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Inverse Laplace and Fourier Transform in Statistics
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I am currently exploring the use of inverse laplace transform and inverse fourier transform in statistics. From what I have read, for a random variable $ X $ with $ f(x) $ and $ F(x) $ as its PDF and CDF, we can use inverse fourier transform to recover its PDF, which is
$$ f(x) = frac12pi int_-infty^infty e^-itx phi(t) dt $$
If I want to calculate the CDF of $ X $ at point a, I can integrate above formula
$$ F(a) = frac12pi int_-infty^a int_-infty^infty e^-itx phi(t) dt dx $$
with $ phi(t) = E[e^itX] $ is a characteristic function of $ X $. From what I have read from Wikipedia, we can use inverse laplace transform to calculate a CDF of a random variable directly by using this formula
$$ F(a) = mathcalL^-1 bigg frac1s E Big[e^-sX Big] bigg $$
$$ F(a) = frac12 pi i int_gamma - i infty^gamma + i infty frace^sas E Big[e^-sX Big] ds $$
I know inverse fourier transform is equal to inverse laplace transform if we take $ s = gamma + it $. I am trying to prove the formula for $ F(a) $ is equal if we use inverse fourier transform and inverse laplace transform, but somehow, it is obvious that they are different, one with one integral and the other with two integrals. Or is there something wrong in the formula that I have stated so both of the are not the same? If it is already correct, kindly need your help to prove it. Thank you.
statistics laplace-transform fourier-transform
asked Jul 31 at 14:22
Ben
334
add a comment |Â
up vote
2
down vote
favorite
I am currently exploring the use of inverse laplace transform and inverse fourier transform in statistics. From what I have read, for a random variable $ X $ with $ f(x) $ and $ F(x) $ as its PDF and CDF, we can use inverse fourier transform to recover its PDF, which is
$$ f(x) = frac12pi int_-infty^infty e^-itx phi(t) dt $$
If I want to calculate the CDF of $ X $ at point a, I can integrate above formula
$$ F(a) = frac12pi int_-infty^a int_-infty^infty e^-itx phi(t) dt dx $$
with $ phi(t) = E[e^itX] $ is a characteristic function of $ X $. From what I have read from Wikipedia, we can use inverse laplace transform to calculate a CDF of a random variable directly by using this formula
$$ F(a) = mathcalL^-1 bigg frac1s E Big[e^-sX Big] bigg $$
$$ F(a) = frac12 pi i int_gamma - i infty^gamma + i infty frace^sas E Big[e^-sX Big] ds $$
I know inverse fourier transform is equal to inverse laplace transform if we take $ s = gamma + it $. I am trying to prove the formula for $ F(a) $ is equal if we use inverse fourier transform and inverse laplace transform, but somehow, it is obvious that they are different, one with one integral and the other with two integrals. Or is there something wrong in the formula that I have stated so both of the are not the same? If it is already correct, kindly need your help to prove it. Thank you.
statistics laplace-transform fourier-transform
asked Jul 31 at 14:22
Ben
334
Notice that your last formula is also a double integral because the expectation is an integral. Regarding the two-sided Laplace transform, see here.
– Maxim
Aug 5 at 0:20
If we count the expectation, cdf obtained from inverse fourier transform has 3 integral (characterictic function is an expectation) and from inverse laplace transform, we only get 2
– Ben
2 days ago
I see, then it seems the question is essentially the same as the one I linked to, one integration is eliminated by $mathcal L[ int_-infty^t f(tau) dtau ] = F(s)/s$.
– Maxim
2 days ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am currently exploring the use of inverse laplace transform and inverse fourier transform in statistics. From what I have read, for a random variable $ X $ with $ f(x) $ and $ F(x) $ as its PDF and CDF, we can use inverse fourier transform to recover its PDF, which is
$$ f(x) = frac12pi int_-infty^infty e^-itx phi(t) dt $$
If I want to calculate the CDF of $ X $ at point a, I can integrate above formula
$$ F(a) = frac12pi int_-infty^a int_-infty^infty e^-itx phi(t) dt dx $$
with $ phi(t) = E[e^itX] $ is a characteristic function of $ X $. From what I have read from Wikipedia, we can use inverse laplace transform to calculate a CDF of a random variable directly by using this formula
$$ F(a) = mathcalL^-1 bigg frac1s E Big[e^-sX Big] bigg $$
$$ F(a) = frac12 pi i int_gamma - i infty^gamma + i infty frace^sas E Big[e^-sX Big] ds $$
I know inverse fourier transform is equal to inverse laplace transform if we take $ s = gamma + it $. I am trying to prove the formula for $ F(a) $ is equal if we use inverse fourier transform and inverse laplace transform, but somehow, it is obvious that they are different, one with one integral and the other with two integrals. Or is there something wrong in the formula that I have stated so both of the are not the same? If it is already correct, kindly need your help to prove it. Thank you.
