Solution to Homework Four

Solution to Homework Four - MAFS 5030 - Quantitative...

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Unformatted text preview: MAFS 5030 - Quantitative Modeling of Derivatives Securities Solution to Homework Four 1. (a) It is easily seen that X 1 ( t + s )- X 1 ( s ) = k bracketleftbigg Z parenleftbigg t + s k 2 parenrightbigg- Z parenleftBig s k 2 parenrightBig bracketrightbigg is normally distributed with mean zero and variance k 2 ( t + s k 2- s k 2 ) = t . Also the increments X 1 ( t i +1 )- X 1 ( t i ) = k [ Z ( t i +1 /k 2 )- Z ( t i /k 2 )] over disjoint time intervals [ t i , t i +1 ] , i = 1 , 2 , ··· , n- 1 are independent and X 1 ( t ) is continuous at X 1 (0) = 0. Hence X 1 ( t ) is a standard Brownian motion. (b) Since Z ( 1 s )- Z ( 1 t + s ) and Z ( 1 t + s ) are independent, it follows that X 2 ( t + s )- X 2 ( s ) = tZ parenleftbigg 1 t + s parenrightbigg- s bracketleftbigg Z parenleftbigg 1 s parenrightbigg- Z parenleftbigg 1 t + s parenrightbiggbracketrightbigg is normally distributed with mean zero and variance t 2 t + s + s 2 ( 1 s- 1 s + t ) = t. To prove that the increments X 2 ( t i +1 )- X 2 ( t i ) over disjoint time intervals [ t i , t i +1 ] , i = 1 , 2 , ··· , n- 1 are independent, since they are normal, it suffices to show that cov( X 2 ( t i +1 )- X 2 ( t i ) , X 2 ( t j +1 )- X 2 ( t j )) = 0, for i < j . This follows from E [[ X 2 ( t i +1 )- X 2 ( t i )][ X 2 ( t j +1 )- X 2 ( t j )]] = t i +1 t j +1 E bracketleftbigg Z parenleftbigg 1 t i +1 parenrightbigg Z parenleftbigg 1 t j +1 parenrightbiggbracketrightbigg- t i +1 t j E bracketleftbigg Z parenleftbigg 1 t i +1 parenrightbigg Z parenleftbigg 1 t j parenrightbiggbracketrightbigg- t i t j +1 E bracketleftbigg Z parenleftbigg 1 t i parenrightbigg Z parenleftbigg 1 t j +1 parenrightbiggbracketrightbigg + t i t j E bracketleftbigg Z parenleftbigg 1 t i parenrightbigg Z parenleftbigg 1 t j parenrightbiggbracketrightbigg = t i +1 t j +1 1 t j +1- t i +1 t j 1 t j- t i t j +1 1 t j +1 + t i t j 1 t j = 0 . To show that X 2 ( t ) is continuous at t = 0, note that √ tZ ( 1 t ) is normally distributed with mean zero and unit variance, we establish E parenleftbigg lim t → + | X 2 ( t ) | parenrightbigg = lim t → + E | X 2 ( t ) | = lim t → + radicalbigg 2 t π integraldisplay ∞ xe- x 2 2 dx = lim t → + radicalbigg 2 t π = 0 , implying P [lim t → 0+ X 2 ( t ) = 0] = 1 . Hence X 2 ( t ) is a standard Brownian mo- tion....
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Solution to Homework Four - MAFS 5030 - Quantitative...

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