hwk1_soln - IEOR E4706 FE (I) HWK Solution 1 1. a. E(Y |X)...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
IEOR E4706 FE (I) HWK Solution 1 1. a. E ( Y | X )= E ( X 2 + Z | X E ( X 2 | X )+ E ( Z | X X 2 . Note that E ( Y E ( E ( Y | X )) = E ( X 2 Var ( X )=1 . b. E ( XY E ( E ( | X )) = E ( X · X 2 E ( X 3 )=0 . c. Cov ( E ( ) E ( X ) E ( Y . Therefore, corr ( X,Y cov ( ) / ( σ X σ Y . 2. For the mean we have E [ S ( t )] = S (0) exp { μt } E [exp { σ W t } ]= S (0) exp { μt } e σ 2 t/ 2 , by using the moment generating function of normal distribution. For the variance we have f rst that E [( S ( t )) 2 S 2 (0) exp { 2 μt } E [exp { 2 σ W t } ] = S 2 (0) exp { 2 μt } e (2 σ ) 2 t/ 2 = S 2 (0) exp { 2 μt } e 2 σ 2 t , and then [ S ( t )] = E [( S ( t )) 2 ] ( E [ S ( t )]) 2 = S 2 (0) exp { 2 μt } e (2 σ ) 2 t/ 2 S 2 (0) ³ exp { μt
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online