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Random Vector Martina Litschmannová [email protected] EA 538 Random Vectors ο§ An k- dimensional random vector is a function X = (π1, β¦ , ππ )π that associates a vector of real numbers with each element is a random variable. For example: ο§ A semiconductor chip is divided into βMβ regions. For the random experiment of finding the number of defects and their locations, let ππ denote the number of defects in ith region. Then π΅ = (π1, β¦ , ππ)π is a discrete random vector. ο§ In a random experiment of selecting a studentβs name, let π»π = height of ith student in inches and ππ = weight of ith student in pounds. Then πΊ = (π»π, ππ )π is a continuous random vector. 2 β dimensional Random Vectors ο§ We're going to focus on 2-dimensional distributions (i.e. random vector consists only of two random variables) but higher dimensions (more than two variables) are also possible. Joint Probability Distribution ο§ Joint distribution for random vector π, π π defines the probability of events defined in terms of both X and Y. Joint cumulative distribution function for πΏ = π, π π is given by πΉ π₯, π¦ = π π < π₯ β§ π < π¦ = π π < π₯; π < π¦ . πΉ π₯, π¦ Joint cumulative distribution function Joint cumulative distribution function for πΏ = π, π π is given by πΉ π₯, π¦ = π π < π₯ β§ π < π¦ = π π < π₯; π < π¦ . Properties of joint CDF: 1. 2. 3. β π₯, π¦ β β2 : 0 β€ πΉ π₯, π¦ β€ 1, lim πΉ π₯, π¦ = lim πΉ π₯, π¦ = 0, π₯βββ lim π₯,π¦ β β,β π¦βββ πΉ π₯, π¦ = 1, 4. πΉ π₯, π¦ is nondecreasing in each variable, 5. πΉ π₯, π¦ is continous from the left in each variable. 6. π π β€ π < π, π β€ π < π = πΉ π, π β πΉ π, π β πΉ π, π + πΉ π, π Discrete Joint Probability Distributions The probability function, also known as the probability mass function for a joint probability distribution is defined such that: π π₯π , π¦π = π π = π₯π β§ π = π¦π = π π = π₯π ; π = π¦π . π π₯π , π¦π = P π = π₯π |π = π¦π β π π = π¦π = P π = π¦π |π = π₯π β π π = π₯π π1 π2 π π₯π , π¦π = 1, π=1 π=1 π1 β₯ 1, π2 β₯ 1 Discrete Joint Probability Distributions Probability Mass Function for a Joint Probability Distribution : π π₯π , π¦π = π π = π₯π β§ π = π¦π = π π = π₯π ; π = π¦π Properties of p.m.f.: 1. π π₯π , π¦π > 0 only for a finite or countable set of values π₯π , π¦π 2. 0 β€ π π₯π , π¦π β€ 1, π1 π2 3. π1 β₯ 1, π2 β₯ 1, π=1 π=1 π π₯π , π¦π = 1, 4. π π₯π , π¦π = P π = π₯π |π = π¦π β π π = π¦π = P π = π¦π |π = π₯π β π π = π₯π , 5. If X, Y are independent π π₯π , π¦π = P π = π₯π β π π = π¦π Table of joint probabilities X\Y x1 x2 y1 p(x1, y1) p(x2, y1) y2 p(x1, y2) p(x2, y2) xn1 p(xn1, y1) p(xn1, y2) ... ... ... ... ... yn2 p(x1, yn2) p(x2, yn2) p(xn1, yn2) 1. A random experiment consists of tossing coin (X) and flipping die (Y). Find probability mass function for a joint probability distribution of a random vector π, π . 