Normalizing a Cauchy Distribution
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
all. I'm trying to check in the following distribution is normalized, and am having a difficult time integrating it. If anyone can give me a clue on where to begin the integration, that be great.
The function in question is the Cauchy probability density function. I.e.,
$$ p(x) = fracapi (a^2 + x^2)$$
Beginning my integration, I have:
$$beginalignint_-infty^infty | p(x) |^2 dx & = int_-infty^infty bigg | fracapi (a^2 + x^2) bigg |^2 dx \
& = bigg ( fracapi bigg )^2 int_-infty^infty bigg | frac1a^2 + x^2 bigg |^2 dx endalign$$
Unfortunately, this is as far I got. Would factoring out the $a^2$ term and using u-substitution for $u=x/a$ work for this?
Thanks.
integration probability-distributions
asked Sep 11 at 3:26
Kosta
4991415
add a comment |
up vote
0
down vote
favorite
all. I'm trying to check in the following distribution is normalized, and am having a difficult time integrating it. If anyone can give me a clue on where to begin the integration, that be great.
The function in question is the Cauchy probability density function. I.e.,
$$ p(x) = fracapi (a^2 + x^2)$$
Beginning my integration, I have:
$$beginalignint_-infty^infty | p(x) |^2 dx & = int_-infty^infty bigg | fracapi (a^2 + x^2) bigg |^2 dx \
& = bigg ( fracapi bigg )^2 int_-infty^infty bigg | frac1a^2 + x^2 bigg |^2 dx endalign$$
Unfortunately, this is as far I got. Would factoring out the $a^2$ term and using u-substitution for $u=x/a$ work for this?
Thanks.
integration probability-distributions
asked Sep 11 at 3:26
Kosta
4991415
you really don't need absolute values, right ?
– Ahmad Bazzi
Sep 11 at 3:37
@Kosta: If you only want to check that the Cauchy distribution is normalized, then you only need to check that $int_-infty^infty fracapi (a^2 + x^2) mathrmdx = 1$. Why did you need to use absolute values?
– Winter Soldier
Sep 11 at 4:17
The question is not at all clear. Why are you squaring $p(x)$?. You want to choose $a$ such that $p$ integrates to $1$.
– Kavi Rama Murthy
Sep 11 at 5:42
@KaviRamaMurthy, I was confused. In Quantum Mechanics, normalizing means integrating $|Psi|^2 = Psi^* Psi$. I tried doing the same for $p(x)$ before realizing that the $|Psi|^2$ is a probability distribution. It was an oversight that I was stuck on for a moment.
– Kosta
Sep 13 at 0:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
all. I'm trying to check in the following distribution is normalized, and am having a difficult time integrating it. If anyone can give me a clue on where to begin the integration, that be great.
The function in question is the Cauchy probability density function. I.e.,
$$ p(x) = fracapi (a^2 + x^2)$$
Beginning my integration, I have:
$$beginalignint_-infty^infty | p(x) |^2 dx & = int_-infty^infty bigg | fracapi (a^2 + x^2) bigg |^2 dx \
& = bigg ( fracapi bigg )^2 int_-infty^infty bigg | frac1a^2 + x^2 bigg |^2 dx endalign$$
Unfortunately, this is as far I got. Would factoring out the $a^2$ term and using u-substitution for $u=x/a$ work for this?
Thanks.
integration probability-distributions
asked Sep 11 at 3:26
Kosta
4991415
all. I'm trying to check in the following distribution is normalized, and am having a difficult time integrating it. If anyone can give me a clue on where to begin the integration, that be great.
The function in question is the Cauchy probability density function. I.e.,
$$ p(x) = fracapi (a^2 + x^2)$$
Beginning my integration, I have:
$$beginalignint_-infty^infty | p(x) |^2 dx & = int_-infty^infty bigg | fracapi (a^2 + x^2) bigg |^2 dx \
& = bigg ( fracapi bigg )^2 int_-infty^infty bigg | frac1a^2 + x^2 bigg |^2 dx endalign$$
Unfortunately, this is as far I got. Would factoring out the $a^2$ term and using u-substitution for $u=x/a$ work for this?
Thanks.
integration probability-distributions
integration probability-distributions
asked Sep 11 at 3:26
Kosta
4991415
asked Sep 11 at 3:26
Kosta
4991415
asked Sep 11 at 3:26
Kosta
4991415
asked Sep 11 at 3:26
Kosta
4991415
asked Sep 11 at 3:26
Kosta
4991415
4991415
you really don't need absolute values, right ?
– Ahmad Bazzi
Sep 11 at 3:37
@Kosta: If you only want to check that the Cauchy distribution is normalized, then you only need to check that $int_-infty^infty fracapi (a^2 + x^2) mathrmdx = 1$. Why did you need to use absolute values?
– Winter Soldier
Sep 11 at 4:17
The question is not at all clear. Why are you squaring $p(x)$?. You want to choose $a$ such that $p$ integrates to $1$.
