CDF and Inverse CDF of Wrapped Cauchy Distribution

Multi tool use
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
The standard wrapped-up Cauchy distribution has the following probability density function:
$$f(x,p)= frac1-p^22pi(1+p^2)-2pcos(x)$$
where x is from $0$ to $2pi$
Can anybody know, what is the CDF and InvCDF of this distribution?
Can you please also suggest any other wrapped distribution?
probability-distributions
add a comment |Â
up vote
0
down vote
favorite
The standard wrapped-up Cauchy distribution has the following probability density function:
$$f(x,p)= frac1-p^22pi(1+p^2)-2pcos(x)$$
where x is from $0$ to $2pi$
Can anybody know, what is the CDF and InvCDF of this distribution?
Can you please also suggest any other wrapped distribution?
probability-distributions
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The standard wrapped-up Cauchy distribution has the following probability density function:
$$f(x,p)= frac1-p^22pi(1+p^2)-2pcos(x)$$
where x is from $0$ to $2pi$
Can anybody know, what is the CDF and InvCDF of this distribution?
Can you please also suggest any other wrapped distribution?
probability-distributions
The standard wrapped-up Cauchy distribution has the following probability density function:
$$f(x,p)= frac1-p^22pi(1+p^2)-2pcos(x)$$
where x is from $0$ to $2pi$
Can anybody know, what is the CDF and InvCDF of this distribution?
Can you please also suggest any other wrapped distribution?
probability-distributions
probability-distributions
edited Sep 10 at 14:20


gammatester
16.1k21529
16.1k21529
asked Sep 10 at 10:42


Wasim Shaikh
1
1
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Wolfram Alpha nows the antiderivate of your PDF. Simply enterint((1-p^2)/(2*Pi*(1+p^2)-2*p*cos(x)), x)
and get a complicated looking expession.
$$frac(p^2 - 1) tanh^-1left(frac(À p^2 + p + pi) tan(x/2)sqrtp^2 - À^2 (p^2 + 1)^2right)sqrtp^2 - pi^2 (p^2 + 1)^2 + c$$
But the dependence on $x$ is in a single $tan(x/2)$ term, so it can be easily inverted.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Wolfram Alpha nows the antiderivate of your PDF. Simply enterint((1-p^2)/(2*Pi*(1+p^2)-2*p*cos(x)), x)
and get a complicated looking expession.
$$frac(p^2 - 1) tanh^-1left(frac(À p^2 + p + pi) tan(x/2)sqrtp^2 - À^2 (p^2 + 1)^2right)sqrtp^2 - pi^2 (p^2 + 1)^2 + c$$
But the dependence on $x$ is in a single $tan(x/2)$ term, so it can be easily inverted.
add a comment |Â
up vote
0
down vote
Wolfram Alpha nows the antiderivate of your PDF. Simply enterint((1-p^2)/(2*Pi*(1+p^2)-2*p*cos(x)), x)
and get a complicated looking expession.
$$frac(p^2 - 1) tanh^-1left(frac(À p^2 + p + pi) tan(x/2)sqrtp^2 - À^2 (p^2 + 1)^2right)sqrtp^2 - pi^2 (p^2 + 1)^2 + c$$
But the dependence on $x$ is in a single $tan(x/2)$ term, so it can be easily inverted.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Wolfram Alpha nows the antiderivate of your PDF. Simply enterint((1-p^2)/(2*Pi*(1+p^2)-2*p*cos(x)), x)
and get a complicated looking expession.
$$frac(p^2 - 1) tanh^-1left(frac(À p^2 + p + pi) tan(x/2)sqrtp^2 - À^2 (p^2 + 1)^2right)sqrtp^2 - pi^2 (p^2 + 1)^2 + c$$
But the dependence on $x$ is in a single $tan(x/2)$ term, so it can be easily inverted.
Wolfram Alpha nows the antiderivate of your PDF. Simply enterint((1-p^2)/(2*Pi*(1+p^2)-2*p*cos(x)), x)
and get a complicated looking expession.
$$frac(p^2 - 1) tanh^-1left(frac(À p^2 + p + pi) tan(x/2)sqrtp^2 - À^2 (p^2 + 1)^2right)sqrtp^2 - pi^2 (p^2 + 1)^2 + c$$
But the dependence on $x$ is in a single $tan(x/2)$ term, so it can be easily inverted.
answered Sep 10 at 14:20


gammatester
16.1k21529
16.1k21529
add a comment |Â
add a comment |Â
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%2f2911792%2fcdf-and-inverse-cdf-of-wrapped-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