 * : 𝑦 * 𝑛(𝑦) − 2𝜈 : 𝜈 * 𝑛 𝑦 = : 𝑦 * 𝑛(𝑦) = 𝐹(𝑌 * ) : %∈) %∈) %∈) %∈) : 𝑦𝑛 𝑦 = 𝜈 𝑦𝑛 𝑦 + 𝜈 * : 𝑦 * 𝑛(𝑦) − 2𝜈 : 𝜈 * 𝑛 𝑦 %∈) 𝑊 𝑌 = : %∈) %∈) %∈) = 𝐹 𝑌 * − 2𝜈 * + 𝜈 * $ 𝑛 𝑦 = 1 = 𝐹 𝑌 * − 𝜈 * !∈# XC 2020

Example 1 Tos oss a coi coin head or tail 1 or 0 𝑛 𝑦 = 1 2 𝜈 = 𝐹 𝑌 = 1 2 𝜏 ( = 𝑊 𝑌 = ⋯ XC 2020

Example 1 Tos oss a coi coin head or tail 1 or 0 𝑛 𝑦 = 1 2 𝜈 = 𝐹 𝑌 = 1 2 𝜏 ( = 𝑊 𝑌 = 1 4 XC 2020

Example 2 Rol oll a dice ce 1, 2, 3, 4, 5, or 6 𝑛 𝑦 = 1 6 𝜈 = 𝐹 𝑌 = 7 2 𝜏 ( = 𝑊 𝑌 = ⋯ XC 2020

Example 2 Rol oll a dice ce 1, 2, 3, 4, 5, or 6 𝑛 𝑦 = 1 6 𝜈 = 𝐹 𝑌 = 7 2 𝜏 ( = 𝑊 𝑌 = 91 6 − 49 4 = 35 12 XC 2020

Quiz 9 What is the expected number of dice tosses needed to get two consecutive six's? Number of tosses 2, 3, 4, … 𝐹 𝑌 = 𝐹 𝑌 1 𝑄 1 + 𝐹 𝑌 2 + 𝐹 𝑌 3 𝑄 3 + 𝐹 𝑌 4 𝑄 4 + 𝐹 𝑌 5 𝑄 5 + 𝐹 𝑌 6 𝑄(6) 𝐹 𝑌 = 5 6 𝐹 𝑌 1 + 1 6 𝐹 𝑌 6 = 5 + 1 6 (5 6 𝐹 𝑌 61 + 1 6 1 + 𝐹 𝑌 6 𝐹(𝑌|66)) 𝐹 𝑌 = 5 + 1 6 [5 + 2 6 1 + 𝐹 𝑌 6 2 + 𝐹 𝑌 6] XC 2020

Is the distribution function a must for calculating expectation? 𝑭 𝒀 𝑮 𝟒 𝑸(𝑮 𝟒 ) 𝑭 𝒀 𝑮 𝟓 𝑸(𝑮 𝟓 ) 𝑭 𝒀 𝑮 𝟑 𝑸(𝑮 𝟑 ) 𝑭 𝒀 𝑮 𝟔 𝑸(𝑮 𝟔 ) 𝑭 𝒀 𝑮 𝟐 𝑸(𝑮 𝟐 ) 𝑭 𝒀 𝑮 𝟕 𝑸(𝑮 𝟕 ) 𝑭 𝒀 XC 2020

Example 3 Consider the general Bernoulli trial process. As usual, we let 𝑌 = 1 if the outcome is a success and 0 if it is a failure. Expect cted value 𝑭(𝒀) M 𝑦𝑛(𝑦) = 1×𝑞 + 0× 1 − 𝑞 = 𝑞 &∈0 Ber Bernoulli t tria ial 𝑞, 𝑌 = 1 𝑛 𝑦 = O Variance ce 𝑊 𝑌 1 − 𝑞, 𝑌 = 0 𝐹 𝑌 ( − 𝜈 ( = ⋯ XC 2020

