Proof of Feynman-Kac theorem
I am reading a proof of Feynman-Kac theorem as done here, where I do not follow one step. Specifically, after the author derived:
$$dY_s = u_x(t-s, W_s)e^{-R(s)}dW_s$$
they seem to have concluded $$Y_t =u(0,W_t)e^{-R(t)}$$
in the next line. But the partial derivative in $x$ magically seemed to have disappeared.
Can anyone point out if this is a mistake or this is how it is supposed to be?
probability-theory stochastic-processes brownian-motion
asked Nov 27 '18 at 1:41
dezdichado
6,1781929
add a comment |
I am reading a proof of Feynman-Kac theorem as done here, where I do not follow one step. Specifically, after the author derived:
$$dY_s = u_x(t-s, W_s)e^{-R(s)}dW_s$$
they seem to have concluded $$Y_t =u(0,W_t)e^{-R(t)}$$
in the next line. But the partial derivative in $x$ magically seemed to have disappeared.
Can anyone point out if this is a mistake or this is how it is supposed to be?
probability-theory stochastic-processes brownian-motion
asked Nov 27 '18 at 1:41
dezdichado
6,1781929
add a comment |
I am reading a proof of Feynman-Kac theorem as done here, where I do not follow one step. Specifically, after the author derived:
$$dY_s = u_x(t-s, W_s)e^{-R(s)}dW_s$$
they seem to have concluded $$Y_t =u(0,W_t)e^{-R(t)}$$
in the next line. But the partial derivative in $x$ magically seemed to have disappeared.
Can anyone point out if this is a mistake or this is how it is supposed to be?
probability-theory stochastic-processes brownian-motion
asked Nov 27 '18 at 1:41
dezdichado
6,1781929
I am reading a proof of Feynman-Kac theorem as done here, where I do not follow one step. Specifically, after the author derived:
$$dY_s = u_x(t-s, W_s)e^{-R(s)}dW_s$$
they seem to have concluded $$Y_t =u(0,W_t)e^{-R(t)}$$
in the next line. But the partial derivative in $x$ magically seemed to have disappeared.
Can anyone point out if this is a mistake or this is how it is supposed to be?
probability-theory stochastic-processes brownian-motion
probability-theory stochastic-processes brownian-motion
asked Nov 27 '18 at 1:41
dezdichado
6,1781929
asked Nov 27 '18 at 1:41
dezdichado
6,1781929
asked Nov 27 '18 at 1:41
dezdichado
6,1781929
asked Nov 27 '18 at 1:41
dezdichado
6,1781929
asked Nov 27 '18 at 1:41
dezdichado
6,1781929
6,1781929
add a comment |
add a comment |
1 Answer
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It's a lot simpler than you thought. The definition of $Y_s$ is given in the beginning of the proof. He doesn't use the expression for $dY_s$ in any way in writing down $Y_t$.
At the beginning he defines $Y_s=e^{-R(s)}u(t-s,W_s)$. Letting $s=t$ gives $Y_t=e^{-R(t)}u(0,W_t)$.
answered Nov 27 '18 at 5:16
Zachary Selk
579311
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It's a lot simpler than you thought. The definition of $Y_s$ is given in the beginning of the proof. He doesn't use the expression for $dY_s$ in any way in writing down $Y_t$.
At the beginning he defines $Y_s=e^{-R(s)}u(t-s,W_s)$. Letting $s=t$ gives $Y_t=e^{-R(t)}u(0,W_t)$.
answered Nov 27 '18 at 5:16
Zachary Selk
579311
add a comment |
It's a lot simpler than you thought. The definition of $Y_s$ is given in the beginning of the proof. He doesn't use the expression for $dY_s$ in any way in writing down $Y_t$.
At the beginning he defines $Y_s=e^{-R(s)}u(t-s,W_s)$. Letting $s=t$ gives $Y_t=e^{-R(t)}u(0,W_t)$.
answered Nov 27 '18 at 5:16
Zachary Selk
579311
add a comment |
It's a lot simpler than you thought. The definition of $Y_s$ is given in the beginning of the proof. He doesn't use the expression for $dY_s$ in any way in writing down $Y_t$.
At the beginning he defines $Y_s=e^{-R(s)}u(t-s,W_s)$. Letting $s=t$ gives $Y_t=e^{-R(t)}u(0,W_t)$.
answered Nov 27 '18 at 5:16
Zachary Selk
579311
It's a lot simpler than you thought. The definition of $Y_s$ is given in the beginning of the proof. He doesn't use the expression for $dY_s$ in any way in writing down $Y_t$.
At the beginning he defines $Y_s=e^{-R(s)}u(t-s,W_s)$. Letting $s=t$ gives $Y_t=e^{-R(t)}u(0,W_t)$.
answered Nov 27 '18 at 5:16
Zachary Selk
579311
answered Nov 27 '18 at 5:16
Zachary Selk
579311
answered Nov 27 '18 at 5:16
Zachary Selk
579311
answered Nov 27 '18 at 5:16
Zachary Selk
579311
579311
add a comment |
add a comment |
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