– Kavi Rama Murthy
Sep 11 at 5:42
@KaviRamaMurthy, I was confused. In Quantum Mechanics, normalizing means integrating $|Psi|^2 = Psi^* Psi$. I tried doing the same for $p(x)$ before realizing that the $|Psi|^2$ is a probability distribution. It was an oversight that I was stuck on for a moment.
– Kosta
Sep 13 at 0:45
add a comment |
you really don't need absolute values, right ?
– Ahmad Bazzi
Sep 11 at 3:37
@Kosta: If you only want to check that the Cauchy distribution is normalized, then you only need to check that $int_-infty^infty fracapi (a^2 + x^2) mathrmdx = 1$. Why did you need to use absolute values?
– Winter Soldier
Sep 11 at 4:17
The question is not at all clear. Why are you squaring $p(x)$?. You want to choose $a$ such that $p$ integrates to $1$.
– Kavi Rama Murthy
Sep 11 at 5:42
@KaviRamaMurthy, I was confused. In Quantum Mechanics, normalizing means integrating $|Psi|^2 = Psi^* Psi$. I tried doing the same for $p(x)$ before realizing that the $|Psi|^2$ is a probability distribution. It was an oversight that I was stuck on for a moment.
– Kosta
Sep 13 at 0:45
you really don't need absolute values, right ?
– Ahmad Bazzi
Sep 11 at 3:37
you really don't need absolute values, right ?
– Ahmad Bazzi
Sep 11 at 3:37
@Kosta: If you only want to check that the Cauchy distribution is normalized, then you only need to check that $int_-infty^infty fracapi (a^2 + x^2) mathrmdx = 1$. Why did you need to use absolute values?
– Winter Soldier
Sep 11 at 4:17
@Kosta: If you only want to check that the Cauchy distribution is normalized, then you only need to check that $int_-infty^infty fracapi (a^2 + x^2) mathrmdx = 1$. Why did you need to use absolute values?
– Winter Soldier
Sep 11 at 4:17
The question is not at all clear. Why are you squaring $p(x)$?. You want to choose $a$ such that $p$ integrates to $1$.
– Kavi Rama Murthy
Sep 11 at 5:42
The question is not at all clear. Why are you squaring $p(x)$?. You want to choose $a$ such that $p$ integrates to $1$.
– Kavi Rama Murthy
Sep 11 at 5:42
@KaviRamaMurthy, I was confused. In Quantum Mechanics, normalizing means integrating $|Psi|^2 = Psi^* Psi$. I tried doing the same for $p(x)$ before realizing that the $|Psi|^2$ is a probability distribution. It was an oversight that I was stuck on for a moment.
– Kosta
Sep 13 at 0:45
@KaviRamaMurthy, I was confused. In Quantum Mechanics, normalizing means integrating $|Psi|^2 = Psi^* Psi$. I tried doing the same for $p(x)$ before realizing that the $|Psi|^2$ is a probability distribution. It was an oversight that I was stuck on for a moment.
– Kosta
Sep 13 at 0:45
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You could write your integral using the reduction formula as
$$displaystyleintdfrac1left(x^2+a^2right)^2,mathrmdx =dfracx2a^2left(x^2+a^2right)+classsteps-nodecssIdsteps-node-4dfrac12a^2displaystyleintdfrac1x^2+a^2,mathrmdx $$
Solve $displaystyleintdfrac1x^2+a^2,mathrmdx$ by using the change of variable $u = fracxa$, which will become
$$displaystyleintdfrac1x^2+a^2,mathrmdx=displaystyleintdfracaa^2u^2+a^2,mathrmdu=classsteps-nodecssIdsteps-node-5dfrac1adisplaystyleintdfrac1u^2+1,mathrmdu=dfracarctanleft(uright)a=dfracarctanleft(fracxaright)a$$
Therefore
$$displaystyleintdfrac1left(x^2+a^2right)^2,mathrmdx=dfracarctanleft(fracxaright)2a^3+dfracx2a^2left(x^2+a^2right)+C$$
Taking the integral limits, you get
$$displaystyleint_-infty^inftydfrac1left(x^2+a^2right)^2,mathrmdx=fracpi2a^3$$
answered Sep 11 at 3:35
Ahmad Bazzi
7,4912724
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You could write your integral using the reduction formula as
$$displaystyleintdfrac1left(x^2+a^2right)^2,mathrmdx =dfracx2a^2left(x^2+a^2right)+classsteps-nodecssIdsteps-node-4dfrac12a^2displaystyleintdfrac1x^2+a^2,mathrmdx $$
Solve $displaystyleintdfrac1x^2+a^2,mathrmdx$ by using the change of variable $u = fracxa$, which will become
$$displaystyleintdfrac1x^2+a^2,mathrmdx=displaystyleintdfracaa^2u^2+a^2,mathrmdu=classsteps-nodecssIdsteps-node-5dfrac1adisplaystyleintdfrac1u^2+1,mathrmdu=dfracarctanleft(uright)a=dfracarctanleft(fracxaright)a$$