Example 3 Consider the general Bernoulli trial process. As usual, we let 𝑌 = 1 if the outcome is a success and 0 if it is a failure. Expect cted value 𝑭(𝒀) M 𝑦𝑛(𝑦) = 1×𝑞 + 0× 1 − 𝑞 = 𝑞 &∈0 Ber Bernoulli t tria ial 𝑞, 𝑌 = 1 𝑛 𝑦 = O Variance ce 𝑊 𝑌 1 − 𝑞, 𝑌 = 0 𝐹 𝑌 ( − 𝜈 ( = 𝑞 − 𝑞 ( XC 2020

Bin Binomia ial 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝐹 𝑌 = 𝑜𝑞 , 𝑊 𝑌 = 𝑜𝑞𝑟 𝑙 𝑞 ! 𝑟 "#! Ge Geometric 𝐹 𝑌 = $ 𝑊 𝑌 = $#1 1 , 𝑄 𝑈 = 𝑜 = 𝑟 "#$ 𝑞 1 % Po Poisson 𝑄 𝑌 = 𝑙 = 𝜇 ! 𝐹 𝑌 = 𝜇 , 𝑊 𝑌 = 𝜇 𝑙! 𝑓 #% Negative Ne binomi omial 𝑣 𝑦, 𝑙, 𝑞 = 𝑦 − 1 𝐹 𝑌 = 𝑙 2 𝑊 𝑌 = 𝑙 2 1 , 𝑙 − 1 𝑞 ! 𝑟 &#! 1 % Hyper Hy ergeo eomet etric ic ! '#! 𝐹 𝑌 = 𝑜 ! 𝑊 𝑌 = "! '#! ('#") & "#& ' , ℎ 𝑂, 𝑙, 𝑜, 𝑦 = ' ' % ('#$) " XC 2020

Binomial Distribution and Poisson Distribution Bi Binomial Dist stribution 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞 $ 𝑟 !"$ 𝐹 𝑌 = 𝑜𝑞 , 𝑊 𝑌 = 𝑜𝑞𝑟 Poisson Po Distribution 𝑄 𝑌 = 𝑙 = 𝜇 $ 𝑙! 𝑓 "' 𝐹 𝑌 = 𝜇 , 𝑊 𝑌 = 𝜇 𝑞 = %5 " , 𝑢 = 1 , 𝑜 → ∞ , 𝑞 → 0 = XC 2020

Binomia Bin ial 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞 ! 𝑟 "#! 𝐹 𝑌 = 𝑜𝑞 Variance ce 𝑊 𝑌 𝐹 𝑌 ( − 𝜈 ( = M 𝑦 ( 𝑛 𝑦 − 𝜈 ( &∈0 " " " 𝑙 ( 𝑜 𝑙 ( 𝑜 𝑜! 𝑙 𝑞 ! 𝑟 "#! = M 𝑙 𝑞 ! 𝑟 "#! = M 𝑦 ( 𝑛 𝑦 = M 𝑙 ( 𝑙! 𝑜 − 𝑙 ! 𝑞 ! 𝑟 "#! M &∈0 !67 !6$ !6$ " " (𝑜 − 1)! (𝑙 − 1 + 1) 𝑜 − 1 (𝑙 − 1)! 𝑜 − 𝑙 ! 𝑞 !#$ 𝑟 "#! = 𝑜𝑞 M 𝑙 − 1 𝑞 !#$ 𝑟 "#! = M 𝑜𝑞𝑙 !6$ !6$ "#$ 𝑜 − 1 "#$ 𝑚 𝑜 − 1 𝑞 8 𝑟 "#$#8 + 𝑜𝑞 M 𝑞 8 𝑟 "#$#8 = 𝑜𝑞 M 𝑚 𝑚 867 867 XC 2020

