S330T208soln

# S330T208soln - STAT 330 TEST 2 SOLUTIONS 1( a ) The...

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Unformatted text preview: STAT 330 TEST 2 SOLUTIONS 1( a ) The marginal p.d.f. of Y is f 2 ( y ) = Z 1 &1 f ( x;y ) dy = Z 1 &1 f 2 ( y j x ) f 1 ( x ) dx by the Product Rule = Z y 1 (1 & x ) ¡ & (1 & x ) & & 1 dx = & Z y (1 & x ) & & 2 dx = & " & (1 & x ) & & 1 & & 1 j y # = & & & 1 h 1 & (1 & y ) & & 1 i ; < y < 1 ( b ) The conditional p.d.f. of X given Y = y is f 1 ( x j y ) = f ( x;y ) f 2 ( y ) = & (1 & x ) & & 2 & & & 1 h 1 & (1 & y ) & & 1 i = ( & & 1) (1 & x ) & & 2 h 1 & (1 & y ) & & 1 i ; < x < y < 1 : 1 2 : ( a ) E [ E ( Y j X )] = Z 1 &1 &Z 1 &1 yf ( y j x ) dy ¡ f 1 ( x ) dx = Z 1 &1 Z 1 &1 yf ( y j x ) f 1 ( x ) dxdy = Z 1 &1 Z 1 &1 yf ( x;y ) dxdy by Product Rule = Z 1 &1 y &Z 1 &1 f ( x; y ) dx ¡ dy = Z 1 &1 yf 2 ( y ) dy = E ( Y ) ( b ) E ( Y ) = E [ E ( Y j P )] = E ( nP ) since Y j P = p v BIN ( n; p ) = nE ( P ) = n ¢ & & + 2 £ : V ar ( Y ) = E [ V ar ( Y j P )] + V ar [ E ( Y j P )] = E [ nP (1 & P )] + V ar ( nP ) since Y j P = p v BIN ( n; p ) = n ¤ E ( P ) & E ¥ P 2 ¦§ + n 2 V ar...
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## This note was uploaded on 04/27/2010 for the course STAT 330 taught by Professor Paulasmith during the Fall '08 term at Waterloo.

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S330T208soln - STAT 330 TEST 2 SOLUTIONS 1( a ) The...

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