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Inverse Laplace Transform to Recover CDF
Clash Royale CLAN TAG#URR8PPP
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I know we can recover CDF of a variable random $ X $ by using inverse laplace transform. But as far as I know, laplace transform is defined on $ [0,infty] $. Does it imply that we can only recover CDF of real positive random variable? How about bilateral laplace transform? Does inverse bilateral laplace transform can recover CDF of real variable random $ X $?
statistics laplace-transform
asked Aug 1 at 14:38
Ben
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up vote
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down vote
favorite
I know we can recover CDF of a variable random $ X $ by using inverse laplace transform. But as far as I know, laplace transform is defined on $ [0,infty] $. Does it imply that we can only recover CDF of real positive random variable? How about bilateral laplace transform? Does inverse bilateral laplace transform can recover CDF of real variable random $ X $?
statistics laplace-transform
asked Aug 1 at 14:38
Ben
334
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know we can recover CDF of a variable random $ X $ by using inverse laplace transform. But as far as I know, laplace transform is defined on $ [0,infty] $. Does it imply that we can only recover CDF of real positive random variable? How about bilateral laplace transform? Does inverse bilateral laplace transform can recover CDF of real variable random $ X $?
statistics laplace-transform
asked Aug 1 at 14:38
Ben
334
I know we can recover CDF of a variable random $ X $ by using inverse laplace transform. But as far as I know, laplace transform is defined on $ [0,infty] $. Does it imply that we can only recover CDF of real positive random variable? How about bilateral laplace transform? Does inverse bilateral laplace transform can recover CDF of real variable random $ X $?
statistics laplace-transform
asked Aug 1 at 14:38
Ben
334
asked Aug 1 at 14:38
Ben
334
asked Aug 1 at 14:38
Ben
334
asked Aug 1 at 14:38
Ben
334
334
add a comment |Â
add a comment |Â
1 Answer
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For the two-sided Laplace transform, one has to explicitly specify the region of convergence for the inverse transform. If the inverse transform of $F(p)$ is $f$ when $p$ lies in the vertical strip $ROC(f)$, then
$$mathcal L !left[ int_-infty^t f(tau) dtau right] =
frac F(p) p, \
p in ROC(f) land operatornameRe p > 0.$$
This property can be used to obtain the cdf from the pdf $f$. If you take $p$ in the left half-plane, the answer will differ by $int_-infty^infty f(tau) dtau = 1$.
answered Aug 5 at 0:19
Maxim
1,935112
Can you prove $ mathcalL [ int_-infty^t f(x) dx ] = fracF(p)p $ is valid for the two-sided laplace transform? Because when I am trying to prove it, using integration by parts, there is this term $ lim_t to -infty - frac1s e^-st int_-infty^t f(x) dx $ which most likely not to be 0. But this term has to be 0 if the formula for cdf to be valid. Or is there any connection between the $ ROC(f) $ and this term?
– Ben
yesterday
If the double integral converges absolutely, the domain $t in mathbb R, tau < t$ can be reparametrized as $tau in mathbb R, t > tau$: $$int_-infty^infty dt ,e^-p t int_-infty^t dtau f(tau)= int_-infty^infty dtau f(tau) int_tau^infty dt ,e^-p t.$$
– Maxim
yesterday
Hmm I still dont get it. Maybe it is a silly question, is $ int_-infty^infty f(tau) dtau $ and $ int_-infty^infty dtau f(tau) $ the same? For several times, I have seen this integral, but not sure what it is
– Ben
yesterday
Yes, $int_-infty^infty f(tau) dtau$ and $int_-infty^infty dtau f(tau)$ are exactly the same. If it's clearer, rewrite the above as $int_-infty^infty (e^-p t int_-infty^t f(tau) dtau) dt = int_-infty^infty (f(tau) int_tau^infty e^-p t dt) dtau$.
– Maxim
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
For the two-sided Laplace transform, one has to explicitly specify the region of convergence for the inverse transform. If the inverse transform of $F(p)$ is $f$ when $p$ lies in the vertical strip $ROC(f)$, then
$$mathcal L !left[ int_-infty^t f(tau) dtau right] =
frac F(p) p, \
p in ROC(f) land operatornameRe p > 0.$$
This property can be used to obtain the cdf from the pdf $f$. If you take $p$ in the left half-plane, the answer will differ by $int_-infty^infty f(tau) dtau = 1$.
answered Aug 5 at 0:19
Maxim
1,935112
Can you prove $ mathcalL [ int_-infty^t f(x) dx ] = fracF(p)p $ is valid for the two-sided laplace transform? Because when I am trying to prove it, using integration by parts, there is this term $ lim_t to -infty - frac1s e^-st int_-infty^t f(x) dx $ which most likely not to be 0. But this term has to be 0 if the formula for cdf to be valid. Or is there any connection between the $ ROC(f) $ and this term?
