Tests of Symmetry Based On the Cramer-Von Mises And Watson Statistics.
Tests of Symmetry Based On the Cramer-Von Mises And Watson Statistics.
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(%) Theorem 3.3.1 below shows that U and U have the same asymptotic null n n distribution. The proof is based on the corresponding proof given by 2 Watson [20] for y . Before stating this theorem, two lemymas necessary n for its proof will be established.
Lemma 3.3. 1: Let F denote the EDF based on a sample of size n from F e $ , n s and let -foo V = n^ f [f (y) + F (-y) - l] dF(y) . . (3.3.1) n -' Ln n J 38 Then, as n tends to infinity, V converges in law to a normal random n variable with mea...n zero and variance 1/3.
Proof : Clearly +«> n n °° ^i=l i=l ^ n -H" 1 = n~^ yf] dF(y) + / dF(y)) - n^ .Ay^. — J 1=1 ^1 n 1 F(-Xi) = n-^ y f / dt +/ dt] - n% n = n~^ y t ~ ^^^i^ + F(-X^)] - n^ i=l [1 - F(X^) + F(-X^)J - 1 " i=l Since F(-X.) = 1 - F(X.) if F e $ , the last expression above equals n^ [l - 2n ^ ^ F(X^)j . (3.3.2) i=l Now, since {F(X^) .FCX^) , . . . ,F(X^) } is distributed as a random sample from a uniform distribution on [0,1], the lemma follows from (3.3.2) and the Central Limit Theorem.
What to read after Tests of Symmetry Based On the Cramer-Von Mises And Watson Statistics.?
You can find similar books in the "Read Also" column, or choose other free books by Hill, David Lawrence, 1949- to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
(%) Theorem 3.3.1 below shows that U and U have the same asymptotic null n n distribution. The proof is based on the corresponding proof given by 2 Watson [20] for y . Before stating this theorem, two lemymas necessary n for its proof will be established.
Lemma 3.3. 1: Let F denote the EDF based on a sample of size n from F e $ , n s and let -foo V = n^ f [f (y) + F (-y) - l] dF(y) . . (3.3.1) n -' Ln n J 38 Then, as n tends to infinity, V converges in law to a normal random n variable with mea...n zero and variance 1/3.
Proof : Clearly +«> n n °° ^i=l i=l ^ n -H" 1 = n~^ yf] dF(y) + / dF(y)) - n^ .Ay^. — J 1=1 ^1 n 1 F(-Xi) = n-^ y f / dt +/ dt] - n% n = n~^ y t ~ ^^^i^ + F(-X^)] - n^ i=l [1 - F(X^) + F(-X^)J - 1 " i=l Since F(-X.) = 1 - F(X.) if F e $ , the last expression above equals n^ [l - 2n ^ ^ F(X^)j . (3.3.2) i=l Now, since {F(X^) .FCX^) , . . . ,F(X^) } is distributed as a random sample from a uniform distribution on [0,1], the lemma follows from (3.3.2) and the Central Limit Theorem.
What to read after Tests of Symmetry Based On the Cramer-Von Mises And Watson Statistics.?
You can find similar books in the "Read Also" column, or choose other free books by Hill, David Lawrence, 1949- to read onlineMoreLess
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