Hitting time distribution with exponential growth
$begingroup$
Let $A_0=A>0$ and let $$dA_t = (rA_t - x)dt + sigma dB_t,$$ where $B_t$ is standard Brownian motion and $r,x$ and $sigma$ are positive constants. Let $T= inf { t: A_t = 0 }$ and $$G(A)=Bbb{E}[exp(-rT)].$$ How do I find the function $G$? I think
begin{equation}
rG(A)=(rA-x)G'(A)+frac{1}{2}sigma^2G''(A). tag{*}label{eq:1}
end{equation}
Is this correct? Is there a closed-form solution? Clearly, $G(0)=1$. It should also be true that $lim_{A to infty}G(A)=0$.
Also, it should be true that $G'(A)<0$ at all $A>0$. This would mean that $G''$ does not change sign, and so $G$ must be strictly convex. Thus, $G'$ is strictly monotone in $A$ going from some unknown negative number $G'(0)$ to $0$ as $A$ goes to infinity.
One way to try and solve eqref{eq:1} could be to change the independent variable to $G'$. If we denote $G'=frac{dG}{dA}$ by $M$, we have
$$frac{dA}{dM} = frac{1}{G''}=frac{frac{1}{2}sigma^2}{rG-(rA-x)M}$$
and
$$frac{dG}{dM} = frac{M}{G''}=frac{frac{1}{2}sigma^2M}{rG-(rA-x)M},$$
which could be a solvable system of first-order equations.
Further, if we denote $rG-(rA-x)M$ by $S$, we can transform the above system in $(A(M),G(M))$ into a system in $(A(M),S(M))$ with
$$frac{dA}{dM} =frac{frac{1}{2}sigma^2}{S}$$
and
$$frac{dS}{dM} =rfrac{dG}{dM}-rfrac{dA}{dM}M - (rA-x)=- (rA-x),$$
where we have used $frac{dG}{dM}=frac{frac{1}{2}sigma^2M}{S}=frac{dA}{dM}M$. This system should be solvable by separation of variables.
In particular, we have
$$
S=Cexpleft(-{frac{frac{1}{2}rA^2-xA}{frac{1}{2}sigma^2}}right),
$$
which can be verified by differentiation. The constant of integration $C$ satisfies $C=r+xM(0)$, where $M(0)=G'(0)$ is the slope of $G$ at $A=0$.
I think I should be able to integrate $S$ to recover $M$ and then integrate $M$ to recover $G$, but I'm stuck.
Any help with this problem will be very much appreciated. Thanks!
ordinary-differential-equations brownian-motion stopping-times
asked Dec 14 '18 at 2:32
J.DoeJ.Doe
63
$endgroup$
add a comment |
$begingroup$
Let $A_0=A>0$ and let $$dA_t = (rA_t - x)dt + sigma dB_t,$$ where $B_t$ is standard Brownian motion and $r,x$ and $sigma$ are positive constants. Let $T= inf { t: A_t = 0 }$ and $$G(A)=Bbb{E}[exp(-rT)].$$ How do I find the function $G$? I think
begin{equation}
rG(A)=(rA-x)G'(A)+frac{1}{2}sigma^2G''(A). tag{*}label{eq:1}
end{equation}
Is this correct? Is there a closed-form solution? Clearly, $G(0)=1$. It should also be true that $lim_{A to infty}G(A)=0$.
Also, it should be true that $G'(A)<0$ at all $A>0$. This would mean that $G''$ does not change sign, and so $G$ must be strictly convex. Thus, $G'$ is strictly monotone in $A$ going from some unknown negative number $G'(0)$ to $0$ as $A$ goes to infinity.
One way to try and solve eqref{eq:1} could be to change the independent variable to $G'$. If we denote $G'=frac{dG}{dA}$ by $M$, we have
$$frac{dA}{dM} = frac{1}{G''}=frac{frac{1}{2}sigma^2}{rG-(rA-x)M}$$
and
$$frac{dG}{dM} = frac{M}{G''}=frac{frac{1}{2}sigma^2M}{rG-(rA-x)M},$$
which could be a solvable system of first-order equations.
Further, if we denote $rG-(rA-x)M$ by $S$, we can transform the above system in $(A(M),G(M))$ into a system in $(A(M),S(M))$ with
$$frac{dA}{dM} =frac{frac{1}{2}sigma^2}{S}$$
and
$$frac{dS}{dM} =rfrac{dG}{dM}-rfrac{dA}{dM}M - (rA-x)=- (rA-x),$$
where we have used $frac{dG}{dM}=frac{frac{1}{2}sigma^2M}{S}=frac{dA}{dM}M$. This system should be solvable by separation of variables.
In particular, we have
$$
S=Cexpleft(-{frac{frac{1}{2}rA^2-xA}{frac{1}{2}sigma^2}}right),
$$
which can be verified by differentiation. The constant of integration $C$ satisfies $C=r+xM(0)$, where $M(0)=G'(0)$ is the slope of $G$ at $A=0$.
I think I should be able to integrate $S$ to recover $M$ and then integrate $M$ to recover $G$, but I'm stuck.