Binomia Bin ial 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞 ! 𝑟 "#! 𝐹 𝑌 = 𝑜𝑞 Variance ce 𝑊 𝑌 𝐹 𝑌 ( − 𝜈 ( = M 𝑦 ( 𝑛 𝑦 − 𝜈 ( &∈0 ∑ &∈0 𝑦 ( 𝑛 𝑦 = 𝑜 𝑜 − 1 𝑞 ( ∑ 86$ 8#$ 𝑞 8#$ 𝑟 "#$#8 + 𝑜𝑞 = 𝑜 𝑜 − 1 𝑞 ( + 𝑜𝑞 "#$ "#( § 𝜈 ( = 𝑜 ( 𝑞 ( § ∑ &∈0 𝑦 ( 𝑛 𝑦 − 𝜈 ( = 𝑜 𝑜 − 1 𝑞 ( + 𝑜𝑞 − 𝑜 ( 𝑞 ( = 𝑜𝑞(1 − 𝑞) § XC 2020

Po Poisson 𝑄 𝑌 = 𝑙 = 𝜇 ! 𝐹 𝑌 = 𝜇 𝑙! 𝑓 #% Variance ce 𝑊 𝑌 𝐹 𝑌 ( − 𝜈 ( = M 𝑦 ( 𝑛 𝑦 − 𝜈 ( &∈0 9: 9: 𝑙 ( 𝜇 ! 𝑙 𝜇 ! 𝑙! 𝑓 #% = M 𝑦 ( 𝑛 𝑦 = M 𝑙! 𝑓 #% M &∈0 !67 !6$ 9: 9: 𝜇 !#$ 𝜇 !#$ (𝑙 − 1)! 𝑓 #% = M (𝑙 − 1)! 𝑓 #% = M 𝜇𝑙 𝜇(𝑙 − 1 + 1) !6$ !6$ 9: 𝜇 8 9: 𝑚 𝜇 8 𝑚! 𝑓 #% + 𝜇 M 𝑚! 𝑓 #% = 𝜇 M 867 867 XC 2020

Po Poisson 𝑄 𝑌 = 𝑙 = 𝜇 ! 𝐹 𝑌 = 𝜇 𝑙! 𝑓 #% Variance ce 𝑊 𝑌 𝐹 𝑌 ( − 𝜈 ( = M 𝑦 ( 𝑛 𝑦 − 𝜈 ( &∈0 9: 𝑚 % & 9: % & 9: % &'( 8! 𝑓 #% + 𝜇 ∑ 867 8! 𝑓 #% = 𝜇 ( ∑ 86$ (8#$)! 𝑓 #% + 𝜇 = 𝜇 ( + 𝜇 ∑ &∈0 𝑦 ( 𝑛 𝑦 = 𝜇 ∑ 867 § 𝜈 ( = 𝜇 ( § ∑ &∈0 𝑦 ( 𝑛 𝑦 − 𝜈 ( = 𝜇 ( + 𝜇 − 𝜇 ( = 𝜇 § XC 2020

Linearity 𝐹(𝑌 + 𝑍) = 𝐹(𝑌) + 𝐹(𝑍) 𝐹(𝑑𝑌) = 𝑑𝐹(𝑌) . 𝐹(𝑏𝑌 + 𝑐) = 𝑏𝐹(𝑌) + 𝑐 XC 2020

Properties of Variance § If 𝑌 is any random variable and 𝑑 is any constant, then 𝑊 𝑑𝑌 = 𝑑 ( 𝑊(𝑌) , 𝑊 𝑌 + 𝑑 = 𝑊(𝑌) . Variance ce 𝑊 𝑌 = 𝐹 𝑌 ( − 𝜈 ( = 𝐹 𝑌 ( − [𝐹 𝑌 ] ( 𝐹(𝑏𝑌 + 𝑐) = 𝑏𝐹(𝑌) + 𝑐 𝑊 𝑑𝑌 = 𝐹 (𝑑𝑌) ( − [𝐹 𝑑𝑌 ] ( XC 2020