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For a perpetual american put option $v(s)$, satisfies the following problem:

$$\frac12\sigma^2S^2\frac{\mathrm d^2V}{\mathrm dS^2}+(r-D)S\frac{\mathrm dV}{\mathrm dS} - rV = 0\quad\text{for }S^*<S<\infty,$$

$$V(S) = E- S\quad\text{for }0\le S<S^*,$$

$$V(S^*) = E- S^*, \quad \frac{\mathrm dV}{\mathrm dS}(S^*) = -1,\quad \lim_{S\to\infty} V(S) = 0.$$

If $V(S) = S^a$ is the solution of the differential equation, I can derive that

$$\tag1 \frac12\sigma^2a^2 + \left(r-D-\frac12\sigma^2\right)a - r = 0.$$

One of the roots of the equation, $a_n$ is negative and for $S > S^*$,

$$ V(S) =-\frac{S^*}{a_n} \left(\frac S{S^*}\right)^{a_n},\quad S^* = \frac{a_n}{a_n-1}E.$$

I am given the following condition holds

$$\tag2 \sigma^2a_n+ \left(r - D - \frac12 \sigma^2\right)= - \sqrt{\left(r - D - \frac12 \sigma^2\right)^2 + 2 r \sigma^2} < 0.$$

Suppose now the parameters $\sigma>0$, $r>0$, and $E>0$ are fixed, but $D$ can vary, so you may regard $a_neg$ as a function of $D$, how can I show the following:

As $D$ tends to $\infty$, $a_n$ tends to $-\frac rD$ so that $S^*$ tends to $0$.

As $D$ tends to $-\infty$, $a_n$ tends to $\frac{2D}{\sigma^2}$ so that $S^*$ tends to $E$.

By differentiating (1) above with respect to $D$ and using (2) deduce that $\frac{\mathrm \partial a_n}{\mathrm \partial D} > 0$ and hence $\frac{\mathrm \partial S^*}{\mathrm \partial D} < 0$.

Please can you help me in proving this property?

Many thanks

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I tried to make this more readable by adding $\LaTeX$ markup, but I can only hope that everything is still what you intended to write. However, what is $a_neg$ supposed to mean? – Hagen von Eitzen Apr 14 '13 at 10:51
@HagenvonEitzen $a_n$ is defined as the negative root of (1). And (2) is a formula for $a_n$. – Did Apr 14 '13 at 11:12
Thanks @HagenvonEitzen – sleepybob Apr 14 '13 at 11:32
up vote 1 down vote accepted

Note that $$ a_n= \frac1{\sigma^2}\left(D+\tfrac12 \sigma^2-r- b\right), $$ where $$ b=\sqrt{\left(D+\tfrac12 \sigma^2-r\right)^2 + 2 r \sigma^2}, $$ is also $$ a_n= \frac{-2r}{D+\frac12 \sigma^2-r+b} $$ When $|D|\to+\infty$, $b\sim|D|$. Thus, the first formula shows that $a_n\sim2D/\sigma^2$ when $D\to-\infty$ and the second formula shows that $a_n\sim -r/D$ when $D\to+\infty$. In particular, $a_n\to-\infty$ when $D\to-\infty$ and $a_n\to0$ when $D\to+\infty$.

Now, consider $a_n$ and $b$ as functions of $D$, the parameters $\sigma^2$ and $r$ being kept constant, and differentiate the first formula above. This yields $$ a_n'(D)=\frac1{\sigma^2}\left(1- \frac{D+\tfrac12 \sigma^2-r}{b(D)}\right)=\frac{-a_n(D)}{\sigma^2b(D)}, $$ from which the variations of the function $D\mapsto a_n(D)$ should be clear.

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You seem to be asking a third question about a function $u$ but you do not define the object. – Did Apr 14 '13 at 11:10
thanks a lot for your help, actually it is the partial differentiation operator, I have amended it. Would you please suggest the way forward, thanks! – sleepybob Apr 14 '13 at 11:36
See revised version. – Did Apr 14 '13 at 11:50
Hi Did, thank you for your help. Please could you elaborate how you reached the step $$ a_n= \frac{-2r}{D+\frac12 \sigma^2-r+b} $$ – sleepybob Apr 20 '13 at 14:52
Using the identity $c-b=(c^2-b^2)/(c+b)$ with $c=D+(\sigma^2/2)-r$. – Did Apr 20 '13 at 17:53

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