integral equation and Fourier transform of “almost” the convolution
I am facing an integral equation where one of the terms looks like this:
$$ V(t) = int_t^{+infty} K(x-t) cdot V(x) , dx $$
where $$K(x) = N(frac{-b-asigma^2}{sigma sqrt{x}}) - d^{-2a}N(frac{-b+asigma^2}{sigma sqrt{x}}) $$ and $N(cdot)$ is the cumulative normal distribution. I am inclined to use the Convolution theorem, but the term inside of the $K$ has flipped signs. Also, looking at the formula for $K$ it is not even (nor odd) since its not defined for $x<0$ (Look at the square root term). Any suggestions as to how to make use of the Convolution theorem or any other better way?
fourier-transform integral-equations
asked Dec 2 at 0:36
dleal
1728
add a comment |
I am facing an integral equation where one of the terms looks like this:
$$ V(t) = int_t^{+infty} K(x-t) cdot V(x) , dx $$
where $$K(x) = N(frac{-b-asigma^2}{sigma sqrt{x}}) - d^{-2a}N(frac{-b+asigma^2}{sigma sqrt{x}}) $$ and $N(cdot)$ is the cumulative normal distribution. I am inclined to use the Convolution theorem, but the term inside of the $K$ has flipped signs. Also, looking at the formula for $K$ it is not even (nor odd) since its not defined for $x<0$ (Look at the square root term). Any suggestions as to how to make use of the Convolution theorem or any other better way?
fourier-transform integral-equations
asked Dec 2 at 0:36
dleal
1728
With $U(t)= V(-t)$ it is $U = U ast K 1_{x > 0}$ which can be Fourier transformed
– reuns
Dec 2 at 0:58
@reuns thank you! I think that is what I'm looking for. I would accept it an an answer
– dleal
Dec 2 at 1:13
add a comment |
I am facing an integral equation where one of the terms looks like this:
$$ V(t) = int_t^{+infty} K(x-t) cdot V(x) , dx $$
where $$K(x) = N(frac{-b-asigma^2}{sigma sqrt{x}}) - d^{-2a}N(frac{-b+asigma^2}{sigma sqrt{x}}) $$ and $N(cdot)$ is the cumulative normal distribution. I am inclined to use the Convolution theorem, but the term inside of the $K$ has flipped signs. Also, looking at the formula for $K$ it is not even (nor odd) since its not defined for $x<0$ (Look at the square root term). Any suggestions as to how to make use of the Convolution theorem or any other better way?
fourier-transform integral-equations
asked Dec 2 at 0:36
dleal
1728
I am facing an integral equation where one of the terms looks like this:
$$ V(t) = int_t^{+infty} K(x-t) cdot V(x) , dx $$
where $$K(x) = N(frac{-b-asigma^2}{sigma sqrt{x}}) - d^{-2a}N(frac{-b+asigma^2}{sigma sqrt{x}}) $$ and $N(cdot)$ is the cumulative normal distribution. I am inclined to use the Convolution theorem, but the term inside of the $K$ has flipped signs. Also, looking at the formula for $K$ it is not even (nor odd) since its not defined for $x<0$ (Look at the square root term). Any suggestions as to how to make use of the Convolution theorem or any other better way?
fourier-transform integral-equations
fourier-transform integral-equations
asked Dec 2 at 0:36
dleal
1728
asked Dec 2 at 0:36
dleal
1728
asked Dec 2 at 0:36
dleal
1728
asked Dec 2 at 0:36
dleal
1728
asked Dec 2 at 0:36
dleal
1728
1728
With $U(t)= V(-t)$ it is $U = U ast K 1_{x > 0}$ which can be Fourier transformed
– reuns
Dec 2 at 0:58
@reuns thank you! I think that is what I'm looking for. I would accept it an an answer
– dleal
Dec 2 at 1:13
add a comment |
With $U(t)= V(-t)$ it is $U = U ast K 1_{x > 0}$ which can be Fourier transformed
– reuns
Dec 2 at 0:58
@reuns thank you! I think that is what I'm looking for. I would accept it an an answer
– dleal
Dec 2 at 1:13
With $U(t)= V(-t)$ it is $U = U ast K 1_{x > 0}$ which can be Fourier transformed
– reuns
Dec 2 at 0:58
With $U(t)= V(-t)$ it is $U = U ast K 1_{x > 0}$ which can be Fourier transformed
– reuns
Dec 2 at 0:58
@reuns thank you! I think that is what I'm looking for. I would accept it an an answer
– dleal
Dec 2 at 1:13
@reuns thank you! I think that is what I'm looking for. I would accept it an an answer
– dleal
Dec 2 at 1:13
add a comment |
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With $U(t)= V(-t)$ it is $U = U ast K 1_{x > 0}$ which can be Fourier transformed
– reuns
Dec 2 at 0:58
@reuns thank you! I think that is what I'm looking for. I would accept it an an answer
– dleal
Dec 2 at 1:13