statistics laplace-transform fourier-transform
asked Jul 31 at 14:22
Ben
334
I am currently exploring the use of inverse laplace transform and inverse fourier transform in statistics. From what I have read, for a random variable $ X $ with $ f(x) $ and $ F(x) $ as its PDF and CDF, we can use inverse fourier transform to recover its PDF, which is
$$ f(x) = frac12pi int_-infty^infty e^-itx phi(t) dt $$
If I want to calculate the CDF of $ X $ at point a, I can integrate above formula
$$ F(a) = frac12pi int_-infty^a int_-infty^infty e^-itx phi(t) dt dx $$
with $ phi(t) = E[e^itX] $ is a characteristic function of $ X $. From what I have read from Wikipedia, we can use inverse laplace transform to calculate a CDF of a random variable directly by using this formula
$$ F(a) = mathcalL^-1 bigg frac1s E Big[e^-sX Big] bigg $$
$$ F(a) = frac12 pi i int_gamma - i infty^gamma + i infty frace^sas E Big[e^-sX Big] ds $$
I know inverse fourier transform is equal to inverse laplace transform if we take $ s = gamma + it $. I am trying to prove the formula for $ F(a) $ is equal if we use inverse fourier transform and inverse laplace transform, but somehow, it is obvious that they are different, one with one integral and the other with two integrals. Or is there something wrong in the formula that I have stated so both of the are not the same? If it is already correct, kindly need your help to prove it. Thank you.
statistics laplace-transform fourier-transform
asked Jul 31 at 14:22
Ben
334
edited Jul 31 at 15:00
asked Jul 31 at 14:22
Ben
334
asked Jul 31 at 14:22
Ben
334
asked Jul 31 at 14:22
Ben
334
334
Notice that your last formula is also a double integral because the expectation is an integral. Regarding the two-sided Laplace transform, see here.
– Maxim
Aug 5 at 0:20
If we count the expectation, cdf obtained from inverse fourier transform has 3 integral (characterictic function is an expectation) and from inverse laplace transform, we only get 2
– Ben
2 days ago
I see, then it seems the question is essentially the same as the one I linked to, one integration is eliminated by $mathcal L[ int_-infty^t f(tau) dtau ] = F(s)/s$.
– Maxim
2 days ago
add a comment |Â
Notice that your last formula is also a double integral because the expectation is an integral. Regarding the two-sided Laplace transform, see here.
– Maxim
Aug 5 at 0:20
If we count the expectation, cdf obtained from inverse fourier transform has 3 integral (characterictic function is an expectation) and from inverse laplace transform, we only get 2
– Ben
2 days ago
I see, then it seems the question is essentially the same as the one I linked to, one integration is eliminated by $mathcal L[ int_-infty^t f(tau) dtau ] = F(s)/s$.
– Maxim
2 days ago
Notice that your last formula is also a double integral because the expectation is an integral. Regarding the two-sided Laplace transform, see here.
– Maxim
Aug 5 at 0:20
Notice that your last formula is also a double integral because the expectation is an integral. Regarding the two-sided Laplace transform, see here.
– Maxim
Aug 5 at 0:20
If we count the expectation, cdf obtained from inverse fourier transform has 3 integral (characterictic function is an expectation) and from inverse laplace transform, we only get 2
– Ben
2 days ago
If we count the expectation, cdf obtained from inverse fourier transform has 3 integral (characterictic function is an expectation) and from inverse laplace transform, we only get 2
– Ben
2 days ago
I see, then it seems the question is essentially the same as the one I linked to, one integration is eliminated by $mathcal L[ int_-infty^t f(tau) dtau ] = F(s)/s$.
– Maxim
2 days ago
I see, then it seems the question is essentially the same as the one I linked to, one integration is eliminated by $mathcal L[ int_-infty^t f(tau) dtau ] = F(s)/s$.
– Maxim
2 days ago
add a comment |Â
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Notice that your last formula is also a double integral because the expectation is an integral. Regarding the two-sided Laplace transform, see here.
– Maxim
Aug 5 at 0:20
If we count the expectation, cdf obtained from inverse fourier transform has 3 integral (characterictic function is an expectation) and from inverse laplace transform, we only get 2
– Ben
2 days ago
I see, then it seems the question is essentially the same as the one I linked to, one integration is eliminated by $mathcal L[ int_-infty^t f(tau) dtau ] = F(s)/s$.
– Maxim
2 days ago