Probability & Statistics Professor Wei Zhu July 23rd (1) Let’s have some fun: A miner is trapped! A miner is trapped in a mine containing 3 doors. • The 1st door leads to a tunnel that will take him to safety after 3 hours. • The 2nd door leads to a tunnel that returns him to the mine after 5 hours. • The 3rd door leads to a tunnel that returns him to the mine after 7 hours. At all times, he is equally likely to choose any one of the doors. E(time to reach safety) ? Theorem (Law of Total Expectation): E(X) = EY (EX|Y [X|Y]) Exercise: Prove this theorem for the situation when X and Y are both discrete random variables. Special Case: If A1 , A2 , ⋯ , Ak is a partition of the whole outcome space (*that is, these events are mutually exclusive and exhaustive), then: k E(X) = ∑ E(X|Ai ) P(Ai ) i=1 (2) *Review: MGF, its second function: The m.g.f. will generate the moments Moment: 1st (population) moment: () = ∫ ∙ () 2nd (population) moment: ( 2 ) = ∫ 2 ∙ () … Kth (population) moment: ( ) = ∫ ∙ () Take the Kth derivative of the () with respect to t, and the set t = 0, we obtain the Kth moment of X as follows: ()| = () =0 2 ()| = ( 2 ) 2 =0 … ()| = ( ) =0 Note: The above general rules can be easily proven using calculus. Exercise: Prove the above general relationships. 1 Example (proof of a special case): When ~(, 2 ) , we want to verify the above equations for k=1 & k=2. ()| =0 1 2 2 = ( +2 ) ∙ ( + 2 ) (using the chain rule) So when t=0 ()| = = () =0 2 () = [ ()] 2 1 2 2 = [( +2 ) ∙ ( + 2 )] (using the result of the Product Rule) 1 2 2 1 2 2 = ( +2 ) ∙ ( + 2 )2 + ( +2 ) ∙ 2 2 ()| = 2 + 2 = ( 2 ) 2 =0 Considering 2 = Var(X) = E( 2 ) − 2 *(3). Review: Joint distribution, and independence Definition. The joint cdf of two random variables X and Y are defined as: , (, ) = (, ) = ( ≤ , ≤ ) Definition. The joint pdf of two discrete random variables X and Y are defined as: , (, ) = (, ) = ( = , = ) Definition. The joint pdf of two continuous random variables X and Y are defined as: 2 , (, ) = (, ) = (, ) Definition. The marginal pdf of the discrete random variable X or Y can be obtained by summation of their joint pdf as the following: () = ∑ (, ) ; () = ∑ (, ) ; Definition. The marginal pdf of the continuous random variable X or Y can be ∞ obtained by integration of the joint pdf as the following: () = ∫−∞ (, ) ; ∞ () = ∫−∞ (, ) ; Definition. The conditional pdf of a random variable X or Y is defined as: 2 (, ) (, ) ; (|) = () () Definition. The joint moment generating function of two random variables X and Y is defined as , (1 , 2 ) = ( 1 +2 ) Note that we can obtain the marginal mgf for X or Y as follows: (1 ) = , (1 , 0) = ( 1 +0∗ ) = ( 1 ); (2 ) = , (0, 2 ) = ( 0∗+2 ∗ ) = ( 2 ∗ ) (|) = Theorem. Two random variables X and Y are independent ⇔ (if and only if) , (, ) = () () ⇔ , (, ) = () () ⇔ , (1 , 2 ) = (1 ) (2 ) Definition. The covariance of two random variables X and Y is defined as (, ) = [( − )( − )]. Theorem. If two random variables X and Y are independent, then we have (, ) = 0. (*Note: However, (, ) = 0 does not necessarily mean that X and Y are independent.) (4) *Definitions: population correlation & sample correlation Definition: The population (Pearson Product Moment) correlation coefficient ρ is defined as: (, ) = √() ∗ () Definition: Let (X1 , Y1), …, (Xn , Yn) be a random sample from a given bivariate population, then the sample (Pearson Product Moment) correlation coefficient r is defined as: = ∑( − ̅)( − ̅) √[∑( − ̅)2 ][∑( − ̅)2 ] 3 Left: Karl Pearson FRS[1] (/ˈpɪərsən/; originally named Carl; 27 March 1857 – 27 April 1936[2]) was an influential English mathematician and biometrician. Right: Sir Francis Galton, FRS (/ˈfrɑːnsɪs ˈɡɔːltən/; 16 February 1822 – 17 January 1911) was a British Scientist and Statistician. (5) *Definition: Bivariate Normal Random Variable (, )~( , 2 ; , 2 ; ) where is the correlation between & The joint p.d.f. of (, ) is 1 1 − 2 − − , (, ) = exp {− [( ) − 2 ( )( ) 2 2(1 − ) 2 √1 − 2 − 2 +( ) ]} Exercise: Please derive the mgf of the bivariate normal distribution. Q5. Let X and Y be random variables with joint pdf 1 1 − 2 − − , (, ) = exp {− [( ) − 2 ( )( ) 2 2(1 − ) 2 √1 − 2 − 2 +( ) ]} Where −∞ < < ∞, −∞ < < ∞. Then X and Y are said to have the bivariate normal distribution. The joint moment generating function for X and Y is 1 (1 , 2 ) = exp [1 + 2 + (12 2 + 21 2 + 22 2 )] 2 . (a) Find the marginal pdf’s of X and Y; (b) Prove that X and Y are independent if and only if ρ = 0. 