Therefore
$$displaystyleintdfrac1left(x^2+a^2right)^2,mathrmdx=dfracarctanleft(fracxaright)2a^3+dfracx2a^2left(x^2+a^2right)+C$$
Taking the integral limits, you get
$$displaystyleint_-infty^inftydfrac1left(x^2+a^2right)^2,mathrmdx=fracpi2a^3$$
answered Sep 11 at 3:35
Ahmad Bazzi
7,4912724
add a comment |
up vote
2
down vote
accepted
You could write your integral using the reduction formula as
$$displaystyleintdfrac1left(x^2+a^2right)^2,mathrmdx =dfracx2a^2left(x^2+a^2right)+classsteps-nodecssIdsteps-node-4dfrac12a^2displaystyleintdfrac1x^2+a^2,mathrmdx $$
Solve $displaystyleintdfrac1x^2+a^2,mathrmdx$ by using the change of variable $u = fracxa$, which will become
$$displaystyleintdfrac1x^2+a^2,mathrmdx=displaystyleintdfracaa^2u^2+a^2,mathrmdu=classsteps-nodecssIdsteps-node-5dfrac1adisplaystyleintdfrac1u^2+1,mathrmdu=dfracarctanleft(uright)a=dfracarctanleft(fracxaright)a$$
Therefore
$$displaystyleintdfrac1left(x^2+a^2right)^2,mathrmdx=dfracarctanleft(fracxaright)2a^3+dfracx2a^2left(x^2+a^2right)+C$$
Taking the integral limits, you get
$$displaystyleint_-infty^inftydfrac1left(x^2+a^2right)^2,mathrmdx=fracpi2a^3$$
answered Sep 11 at 3:35
Ahmad Bazzi
7,4912724
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You could write your integral using the reduction formula as
$$displaystyleintdfrac1left(x^2+a^2right)^2,mathrmdx =dfracx2a^2left(x^2+a^2right)+classsteps-nodecssIdsteps-node-4dfrac12a^2displaystyleintdfrac1x^2+a^2,mathrmdx $$
Solve $displaystyleintdfrac1x^2+a^2,mathrmdx$ by using the change of variable $u = fracxa$, which will become
$$displaystyleintdfrac1x^2+a^2,mathrmdx=displaystyleintdfracaa^2u^2+a^2,mathrmdu=classsteps-nodecssIdsteps-node-5dfrac1adisplaystyleintdfrac1u^2+1,mathrmdu=dfracarctanleft(uright)a=dfracarctanleft(fracxaright)a$$
Therefore
$$displaystyleintdfrac1left(x^2+a^2right)^2,mathrmdx=dfracarctanleft(fracxaright)2a^3+dfracx2a^2left(x^2+a^2right)+C$$
Taking the integral limits, you get
$$displaystyleint_-infty^inftydfrac1left(x^2+a^2right)^2,mathrmdx=fracpi2a^3$$
answered Sep 11 at 3:35
Ahmad Bazzi
7,4912724
You could write your integral using the reduction formula as
$$displaystyleintdfrac1left(x^2+a^2right)^2,mathrmdx =dfracx2a^2left(x^2+a^2right)+classsteps-nodecssIdsteps-node-4dfrac12a^2displaystyleintdfrac1x^2+a^2,mathrmdx $$
Solve $displaystyleintdfrac1x^2+a^2,mathrmdx$ by using the change of variable $u = fracxa$, which will become
$$displaystyleintdfrac1x^2+a^2,mathrmdx=displaystyleintdfracaa^2u^2+a^2,mathrmdu=classsteps-nodecssIdsteps-node-5dfrac1adisplaystyleintdfrac1u^2+1,mathrmdu=dfracarctanleft(uright)a=dfracarctanleft(fracxaright)a$$
Therefore
$$displaystyleintdfrac1left(x^2+a^2right)^2,mathrmdx=dfracarctanleft(fracxaright)2a^3+dfracx2a^2left(x^2+a^2right)+C$$
Taking the integral limits, you get
$$displaystyleint_-infty^inftydfrac1left(x^2+a^2right)^2,mathrmdx=fracpi2a^3$$
answered Sep 11 at 3:35
Ahmad Bazzi
7,4912724
answered Sep 11 at 3:35
Ahmad Bazzi
7,4912724
answered Sep 11 at 3:35
Ahmad Bazzi
7,4912724
answered Sep 11 at 3:35
Ahmad Bazzi
7,4912724
7,4912724
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2912709%2fnormalizing-a-cauchy-distribution%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
you really don't need absolute values, right ?
– Ahmad Bazzi
Sep 11 at 3:37
@Kosta: If you only want to check that the Cauchy distribution is normalized, then you only need to check that $int_-infty^infty fracapi (a^2 + x^2) mathrmdx = 1$. Why did you need to use absolute values?
– Winter Soldier
Sep 11 at 4:17
The question is not at all clear. Why are you squaring $p(x)$?. You want to choose $a$ such that $p$ integrates to $1$.
– Kavi Rama Murthy
Sep 11 at 5:42
@KaviRamaMurthy, I was confused. In Quantum Mechanics, normalizing means integrating $|Psi|^2 = Psi^* Psi$. I tried doing the same for $p(x)$ before realizing that the $|Psi|^2$ is a probability distribution. It was an oversight that I was stuck on for a moment.
– Kosta
Sep 13 at 0:45