– Ben
yesterday
If the double integral converges absolutely, the domain $t in mathbb R, tau < t$ can be reparametrized as $tau in mathbb R, t > tau$: $$int_-infty^infty dt ,e^-p t int_-infty^t dtau f(tau)= int_-infty^infty dtau f(tau) int_tau^infty dt ,e^-p t.$$
– Maxim
yesterday
Hmm I still dont get it. Maybe it is a silly question, is $ int_-infty^infty f(tau) dtau $ and $ int_-infty^infty dtau f(tau) $ the same? For several times, I have seen this integral, but not sure what it is
– Ben
yesterday
Yes, $int_-infty^infty f(tau) dtau$ and $int_-infty^infty dtau f(tau)$ are exactly the same. If it's clearer, rewrite the above as $int_-infty^infty (e^-p t int_-infty^t f(tau) dtau) dt = int_-infty^infty (f(tau) int_tau^infty e^-p t dt) dtau$.
– Maxim
yesterday
add a comment |Â
up vote
1
down vote
For the two-sided Laplace transform, one has to explicitly specify the region of convergence for the inverse transform. If the inverse transform of $F(p)$ is $f$ when $p$ lies in the vertical strip $ROC(f)$, then
$$mathcal L !left[ int_-infty^t f(tau) dtau right] =
frac F(p) p, \
p in ROC(f) land operatornameRe p > 0.$$
This property can be used to obtain the cdf from the pdf $f$. If you take $p$ in the left half-plane, the answer will differ by $int_-infty^infty f(tau) dtau = 1$.
answered Aug 5 at 0:19
Maxim
1,935112
Can you prove $ mathcalL [ int_-infty^t f(x) dx ] = fracF(p)p $ is valid for the two-sided laplace transform? Because when I am trying to prove it, using integration by parts, there is this term $ lim_t to -infty - frac1s e^-st int_-infty^t f(x) dx $ which most likely not to be 0. But this term has to be 0 if the formula for cdf to be valid. Or is there any connection between the $ ROC(f) $ and this term?
– Ben
yesterday
If the double integral converges absolutely, the domain $t in mathbb R, tau < t$ can be reparametrized as $tau in mathbb R, t > tau$: $$int_-infty^infty dt ,e^-p t int_-infty^t dtau f(tau)= int_-infty^infty dtau f(tau) int_tau^infty dt ,e^-p t.$$
– Maxim
yesterday
Hmm I still dont get it. Maybe it is a silly question, is $ int_-infty^infty f(tau) dtau $ and $ int_-infty^infty dtau f(tau) $ the same? For several times, I have seen this integral, but not sure what it is
– Ben
yesterday
Yes, $int_-infty^infty f(tau) dtau$ and $int_-infty^infty dtau f(tau)$ are exactly the same. If it's clearer, rewrite the above as $int_-infty^infty (e^-p t int_-infty^t f(tau) dtau) dt = int_-infty^infty (f(tau) int_tau^infty e^-p t dt) dtau$.
– Maxim
yesterday
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For the two-sided Laplace transform, one has to explicitly specify the region of convergence for the inverse transform. If the inverse transform of $F(p)$ is $f$ when $p$ lies in the vertical strip $ROC(f)$, then
$$mathcal L !left[ int_-infty^t f(tau) dtau right] =
frac F(p) p, \
p in ROC(f) land operatornameRe p > 0.$$
This property can be used to obtain the cdf from the pdf $f$. If you take $p$ in the left half-plane, the answer will differ by $int_-infty^infty f(tau) dtau = 1$.
answered Aug 5 at 0:19
Maxim
1,935112
For the two-sided Laplace transform, one has to explicitly specify the region of convergence for the inverse transform. If the inverse transform of $F(p)$ is $f$ when $p$ lies in the vertical strip $ROC(f)$, then
$$mathcal L !left[ int_-infty^t f(tau) dtau right] =
frac F(p) p, \
p in ROC(f) land operatornameRe p > 0.$$
This property can be used to obtain the cdf from the pdf $f$. If you take $p$ in the left half-plane, the answer will differ by $int_-infty^infty f(tau) dtau = 1$.
answered Aug 5 at 0:19
Maxim
1,935112
edited Aug 5 at 0:25
answered Aug 5 at 0:19
Maxim
1,935112
answered Aug 5 at 0:19
Maxim
1,935112
answered Aug 5 at 0:19
Maxim
1,935112
1,935112
Can you prove $ mathcalL [ int_-infty^t f(x) dx ] = fracF(p)p $ is valid for the two-sided laplace transform? Because when I am trying to prove it, using integration by parts, there is this term $ lim_t to -infty - frac1s e^-st int_-infty^t f(x) dx $ which most likely not to be 0. But this term has to be 0 if the formula for cdf to be valid. Or is there any connection between the $ ROC(f) $ and this term?
