Asian Option pricing payoff inequality [closed]
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I'm trying to prove this inequality:
$(exp(frac{1}{n} sum_{i=1}^n log(S_{t_i})) -K)^+ le ((frac{1}{n} sum_{i=1}^n S_{t_i}) -K)^+ $
I can observe the inequality when I use fixed numbers, but I would like to establish it for any set of positive numbers.
derivatives finance
asked Dec 26 '18 at 13:56
user608881user608881
283
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closed as off-topic by RRL, metamorphy, José Carlos Santos, Davide Giraudo, Leucippus Dec 27 '18 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, metamorphy, José Carlos Santos, Davide Giraudo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm trying to prove this inequality:
$(exp(frac{1}{n} sum_{i=1}^n log(S_{t_i})) -K)^+ le ((frac{1}{n} sum_{i=1}^n S_{t_i}) -K)^+ $
I can observe the inequality when I use fixed numbers, but I would like to establish it for any set of positive numbers.
derivatives finance
asked Dec 26 '18 at 13:56
user608881user608881
283
$endgroup$
closed as off-topic by RRL, metamorphy, José Carlos Santos, Davide Giraudo, Leucippus Dec 27 '18 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, metamorphy, José Carlos Santos, Davide Giraudo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Does this follow from Jensen's inequality?
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– ODF
Dec 26 '18 at 14:01
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It quite looks similar, given that the left hand-side is the payoff for the geometric Asian call option and the right hand-side is the payoff for the arithmetic Asian call option.
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– user608881
Dec 26 '18 at 14:09
add a comment |
$begingroup$
I'm trying to prove this inequality:
$(exp(frac{1}{n} sum_{i=1}^n log(S_{t_i})) -K)^+ le ((frac{1}{n} sum_{i=1}^n S_{t_i}) -K)^+ $
I can observe the inequality when I use fixed numbers, but I would like to establish it for any set of positive numbers.
derivatives finance
asked Dec 26 '18 at 13:56
user608881user608881
283
$endgroup$
I'm trying to prove this inequality:
$(exp(frac{1}{n} sum_{i=1}^n log(S_{t_i})) -K)^+ le ((frac{1}{n} sum_{i=1}^n S_{t_i}) -K)^+ $
I can observe the inequality when I use fixed numbers, but I would like to establish it for any set of positive numbers.
derivatives finance
derivatives finance
asked Dec 26 '18 at 13:56
user608881user608881
283
asked Dec 26 '18 at 13:56
user608881user608881
283
asked Dec 26 '18 at 13:56
user608881user608881
283
asked Dec 26 '18 at 13:56
user608881user608881
283
asked Dec 26 '18 at 13:56
user608881user608881
283
283
closed as off-topic by RRL, metamorphy, José Carlos Santos, Davide Giraudo, Leucippus Dec 27 '18 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, metamorphy, José Carlos Santos, Davide Giraudo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, metamorphy, José Carlos Santos, Davide Giraudo, Leucippus Dec 27 '18 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, metamorphy, José Carlos Santos, Davide Giraudo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Does this follow from Jensen's inequality?
$endgroup$
– ODF
Dec 26 '18 at 14:01
$begingroup$
It quite looks similar, given that the left hand-side is the payoff for the geometric Asian call option and the right hand-side is the payoff for the arithmetic Asian call option.
$endgroup$
– user608881
Dec 26 '18 at 14:09
add a comment |
1
$begingroup$
Does this follow from Jensen's inequality?
$endgroup$
– ODF
Dec 26 '18 at 14:01
$begingroup$
It quite looks similar, given that the left hand-side is the payoff for the geometric Asian call option and the right hand-side is the payoff for the arithmetic Asian call option.
$endgroup$
– user608881
Dec 26 '18 at 14:09
1
1
$begingroup$
Does this follow from Jensen's inequality?
$endgroup$
– ODF
Dec 26 '18 at 14:01
$begingroup$
Does this follow from Jensen's inequality?
$endgroup$
– ODF
Dec 26 '18 at 14:01
$begingroup$
It quite looks similar, given that the left hand-side is the payoff for the geometric Asian call option and the right hand-side is the payoff for the arithmetic Asian call option.
$endgroup$
– user608881
Dec 26 '18 at 14:09
$begingroup$
It quite looks similar, given that the left hand-side is the payoff for the geometric Asian call option and the right hand-side is the payoff for the arithmetic Asian call option.
