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Unformatted text preview: 100 PROPERTIES OF EXPECTATIONS Let X be a r.v. a,b be constants Then (a) ( ) ( ) b x E a b aX E + = + (b) ( ) ( ) X Var a b aX Var 2 = + If n X X X ...... , 2 1 are any n rvs, ( ) ( ) ( ) ( ) n 2 1 n 2 1 X E .... X E X E X ....... X X E + + + = + + + But if n are X ,..... X n 1 indep rvs then ( ) ( ) ( ) ( ) n 2 1 n 2 1 X Var .... X Var X Var X ..... X X Var + + + = + + + In particular if X,Y are independent ( ) ( ) ( ) ( ) Y Var X Var Y X Var Y X Var + = = + Please note : whether we add X and Y or subtract Y from X, we always must add their variances. If X,Y are two rvs, we define their covariance ( ) ( )( ) [ ] ( ) ( ) Y E , X E Where Y X E Y , X COV 2 1 2 1 = =   = Th. If X,Y are indep, ( ) ( ) ( ) ( ) Y , X COV and Y E X E XY E = = 101 Sample Mean Let n 2 1 X ..... X , X be n indep rv s each having the same mean and same variance . 2 We define n X ... X X X n 2 1 + + + = X is called the mean of the rvs . X ..... X n 1 Please note that X is also a rv. Theorem 1. ( ) = X E 2. ( ) . n X Var 2 = Proof (i) ( ) ( ) ( ) ( ) [ ] n 2 1 X E .... X E X E n 1 X E + + + = = & + + + = times n n ..... 1 (2) ( ) ( ) ( ) ( ) [ ] n 2 1 2 X Var .... X Var X Var n 1 X Var + + + = (as the variables are independent) n n n times n n 2 2 2 2 2 2 2 .. 1 = = & + + + = 102 Sample variance Let n 1 X ... X be n indep rv s each having the same mean and same variance 2 . Let n X X X X n 2 1 + + = be their sample mean. We define the sample variance as ( ) 2 i n 1 i 2 X X 1 n 1 S = & = Note 2 S is also a r.v. ( ) 2 2 = S E Proof. Read it on page 179. Simulation To simulate the values taken by a continuous r.v. X, we have to use the following theorem. Theorem Let X be a continuous r.v. with density f(x) and cumulative distribution function F(x). Let ) ( X F U = . Then U is a r.v. having uniform distribution on (0,1). In other words, U is a random number. Thus to simulate the value taken by X, we take a random no U from the table 7 (Now you must put a decimal point before the no) And solve for X, the equation ( ) U X F = 103 Example 24 Let X have uniform density on ( ) , . Simulate the values of X using the 3digit random numbers. 937, 133, 753, 503, .. Solution Since X has uniform density on ( ) , its density is ( ) & < < = elsewhere n x f 1 Thus the cumulative distribution function is ( ) & > < = x x x x F x 1 ( ) & & =   = X means...
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 Spring '11
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