1. A random experiment consists of tossing coin (X) and flipping die (Y). Find probability mass function for a joint probability distribution of a random vector π, π . X/Y 0 (head) 1 (tail) 1 1/12 1/12 2 1/12 1/12 3 1/12 1/12 4 1/12 1/12 5 1/12 1/12 6 1/12 1/12 1. A random experiment consists of tossing coin (X) and flipping die (Y). Find probability mass function for a joint probability distribution of a random vector π, π . X/Y 0 (head) 1 (tail) 1 1/12 1/12 2 1/12 1/12 3 1/12 1/12 4 1/12 1/12 5 1/12 1/12 6 1/12 1/12 1 control cell 2. Probability mass function for a joint probability distribution of a random vector π, π is given as: X/Y -1 0 1 -2 0,13 0 0,23 0 0,11 0,11 0,01 1 0,07 0 0 2 0 0,33 0,01 1 Find: a) π 0; 1 b) π 1; 0 c) π 3; 1 d) π π < 1,3; π < β0,6 e) π π < 1,3; π > β0,6 f) πΉ 0,7; 1,3 Continous Joint Probability Distributions πΉ π₯, π¦ = π¦ π₯ π ββ ββ π , π‘ ππ ππ‘, where π π₯, π¦ is Joint Probability Density Function. Properties of Joint Probability Density Function: 1. π π₯, π¦ β₯ 0, 2. β β π ββ ββ 3. If π2 πΉ π₯,π¦ ππ₯ππ¦ π₯, π¦ ππ₯ ππ¦ = 1, exists, pak π π₯, π¦ = 4. π π β€ π < π, π β€ π < π = π2 πΉ π₯,π¦ ππ₯ππ¦ π π π π π , π₯, π¦ ππ₯ ππ¦. 3. Find the constant c so that function π π₯, π¦ = π π₯+π¦ 0 π₯, π¦ β 0; 1 × 0; 1 π₯, π¦ β 0; 1 × 0; 1 can be a joint probability density function of a random vector π, π π . That the function π π₯, π¦ can be a joint probability density function of a random vector π, π π , it has be satisfy condition β β π ββ ββ 1 1 π 0 0 1 1 π 0 0 π₯, π¦ ππ₯ ππ¦ = 1. π₯ + π¦ ππ₯ ππ¦ = 1 π₯ + π¦ ππ₯ ππ¦ = π =π 1 π₯2 0 2 1 + π₯π¦ 1 1 + 0 2 0 ππ¦ = π π¦ ππ¦ = π 1 π¦ 2 1 π₯2 0 2 + 1 + π₯π¦ 1 π¦2 2 0 0 ππ¦ = =π β π=1 Marginal probability distributions Obtained by summing or integrating the joint probability distribution over the values of the other random variable. ο§ Discrete Random Vector ππ π₯π = π¦π π π₯π , π¦π , π β₯ 1, ππ π¦π = π₯π π π₯π , π¦π , π β₯ 1. ο§ Continous Random Vector ππ π₯ = ππ π¦ = β π ββ β π ββ π₯, π¦ ππ¦, π₯ β β, π₯, π¦ ππ₯ , π¦ β β. 4. A random experiment consists of tossing coin (X) and flipping die (Y). Probability mass function for a joint probability distribution of a random vector π, π is given as: X/Y 0 (head) 1 (tail) 1 1/12 1/12 2 1/12 1/12 3 1/12 1/12 4 1/12 1/12 5 1/12 1/12 6 1/12 1/12 1 Find Marginal Probability Mass Functions ππ π₯π and ππ π¦π . 4. A random experiment consists of tossing coin (X) and flipping die (Y). Probability mass function for a joint probability distribution of a random vector π, π is given as: X/Y 0 (head) 1 (tail) ππ π¦π 1 1/12 1/12 2/12 2 1/12 1/12 2/12 3 1/12 1/12 2/12 4 1/12 1/12 2/12 5 1/12 1/12 2/12 6 ππ π₯π 1/12 6/12 1/12 6/12 2/12 1 Find Marginal Probability Mass Functions ππ π₯π and ππ π¦π . 