$endgroup$
– user608881
Dec 26 '18 at 14:09
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You need the AM-GM inequality:
$$sqrt[n]{x_1 x_2 cdots x_n} leq frac{x_1 +x_2 + cdots + x_n}{n} $$
(see, for example:https://artofproblemsolving.com/wiki/index.php/Arithmetic_Mean-Geometric_Mean_Inequality )
Then you have for the exponential term (since nothing happens with the $K$ term or the positive-part term I'll ignore them):
$$ begin{eqnarray}
exp left( frac{1}{n}sum_{i=1}^n log (S_{t_i}) right) &=& exp^{log S_{t_1}^{1/n}} cdot exp^{log S_{t_2}^{1/n}} cdots exp^{log S_{t_n}^{1/n}} \
& = & S_{t_1}^{1/n} cdot S_{t_2}^{1/n} cdots S_{t_n}^{1/n} \
& leq & frac{sum_{i=1}^n S_{t_i}}{n}
end{eqnarray}$$
answered Dec 26 '18 at 16:16
postmortespostmortes
2,04131119
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You need the AM-GM inequality:
$$sqrt[n]{x_1 x_2 cdots x_n} leq frac{x_1 +x_2 + cdots + x_n}{n} $$
(see, for example:https://artofproblemsolving.com/wiki/index.php/Arithmetic_Mean-Geometric_Mean_Inequality )
Then you have for the exponential term (since nothing happens with the $K$ term or the positive-part term I'll ignore them):
$$ begin{eqnarray}
exp left( frac{1}{n}sum_{i=1}^n log (S_{t_i}) right) &=& exp^{log S_{t_1}^{1/n}} cdot exp^{log S_{t_2}^{1/n}} cdots exp^{log S_{t_n}^{1/n}} \
& = & S_{t_1}^{1/n} cdot S_{t_2}^{1/n} cdots S_{t_n}^{1/n} \
& leq & frac{sum_{i=1}^n S_{t_i}}{n}
end{eqnarray}$$
answered Dec 26 '18 at 16:16
postmortespostmortes
2,04131119
$endgroup$
add a comment |
$begingroup$
You need the AM-GM inequality:
$$sqrt[n]{x_1 x_2 cdots x_n} leq frac{x_1 +x_2 + cdots + x_n}{n} $$
(see, for example:https://artofproblemsolving.com/wiki/index.php/Arithmetic_Mean-Geometric_Mean_Inequality )
Then you have for the exponential term (since nothing happens with the $K$ term or the positive-part term I'll ignore them):
$$ begin{eqnarray}
exp left( frac{1}{n}sum_{i=1}^n log (S_{t_i}) right) &=& exp^{log S_{t_1}^{1/n}} cdot exp^{log S_{t_2}^{1/n}} cdots exp^{log S_{t_n}^{1/n}} \
& = & S_{t_1}^{1/n} cdot S_{t_2}^{1/n} cdots S_{t_n}^{1/n} \
& leq & frac{sum_{i=1}^n S_{t_i}}{n}
end{eqnarray}$$
answered Dec 26 '18 at 16:16
postmortespostmortes
2,04131119
$endgroup$
add a comment |
$begingroup$
You need the AM-GM inequality:
$$sqrt[n]{x_1 x_2 cdots x_n} leq frac{x_1 +x_2 + cdots + x_n}{n} $$
(see, for example:https://artofproblemsolving.com/wiki/index.php/Arithmetic_Mean-Geometric_Mean_Inequality )
Then you have for the exponential term (since nothing happens with the $K$ term or the positive-part term I'll ignore them):
$$ begin{eqnarray}
exp left( frac{1}{n}sum_{i=1}^n log (S_{t_i}) right) &=& exp^{log S_{t_1}^{1/n}} cdot exp^{log S_{t_2}^{1/n}} cdots exp^{log S_{t_n}^{1/n}} \
& = & S_{t_1}^{1/n} cdot S_{t_2}^{1/n} cdots S_{t_n}^{1/n} \
& leq & frac{sum_{i=1}^n S_{t_i}}{n}
end{eqnarray}$$
answered Dec 26 '18 at 16:16
postmortespostmortes
2,04131119
$endgroup$
You need the AM-GM inequality:
$$sqrt[n]{x_1 x_2 cdots x_n} leq frac{x_1 +x_2 + cdots + x_n}{n} $$
(see, for example:https://artofproblemsolving.com/wiki/index.php/Arithmetic_Mean-Geometric_Mean_Inequality )
Then you have for the exponential term (since nothing happens with the $K$ term or the positive-part term I'll ignore them):
$$ begin{eqnarray}
exp left( frac{1}{n}sum_{i=1}^n log (S_{t_i}) right) &=& exp^{log S_{t_1}^{1/n}} cdot exp^{log S_{t_2}^{1/n}} cdots exp^{log S_{t_n}^{1/n}} \
& = & S_{t_1}^{1/n} cdot S_{t_2}^{1/n} cdots S_{t_n}^{1/n} \
& leq & frac{sum_{i=1}^n S_{t_i}}{n}
end{eqnarray}$$
answered Dec 26 '18 at 16:16
postmortespostmortes
2,04131119
answered Dec 26 '18 at 16:16
postmortespostmortes
2,04131119
answered Dec 26 '18 at 16:16
postmortespostmortes
2,04131119
answered Dec 26 '18 at 16:16
postmortespostmortes
2,04131119
2,04131119
add a comment |
add a comment |
1
$begingroup$
Does this follow from Jensen's inequality?
$endgroup$
– ODF
Dec 26 '18 at 14:01
$begingroup$
It quite looks similar, given that the left hand-side is the payoff for the geometric Asian call option and the right hand-side is the payoff for the arithmetic Asian call option.
$endgroup$
– user608881
Dec 26 '18 at 14:09