Any help with this problem will be very much appreciated. Thanks!
ordinary-differential-equations brownian-motion stopping-times
asked Dec 14 '18 at 2:32
J.DoeJ.Doe
63
$endgroup$
add a comment |
$begingroup$
Let $A_0=A>0$ and let $$dA_t = (rA_t - x)dt + sigma dB_t,$$ where $B_t$ is standard Brownian motion and $r,x$ and $sigma$ are positive constants. Let $T= inf { t: A_t = 0 }$ and $$G(A)=Bbb{E}[exp(-rT)].$$ How do I find the function $G$? I think
begin{equation}
rG(A)=(rA-x)G'(A)+frac{1}{2}sigma^2G''(A). tag{*}label{eq:1}
end{equation}
Is this correct? Is there a closed-form solution? Clearly, $G(0)=1$. It should also be true that $lim_{A to infty}G(A)=0$.
Also, it should be true that $G'(A)<0$ at all $A>0$. This would mean that $G''$ does not change sign, and so $G$ must be strictly convex. Thus, $G'$ is strictly monotone in $A$ going from some unknown negative number $G'(0)$ to $0$ as $A$ goes to infinity.
One way to try and solve eqref{eq:1} could be to change the independent variable to $G'$. If we denote $G'=frac{dG}{dA}$ by $M$, we have
$$frac{dA}{dM} = frac{1}{G''}=frac{frac{1}{2}sigma^2}{rG-(rA-x)M}$$
and
$$frac{dG}{dM} = frac{M}{G''}=frac{frac{1}{2}sigma^2M}{rG-(rA-x)M},$$
which could be a solvable system of first-order equations.
Further, if we denote $rG-(rA-x)M$ by $S$, we can transform the above system in $(A(M),G(M))$ into a system in $(A(M),S(M))$ with
$$frac{dA}{dM} =frac{frac{1}{2}sigma^2}{S}$$
and
$$frac{dS}{dM} =rfrac{dG}{dM}-rfrac{dA}{dM}M - (rA-x)=- (rA-x),$$
where we have used $frac{dG}{dM}=frac{frac{1}{2}sigma^2M}{S}=frac{dA}{dM}M$. This system should be solvable by separation of variables.
In particular, we have
$$
S=Cexpleft(-{frac{frac{1}{2}rA^2-xA}{frac{1}{2}sigma^2}}right),
$$
which can be verified by differentiation. The constant of integration $C$ satisfies $C=r+xM(0)$, where $M(0)=G'(0)$ is the slope of $G$ at $A=0$.
I think I should be able to integrate $S$ to recover $M$ and then integrate $M$ to recover $G$, but I'm stuck.
Any help with this problem will be very much appreciated. Thanks!
ordinary-differential-equations brownian-motion stopping-times
asked Dec 14 '18 at 2:32
J.DoeJ.Doe
63
$endgroup$
Let $A_0=A>0$ and let $$dA_t = (rA_t - x)dt + sigma dB_t,$$ where $B_t$ is standard Brownian motion and $r,x$ and $sigma$ are positive constants. Let $T= inf { t: A_t = 0 }$ and $$G(A)=Bbb{E}[exp(-rT)].$$ How do I find the function $G$? I think
begin{equation}
rG(A)=(rA-x)G'(A)+frac{1}{2}sigma^2G''(A). tag{*}label{eq:1}
end{equation}
Is this correct? Is there a closed-form solution? Clearly, $G(0)=1$. It should also be true that $lim_{A to infty}G(A)=0$.
Also, it should be true that $G'(A)<0$ at all $A>0$. This would mean that $G''$ does not change sign, and so $G$ must be strictly convex. Thus, $G'$ is strictly monotone in $A$ going from some unknown negative number $G'(0)$ to $0$ as $A$ goes to infinity.
One way to try and solve eqref{eq:1} could be to change the independent variable to $G'$. If we denote $G'=frac{dG}{dA}$ by $M$, we have
$$frac{dA}{dM} = frac{1}{G''}=frac{frac{1}{2}sigma^2}{rG-(rA-x)M}$$
and
$$frac{dG}{dM} = frac{M}{G''}=frac{frac{1}{2}sigma^2M}{rG-(rA-x)M},$$
which could be a solvable system of first-order equations.
Further, if we denote $rG-(rA-x)M$ by $S$, we can transform the above system in $(A(M),G(M))$ into a system in $(A(M),S(M))$ with
$$frac{dA}{dM} =frac{frac{1}{2}sigma^2}{S}$$
and
$$frac{dS}{dM} =rfrac{dG}{dM}-rfrac{dA}{dM}M - (rA-x)=- (rA-x),$$
where we have used $frac{dG}{dM}=frac{frac{1}{2}sigma^2M}{S}=frac{dA}{dM}M$. This system should be solvable by separation of variables.
In particular, we have
$$
S=Cexpleft(-{frac{frac{1}{2}rA^2-xA}{frac{1}{2}sigma^2}}right),
$$
which can be verified by differentiation. The constant of integration $C$ satisfies $C=r+xM(0)$, where $M(0)=G'(0)$ is the slope of $G$ at $A=0$.
I think I should be able to integrate $S$ to recover $M$ and then integrate $M$ to recover $G$, but I'm stuck.
Any help with this problem will be very much appreciated. Thanks!
ordinary-differential-equations brownian-motion stopping-times
ordinary-differential-equations brownian-motion stopping-times
asked Dec 14 '18 at 2:32
J.DoeJ.Doe
63
asked Dec 14 '18 at 2:32
J.DoeJ.Doe
63
edited Dec 31 '18 at 0:50
J.Doe
asked Dec 14 '18 at 2:32
J.DoeJ.Doe
63
asked Dec 14 '18 at 2:32
J.DoeJ.Doe
63
asked Dec 14 '18 at 2:32
J.DoeJ.Doe
63
63
add a comment |
add a comment |
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