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STATSLIDE8 - SAMPLE STATISTICS A random sample x1 x2 xn...

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SAMPLE STATISTICS A random sample x 1 , x 2 , . . . , x n from a distribution f ( x ) is a set of independently and iden- tically variables with x i f ( x ) for all i . Their joint p.d.f is f ( x 1 , x 2 , . . . , x n ) = f ( x 1 ) f ( x 2 ) · · · f ( x 2 ) = n i =1 f ( x i ) . The sample moments provide estimates of the moments of f ( x ). We need to know how they are distributed The mean ¯ x of a random sample is an unbiased estimate of the population moment µ = E ( x ), since E x ) = E x i n = 1 n E ( x i ) = n n µ = µ. The variance of a sum of independent variables is the sum of their variances, since covariances are zero. Therefore V x ) = V x i n = 1 n 2 V ( x i ) = n n 2 σ 2 = σ 2 n . Observe that V x ) 0 as n → ∞ . Since E x ) = µ , the estimates become increas- ingly concentrated around the true population parameter. Such an estimate is said to be consistent. 1
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The sample variance is not an unbiased estimate of σ 2 = V ( x ), since E ( s 2 ) = E 1 n ( x i ¯ x ) 2 = E 1 n ( x i µ ) + ( µ ¯ x ) 2 = E 1 n ( x i µ ) 2 + 2( x i µ )( µ ¯ x ) + ( µ ¯ x ) 2 = V ( x ) 2 E { x µ ) 2 } + E { x µ ) 2 } = V ( x ) V x ) . Here, we have used the result that E 1 n ( x i µ )( µ ¯ x ) = E { ( µ ¯ x ) 2 } = V x ) . It follows that E ( s 2 ) = V ( x ) V x ) = σ 2 σ 2 n = σ 2 ( n 1) n . Therefore, s 2 is a biased estimator of the population variance. For an unbiased estimate, we should use ˆ σ 2 = s 2 n n 1 = ( x i ¯ x ) 2 n 1 . However, s 2 is still a consistent estimator, since E ( s 2 ) σ 2 as n → ∞ and also V ( s 2 ) 0. The value of V ( s 2 ) depends on the distribution of underlying population, which is often assumed to be a normal. 2
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Theorem. Let x 1 , x 2 , . . . , x n be a random sample from the normal population N ( µ, σ 2 ). Then, y = a i x i is normally distributed with E ( y ) = a i E ( x i ) = µ a i and V ( y ) = a 2 i V ( x i ) = σ 2 a 2 i . Any linear function of a set of normally distributed variables is normally distributed. If x i N ( µ, σ 2 ); i = 1 , . . . , n is a normal random sample then ¯ x N ( µ, σ 2 /n ) . Let µ = [ µ 1 , µ 2 , . . . , µ n ] = E ( x ) be the expected value of x = [ x 1 , x 2 , . . . , x n ] and let Σ = [ σ ij ; i, j = 1 , 2 , . . . , n ] be the variance–covariance matrix. If a = [ a 1 , a 2 , . . . , a n ] is a constant vector, then a x N ( a µ, a Σ a ) is a normally distributed with a mean of E ( a x ) = a µ = a i µ i and a variance of V ( ax ) = a Σ a = i j a i a j σ ij = i a 2 i σ ii + i j = i a i a j σ ij . 3
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Let ι = [1 , 1 , . . . , 1] . Then, if x = [ x 1 , x 2 , . . . , x n ] has x i N ( µ, σ 2 ) for all i , there is x N ( µι, σ 2 I n ), where µι = [ µ, µ, . . . , µ ] and I n is an identity matrix of order n . Writing this explicitly, we have x = x 1 x 2 .
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