4. A random experiment consists of tossing coin (X) and flipping die (Y). Probability mass function for a joint probability distribution of a random vector π, π is given as: X/Y 0 (head) 1 (tail) ππ π¦π 1 1/12 1/12 2/12 2 1/12 1/12 2/12 3 1/12 1/12 2/12 4 1/12 1/12 2/12 5 1/12 1/12 2/12 6 ππ π₯π 1/12 6/12 1/12 6/12 2/12 1 Marginal Probability Mass Functions ππ π₯π and ππ π¦π . Y ππ π¦π X ππ π₯π 1 2/12 2 2/12 0 (head) 6/12 3 2/12 1 (tail) 6/12 4 2/12 5 2/12 6 2/12 4. A random experiment consists of tossing coin (X) and flipping die (Y). Probability mass function for a joint probability distribution of a random vector π, π is given as: X/Y 0 (head) 1 (tail) ππ π¦π 1 1/12 1/12 1/6 2 1/12 1/12 1/6 3 1/12 1/12 1/6 4 1/12 1/12 1/6 5 1/12 1/12 1/6 6 ππ π₯π 1/12 1/2 1/12 1/2 1/6 1 Marginal Probability Mass Functions ππ π₯π and ππ π¦π . Y ππ π¦π X ππ π₯π 1 1/6 2 1/6 0 (head) 1/2 3 1/6 1 (tail) 1/2 4 1/6 5 1/6 6 1/6 5. A certain farm produces two kinds of eggs on any given day; organic and non-organic. Let these two kinds of eggs be represented by the random variables X and Y respectively. Given that the joint probability density function of these variables is given by 2 π π₯; π¦ = 3 π₯ + 2π¦ 0 0 β€ π₯ β€ 1; 0 β€ π¦ β€ 1 πππ ππ€βπππ. Find: a) marginal density functions ππ π₯ and ππ π¦ . 5. A certain farm produces two kinds of eggs on any given day; organic and non-organic. Let these two kinds of eggs be represented by the random variables X and Y respectively. Given that the joint probability density function of these variables is given by 2 π π₯; π¦ = 3 π₯ + 2π¦ 0 Find: b) π π < 1 ;π 2 β€ 1 2 0 β€ π₯ β€ 1; 0 β€ π¦ β€ 1 πππ ππ€βπππ. 5. A certain farm produces two kinds of eggs on any given day; organic and non-organic. Let these two kinds of eggs be represented by the random variables X and Y respectively. Given that the joint probability density function of these variables is given by 2 π π₯; π¦ = 3 π₯ + 2π¦ 0 Find: c) π π < 1 ;π 2 > 1 4 0 β€ π₯ β€ 1; 0 β€ π¦ β€ 1 πππ ππ€βπππ. Conditional probability distributions Conditional Probability Distributions arise from joint probability distributions where by we need to know that probability of one event given that the other event has happened, and the random variables behind these events are joint. ο§ Discrete Random Vector π π₯π |π¦π π π₯π ,π¦π = π π¦ π π π π₯π ,π¦π π π¦π |π₯π = ππ π₯π , ππ π¦π β 0, , ππ π₯π β 0. ο§ Continous Random Vector π π₯|π¦ = π π¦|π₯ = π π₯,π¦ ππ π¦ π π₯,π¦ ππ π₯ , ππ π¦ β 0, , ππ π₯ β 0. 6. Joint and marginal probability distribution of a random vector π, π is given as: X/Y -1 0 1 ππ π¦π Find: a) π π = 0|π b) π π = 0|π c) π π = 0|π d) π π = 0|π e) π π₯|π¦ f) π π¦|π₯ =1 =1 =2 =2 -2 0,13 0 0,23 0,36 0 0,11 0,11 0,01 0,23 1 0,07 0 0 0,07 2 0 0,33 0,01 0,34 ππ π₯π 0,31 0,44 0,25 1 7. Joint probability density function of π, π is given by 2 π π₯; π¦ = 3 π₯ + 2π¦ 0 0 β€ π₯ β€ 1; 0 β€ π¦ β€ 1 Find: a) conditional density function π π¦|π₯ , π π¦|π₯ = π π₯, π¦ ππ π₯ 2 ππ π₯ = 3 π₯ + 1 0 0β€π₯β€1 πππ ππ€βπππ πππ ππ€βπππ. 7. Joint probability density function of π, π is given by 2 π π₯; π¦ = 3 π₯ + 2π¦ 0 0 β€ π₯ β€ 1; 0 β€ π¦ β€ 1 Find: b) conditional