# From the above we can obtain s 1 s 2 s 3 s 3 e r σ 2

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From the above, we can obtain: S 1 S 2 S 3 = ( S 0 ) 3 e ( r - σ 2 2 )6 e σ ( W 1 + W 2 + W 3 ) ( S 1 S 2 S 3 ) 1 3 = ( S 0 ) e ( r - σ 2 2 )2 e σ 3 ( W 1 + W 2 + W 3 ) E bracketleftBig ( ( S 1 S 2 S 3 ) 1 / 3 - K ) + bracketrightBig = E bracketleftbigg max parenleftbigg ( S 1 S 2 S 3 ) 1 3 - K, 0 parenrightbiggbracketrightbigg = E bracketleftbigg max parenleftbigg S 0 e ( r - σ 2 2 )2 e σ 3 ( W 1 + W 2 + W 3 ) - K, 0 parenrightbiggbracketrightbigg = E bracketleftbigg ( S 0 e ( r - σ 2 2 )2 e σ 3 ( W 1 + W 2 + W 3 ) - K ) 1 ( S 1 S 2 S 3 ) 1 3 >K bracketrightbigg = S 0 e 2 r - σ 2 E bracketleftbigg e σ 3 ( W 1 + W 2 + W 3 ) 1 ( S 1 S 2 S 3 ) 1 3 >K bracketrightbigg - Ke - r 3 E bracketleftbigg 1 ( S 1 S 2 S 3 ) 1 3 >K bracketrightbigg

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Now, σ 3 ( W 1 + W 2 + W 3 ) is distributed normally with mean 0 and variance: V ar [ σ 3 ( W 1 + W 2 + W 3 )] = σ 2 9 V ar [3 W 1 + 2( W 2 - W 1 ) + ( W 3 - W 2 )] = σ 2 9 (9 V ar [ W 1 ] + 4 V ar [ W 2 - W 1 ] + V ar [ W 3 - W 2 ]) = 14 σ 2 9 (2) Also, the condition ( S 1 S 2 S 3 ) 1 3 > K is equivalent to: S 0 e ( r - σ 2 2 )2 e σ 3 ( W 1 + W 2 + W 3 ) > K e σ 3 ( W 1 + W 2 + W 3 ) > K S 0 e - ( r - σ 2 2 )2 = A σ 3 ( W 1 + W 2 + W 3 ) > log parenleftbigg K S 0 e - ( r - σ 2 2 )2 parenrightbigg = B So, putting it all together now: S 0 e - r - σ 2 E bracketleftbigg e σ 3 ( W 1 + W 2 + W 3 ) 1 ( S 1 S 2 S 3 ) 1 3 >K bracketrightbigg - Ke - r 3 E bracketleftbigg 1 ( S 1 S 2 S 3 ) 1 3 >K bracketrightbigg = S 0 e - r - σ 2 E bracketleftbigg e σ 3 ( W 1 + W 2 + W 3 ) 1 e σ 3 ( W 1 + W 2 + W 3 ) >A bracketrightbigg - Ke - 3 r E bracketleftbigg 1 σ 3 ( W 1 + W 2 + W 3 ) >B bracketrightbigg = S 0 e - r - σ 2 exp parenleftbigg σ 2 7 9 parenrightbigg N parenleftbigg 14 9 σ 2 - log( A ) σ radicalBig 14 9 parenrightbigg - Ke - 3 r parenleftbigg 1 - N parenleftbigg B σ radicalBig 14 9 parenrightbiggparenrightbigg
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