174%202011%20Summer%20Quiz_3

# 174%202011%20Summer%20Quiz_3 - the geometric average of two...

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174 Quiz 3 Solutions Gabe Merton August 31, 2011 60 52 . 03 70 . 23 45 . 12 60 . 91 60 . 91 82 . 22 (1 - p ) 2 1 Part a We can determine u and d from the prices in the tree. We need these to calculate p * . 1

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u = 70 . 23 / 60 = 1 . 1705 d = 52 . 03 / 60 = 0 . 8672 p * = e ( r - δ ) h - d u - d = e (0 . 03)(0 . 25) - 0 . 8672 1 . 1705 - 0 . 8672 = 0 . 4627 Now construct a table of the various possible payoﬀs and probabilities. ( A ) ( B ) = max (0 , ( A ) - 60) ( C ) ( D ) = ( B ) × ( C ) Arith Avg Price Payoﬀ Prob Weighted Payoﬀ 76.225 16.225 0.2141 3.4732 65.57 5.57 0.2486 1.3847 56.47 0 0.2486 0 48.575 0 0.2887 0 - 1.00 4.8579 The present value of the risk-neutral weighted average of the scenarios is, c = e - (0 . 06)(0 . 5) (4 . 8579) = (0 . 97)(4 . 8679) = 4 . 7122 2 Part b We recylce some of the calculations in part a (speciﬁcally the calculation of p * and the various probabilities). We just need a new table with geometric averages. Recall that
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Unformatted text preview: the geometric average of two numbers a and b is √ ab . ( A ) ( B ) ( C ) = max (0 , ( B )-( A )) ( D ) ( D ) = ( C ) × ( D ) Price Geo Avg Strike Payoﬀ Prob Weighted Payoﬀ 82.22 75.99 0.2141 60.91 65.40 4.49 0.2486 1.1162 60.91 56.30 0.2486 45.12 48.45 3.33 0.2887 1.102 ∑--1.00 2.0777 The price is the present value of the risk-neutral weighted average of the scenarios. p = e-(0 . 06)(0 . 5) (2 . 0777) = (0 . 97)(2 . 0777) = 2 . 0154 2...
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## This note was uploaded on 09/20/2011 for the course MATH Math 174 taught by Professor Kong during the Summer '10 term at UCLA.

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174%202011%20Summer%20Quiz_3 - the geometric average of two...

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