Can anyone help me solve this problem. I have no idea where to even start on it. Link inside stock option problem
Stock Option induction problem
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0I can't post an image inside because I am a new user. It gives me an error when I do. I do not know how to type it up so I thought a link to an image would be the best way to view the problem instead of me transposing it poorly. – 2011-02-28
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0@demondeac11: The typing uses simple LaTeX syntax. If you plan on frequenting this site, it might be a good idea to learn at least the basics: they are pretty straightforward. – 2011-02-28
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1@demondeac11: As to where to get started, I would suggest remembering the definition of "Expected value" and applying it to the definition you have. – 2011-02-28
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0Could you be more specific? I am still lost. – 2011-02-28
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0Is $F$ an arbitrary distribution? – 2011-02-28
1 Answers
You are asked to prove that $$ \int_{ - \infty }^\infty {V_{n - 1} (s + x)dF(x)} > s - c, $$ for all $n \geq 1$. For $n=1$, substituting from the definition of $V_0$, you need to show that $$ \int_{ - \infty }^\infty {\max \lbrace s + x - c,0\rbrace dF(x)} > s - c. $$ For this purpose, first note that $$ \max \lbrace s + x - c,0\rbrace \ge s + x - c. $$ Then, by linearity of the integral, you can consider the sum $$ \int_{ - \infty }^\infty {(s - c)dF(x)} + \int_{ - \infty }^\infty {xdF(x)} , $$ from which the assertion for $n=1$ follows.
To complete the inductive proof, substituting from the definition of $V_n$, you need to show that $$ \int_{ - \infty }^\infty {\max \bigg\lbrace s + x - c,\int_{ - \infty }^\infty {V_{n - 1} (s + x + u)dF(u)} \bigg\rbrace dF(x)} > s - c, $$ under the induction hypothesis that, for any $s \in \mathbb{R}$, $$ \int_{ - \infty }^\infty {V_{n - 1} (s + x)dF(x)} > s - c. $$ For this purpose, recall the end of the proof for $n=1$.