4 (Here ρ is indeed, the <population> correlation coefficient between X and Y.) (c) Find the distribution of( + ). (d) Find the conditional pdf of f(x|y), and f(y|x) Solution: (a) The moment generating function of X can be given by 1 () = (, 0) = [ + 2 2 ]. 2 Similarly, the moment generating function of Y can be given by 1 () = (, 0) = [ + 2 2 ]. 2 Thus, X and Y are both marginally normal distributed, i.e., ~( , 2 ), and ~( , 2 ). The pdf of X is () = The pdf of Y is () = 1 √2 1 √2 [− [− ( − )2 22 ( − )2 22 ]. ]. (b) If = 0, then 1 (1 , 2 ) = exp [ 1 + 2 + (2 12 + 2 22 )] = (1 , 0) ∙ (0, 2 ) 2 Therefore, X and Y are independent. If X and Y are independent, then 1 (1 , 2 ) = (1 , 0) ∙ (0, 2 ) = exp [ 1 + 2 + (2 12 + 2 22 )] 2 1 2 2 = exp [ 1 + 2 + ( 1 + 2 1 2 + 2 22 )] 2 Therefore, = 0 (c) + () = [ (+) ] = [ + ] Recall that (1 , 2 ) = [ 1 +2 ], therefore we can obtain + ()by 1 = 2 = in (1 , 2 ) That is, 5 1 + () = (, ) = exp [ + + (2 2 + 2 2 + 2 2 )] 2 1 2 = exp [( + ) + ( + 2 + 2 ) 2 ] 2 ∴ + ~( = + , 2 = 2 + 2 + 2 ) (d) The conditional distribution of X given Y=y is given by (|) = (, ) 1 = − () √2 √1 − 2 2 ( − − ( − )) { Similarly, we have the conditional distribution of Y given X=x is (|) = (, ) 1 = − () √2 √1 − 2 } 2 ( − − ( − )) 2(1 − 2 )2 { Therefore: . 2(1 − 2 )2 . } ( − ), (1 − 2 )2 ) | = ~ ( + ( − ), (1 − 2 )2 ) | = ~ ( + Exercise: 1. Linear transformation : Let X ~ N ( , 2 ) and Y a X b , where a&b are constants, what is the distribution of Y? 2. The Z-Score Distribution: Let X ~ N ( , 2 ) , and Z X a 1 ,b What is the distribution of Z? 3. Distribution of the Sample Mean: If X 1 , X 2 , i .i . d . , X n ~ N ( , 2 ) , prove that X ~ N ( , 2 n ). 6 4. Some musing over the weekend: We know from the bivariate normal distribution that if X and Y follow a joint bivariate nornmal distribution, then each variable (X or Y) is univariate normal, and furthermore, their sum, (X+Y) also follows a univariate normal distribution. Now my question is, do you think the sum of any two (univariate) normal random variables (*even for those who do not have a joint bivriate normal distribution), would always follow a univariate normal distribution? Prove you claim if you answer is Yes, and provide at least one counter example if your answer is no. Exercise -- Solutions: 1. Linear transformation : Let X ~ N ( , 2 ) and Y a X b , where a&b are constants, what is the distribution of Y? Solution: M Y (t ) E (e tY ) E[e t ( aX b ) ] E (e atX bt ) E (e atX e bt ) e E (e bt atX ) e e bt at a 2 2t 2 2 exp[( a b)t a 2 2 t 2 ] 2 Thus, Y ~ N (a b, a 2 2 ) 2. Distribution of the Z-score: Let X ~ N ( , 2 ) , and Z X a 1 ,b What is the distribution of Z? Solution (1), the mgf approach: M Z (t ) e t 1 1 1 1 t ( t ) 2 ( t )2 t2 1 M X ( t) e e 2 e 2 → m.g.f. for N (0,1) Thus, Z X ~ N (0,1) 7 Now with one standard normal table, we will be able to calculate the probability of any normal random variable: P ( a X b) P ( P( a a X Z b b ) ) Solution (2), the c.d.f. approach: FZ ( z ) P( Z z ) P( X z) P( X z ) FX ( z ) f Z ( z) d d FZ ( z ) FX ( z ) f X ( z ) dz dz 1 e 2 ( z ) 2 2 2 z2 1 2 e → the p.d.f. for N(0,1) 2 Solution (3), the p.d.f. approach: 1 f X ( x) 2 e ( x )2 2 2 x z Let the Jacobian be J: J dx d ( z ) dz dz 1 f z ( z ) | J | f x ( x) e 2 Z X ( x )2 2 2 1 e 2 X ~ N ( , n 2 1 z2 e 2 ~ N (0,1) 3. Distribution of the Sample Mean: If X 1 , X 2 , 2 ( z )2 2 2 i .i . d . , X n ~ N ( , 2 ) , then ). Solution: M X (t ) E (e tX ) E (e t X 1 X 2 X n n ) 8 = E (et * ( X1 X n ) ), where t * t / n, M X1 X n (t * ) M X1 (t * )M X n (t * ) (e e X ~ N ( , 1 2 t * 2t *2 )n t 1 t n n 2 ( ) 2 n 2 n e t 1 2 2 t 2 n 2 n ) 9

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