– Ben
yesterday
If the double integral converges absolutely, the domain $t in mathbb R, tau < t$ can be reparametrized as $tau in mathbb R, t > tau$: $$int_-infty^infty dt ,e^-p t int_-infty^t dtau f(tau)= int_-infty^infty dtau f(tau) int_tau^infty dt ,e^-p t.$$
– Maxim
yesterday
Hmm I still dont get it. Maybe it is a silly question, is $ int_-infty^infty f(tau) dtau $ and $ int_-infty^infty dtau f(tau) $ the same? For several times, I have seen this integral, but not sure what it is
– Ben
yesterday
Yes, $int_-infty^infty f(tau) dtau$ and $int_-infty^infty dtau f(tau)$ are exactly the same. If it's clearer, rewrite the above as $int_-infty^infty (e^-p t int_-infty^t f(tau) dtau) dt = int_-infty^infty (f(tau) int_tau^infty e^-p t dt) dtau$.
– Maxim
yesterday
add a comment |Â
Can you prove $ mathcalL [ int_-infty^t f(x) dx ] = fracF(p)p $ is valid for the two-sided laplace transform? Because when I am trying to prove it, using integration by parts, there is this term $ lim_t to -infty - frac1s e^-st int_-infty^t f(x) dx $ which most likely not to be 0. But this term has to be 0 if the formula for cdf to be valid. Or is there any connection between the $ ROC(f) $ and this term?
– Ben
yesterday
If the double integral converges absolutely, the domain $t in mathbb R, tau < t$ can be reparametrized as $tau in mathbb R, t > tau$: $$int_-infty^infty dt ,e^-p t int_-infty^t dtau f(tau)= int_-infty^infty dtau f(tau) int_tau^infty dt ,e^-p t.$$
– Maxim
yesterday
Hmm I still dont get it. Maybe it is a silly question, is $ int_-infty^infty f(tau) dtau $ and $ int_-infty^infty dtau f(tau) $ the same? For several times, I have seen this integral, but not sure what it is
– Ben
yesterday
Yes, $int_-infty^infty f(tau) dtau$ and $int_-infty^infty dtau f(tau)$ are exactly the same. If it's clearer, rewrite the above as $int_-infty^infty (e^-p t int_-infty^t f(tau) dtau) dt = int_-infty^infty (f(tau) int_tau^infty e^-p t dt) dtau$.
– Maxim
yesterday
Can you prove $ mathcalL [ int_-infty^t f(x) dx ] = fracF(p)p $ is valid for the two-sided laplace transform? Because when I am trying to prove it, using integration by parts, there is this term $ lim_t to -infty - frac1s e^-st int_-infty^t f(x) dx $ which most likely not to be 0. But this term has to be 0 if the formula for cdf to be valid. Or is there any connection between the $ ROC(f) $ and this term?
– Ben
yesterday
Can you prove $ mathcalL [ int_-infty^t f(x) dx ] = fracF(p)p $ is valid for the two-sided laplace transform? Because when I am trying to prove it, using integration by parts, there is this term $ lim_t to -infty - frac1s e^-st int_-infty^t f(x) dx $ which most likely not to be 0. But this term has to be 0 if the formula for cdf to be valid. Or is there any connection between the $ ROC(f) $ and this term?
– Ben
yesterday
If the double integral converges absolutely, the domain $t in mathbb R, tau < t$ can be reparametrized as $tau in mathbb R, t > tau$: $$int_-infty^infty dt ,e^-p t int_-infty^t dtau f(tau)= int_-infty^infty dtau f(tau) int_tau^infty dt ,e^-p t.$$
– Maxim
yesterday
If the double integral converges absolutely, the domain $t in mathbb R, tau < t$ can be reparametrized as $tau in mathbb R, t > tau$: $$int_-infty^infty dt ,e^-p t int_-infty^t dtau f(tau)= int_-infty^infty dtau f(tau) int_tau^infty dt ,e^-p t.$$
– Maxim
yesterday
Hmm I still dont get it. Maybe it is a silly question, is $ int_-infty^infty f(tau) dtau $ and $ int_-infty^infty dtau f(tau) $ the same? For several times, I have seen this integral, but not sure what it is
– Ben
yesterday
Hmm I still dont get it. Maybe it is a silly question, is $ int_-infty^infty f(tau) dtau $ and $ int_-infty^infty dtau f(tau) $ the same? For several times, I have seen this integral, but not sure what it is
– Ben
yesterday
Yes, $int_-infty^infty f(tau) dtau$ and $int_-infty^infty dtau f(tau)$ are exactly the same. If it's clearer, rewrite the above as $int_-infty^infty (e^-p t int_-infty^t f(tau) dtau) dt = int_-infty^infty (f(tau) int_tau^infty e^-p t dt) dtau$.
– Maxim
yesterday
Yes, $int_-infty^infty f(tau) dtau$ and $int_-infty^infty dtau f(tau)$ are exactly the same. If it's clearer, rewrite the above as $int_-infty^infty (e^-p t int_-infty^t f(tau) dtau) dt = int_-infty^infty (f(tau) int_tau^infty e^-p t dt) dtau$.
– Maxim
yesterday
add a comment |Â
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