density function π π₯|π¦ , π π₯|π¦ = π π₯, π¦ ππ π¦ 1 ππ π¦ = 3 1 + 4π¦ 0 0β€π¦β€1 πππ ππ€βπππ πππ ππ€βπππ. Conditional expected value (expectation) Conditional expectation is the expected value of a real random variable with respect to a conditional probability distribution. Discrete random vector: ο§ πΈ π|π = π¦ = (π₯) π₯ β π π₯|π = π¦ ο§ πΈ π|π = π₯ = (π¦) π¦ β π π¦|π = π₯ Continous random vector: ο§ πΈ π|π = π¦ = ο§ πΈ π|π = π₯ = β π₯ ββ β π¦ ββ β π π₯|π = π¦ ππ₯ β π π¦|π = π₯ ππ¦ Conditional variance Conditional variance is the variance of a conditional probability distribution. π· π|π = π¦ = πΈ π β πΈ π|π = π¦ π· π|π = π₯ = πΈ π β πΈ π|π = π₯ 2 |π = π¦ = πΈ π 2 |π = π¦ β πΈ π|π = π¦ 2 2 |π = π₯ = πΈ π 2 |π = π₯ β πΈ π|π = π₯ 2 8. Joint and marginal probability distribution of a random vector π, π is given as: X/Y -1 0 1 ππ π¦π Find: a) πΈ π|π = 1 b) π· π|π = 1 -2 0,13 0 0,23 0,36 0 0,11 0,11 0,01 0,23 1 0,07 0 0 0,07 2 0 0,33 0,01 0,34 ππ π₯π 0,31 0,44 0,25 1 9. Joint probability density function of π, π is given by 2 π π₯; π¦ = 3 π₯ + 2π¦ 0 Find: a) E π|π = 0,5 , b) D π|π = 0,5 . 0 β€ π₯ β€ 1; 0 β€ π¦ β€ 1 πππ ππ€βπππ. Independence Two random variables X and Y are independent if ο§ Discrete Random Variables βπ, π: π π₯π , π¦π = ππ π₯π β ππ π¦π ο§ Continous Random Vector βπ₯, π¦: π π₯, π¦ = ππ π₯ β ππ π¦ . Measures of Linear Independence Covariance: πππ = πππ£ π, π = πΈ π β πΈπ π β πΈπ ο§ πππ β ββ, β Correlation coefficient: πππ πππ£ π, π = ππ ππ ο§ πππ β β1; 1 ο§ πππ is a scaled version of covariance = πΈ ππ β πΈπ β πΈπ Covariance Covariance: πππ = πππ£ π, π = πΈ π β πΈπ π β πΈπ ο§ πππ β ββ, β Covariance matrix: π·π πππ£ π, π πππ£ π, π π·π = πΈ ππ β πΈπ β πΈπ Correlation Correlation: πππ ο§ ο§ ο§ ο§ πππ πππ πππ πππ πππ£ π, π = ππ ππ β β1; 1 > 0 β¦ π, π are positively correlated < 0 β¦ π, π are negatively correlated = 0 β¦ π, π are uncorrelated Correlation matrix: 1 πππ πππ 1 Correlation π π, π =1,000 π π, π =0,967 π π, π π π, π = -1,000 π π, π =0,000 π π, π =0,934 =0,857 π π, π =-0,143 π π, π =0,608 10. Joint and marginal probability distribution of a random vector π, π is given as: X/Y -1 0 1 ππ π¦π -2 0,13 0 0,23 0,36 0 0,11 0,11 0,01 0,23 1 0,07 0 0 0,07 2 0 0,33 0,01 0,34 Find: a) πΈ π , πΈ π b) π· π , π· π c) π π , π π d) πππ£(π, π) e) π π, π f) Are random variable X, Y independent? g) Are random variable X, Y linear independent? ππ π₯π 0,31 0,44 0,25 1 11. Joint probability density function of π, π is given by 2 π π₯; π¦ = 3 π₯ + 2π¦ 0 0 β€ π₯ β€ 1; 0 β€ π¦ β€ 1 πππ ππ€βπππ. Marginal density functions are: 2 ππ π₯ = 3 π₯ + 1 0 0β€π₯β€1 , πππ ππ€βπππ 1 ππ π¦ = 3 1 + 4π¦ 0 Find: a) πΈ π , πΈ π b) π· π , π· π c) π π , π π d) πππ£(π, π) e) π π, π f) Are random variable X, Y independent? g) Are random variable X, Y linear independent? 0β€π¦β€1 πππ ππ€βπππ Study materials : ο§ http://homel.vsb.cz/~bri10/Teaching/Bris%20Prob%20&%20Stat.pdf (p. 64 - p.70)