Is $F(x)=frac1pitan^-1(x),-infty<x<infty$ a distribution function?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Check whether $F(x)=frac1pitan^-1(x),-infty<x<infty$ is a distribution function?
What I attempted:- There are three conditions for a function $F(x)$ to be a distribution function.
beginequation
beginaligned
&(1)mboxIt should be non-decreasing\
&(2)mboxIt should be right continuous \
&(3)mbox$F(-infty)=0$ and $F(infty)=1$
endaligned
endequation
$(2)$ $tan^-1x$ is continuous over the real line.
$(1)$ $F'(x)=frac1pifrac11+x^2>0 quad forall xin R$
$(3)$ $F(infty)=lim_xto inftyfrac1pitan^-1x=frac1pitimes fracpi2=frac12not=1$. Am I correct?
Thus it couldn't be a distribution function.
probability-theory
edited Aug 12 at 8:07
pointguard0
1,271821
asked Aug 12 at 7:44
Bhargob
456214
add a comment |Â
up vote
0
down vote
favorite
Check whether $F(x)=frac1pitan^-1(x),-infty<x<infty$ is a distribution function?
What I attempted:- There are three conditions for a function $F(x)$ to be a distribution function.
beginequation
beginaligned
&(1)mboxIt should be non-decreasing\
&(2)mboxIt should be right continuous \
&(3)mbox$F(-infty)=0$ and $F(infty)=1$
endaligned
endequation
$(2)$ $tan^-1x$ is continuous over the real line.
$(1)$ $F'(x)=frac1pifrac11+x^2>0 quad forall xin R$
$(3)$ $F(infty)=lim_xto inftyfrac1pitan^-1x=frac1pitimes fracpi2=frac12not=1$. Am I correct?
Thus it couldn't be a distribution function.
probability-theory
edited Aug 12 at 8:07
pointguard0
1,271821
asked Aug 12 at 7:44
Bhargob
456214
Sure you were not asked about $F(x)=frac12+frac1pitan^-1x$ instead?
– Did
Aug 12 at 15:18
Yes, the question was asked about that function only which I presented here. Of course, the function that you have revealed is a distribution function of the Cauchy Distribution.
– Bhargob
Aug 12 at 18:17
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Check whether $F(x)=frac1pitan^-1(x),-infty<x<infty$ is a distribution function?
What I attempted:- There are three conditions for a function $F(x)$ to be a distribution function.
beginequation
beginaligned
&(1)mboxIt should be non-decreasing\
&(2)mboxIt should be right continuous \
&(3)mbox$F(-infty)=0$ and $F(infty)=1$
endaligned
endequation
$(2)$ $tan^-1x$ is continuous over the real line.
$(1)$ $F'(x)=frac1pifrac11+x^2>0 quad forall xin R$
$(3)$ $F(infty)=lim_xto inftyfrac1pitan^-1x=frac1pitimes fracpi2=frac12not=1$. Am I correct?
Thus it couldn't be a distribution function.
probability-theory
edited Aug 12 at 8:07
pointguard0
1,271821
asked Aug 12 at 7:44
Bhargob
456214
Check whether $F(x)=frac1pitan^-1(x),-infty<x<infty$ is a distribution function?
What I attempted:- There are three conditions for a function $F(x)$ to be a distribution function.
beginequation
beginaligned
&(1)mboxIt should be non-decreasing\
&(2)mboxIt should be right continuous \
&(3)mbox$F(-infty)=0$ and $F(infty)=1$
endaligned
endequation
$(2)$ $tan^-1x$ is continuous over the real line.
$(1)$ $F'(x)=frac1pifrac11+x^2>0 quad forall xin R$
$(3)$ $F(infty)=lim_xto inftyfrac1pitan^-1x=frac1pitimes fracpi2=frac12not=1$. Am I correct?
Thus it couldn't be a distribution function.
probability-theory
edited Aug 12 at 8:07
pointguard0
1,271821
asked Aug 12 at 7:44
Bhargob
456214
edited Aug 12 at 8:07
pointguard0
1,271821
edited Aug 12 at 8:07
pointguard0
1,271821
edited Aug 12 at 8:07
pointguard0
1,271821
1,271821
asked Aug 12 at 7:44
Bhargob
456214
asked Aug 12 at 7:44
Bhargob
456214
asked Aug 12 at 7:44
Bhargob
456214
456214
Sure you were not asked about $F(x)=frac12+frac1pitan^-1x$ instead?
– Did
Aug 12 at 15:18
Yes, the question was asked about that function only which I presented here. Of course, the function that you have revealed is a distribution function of the Cauchy Distribution.
– Bhargob
Aug 12 at 18:17
add a comment |Â
Sure you were not asked about $F(x)=frac12+frac1pitan^-1x$ instead?
– Did
Aug 12 at 15:18
Yes, the question was asked about that function only which I presented here. Of course, the function that you have revealed is a distribution function of the Cauchy Distribution.
– Bhargob
Aug 12 at 18:17
Sure you were not asked about $F(x)=frac12+frac1pitan^-1x$ instead?
– Did
Aug 12 at 15:18
Sure you were not asked about $F(x)=frac12+frac1pitan^-1x$ instead?
– Did
Aug 12 at 15:18
Yes, the question was asked about that function only which I presented here. Of course, the function that you have revealed is a distribution function of the Cauchy Distribution.
– Bhargob
Aug 12 at 18:17
Yes, the question was asked about that function only which I presented here. Of course, the function that you have revealed is a distribution function of the Cauchy Distribution.
– Bhargob
Aug 12 at 18:17
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Yes, you are right, it fails the condition that $F(infty)=1$, hence it can't be a CDF.
Also, $F$ takes negative values at negative $x$, for example, $F(-1)<0$.
answered Aug 12 at 7:53
Siong Thye Goh
78.7k134997
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, you are right, it fails the condition that $F(infty)=1$, hence it can't be a CDF.
Also, $F$ takes negative values at negative $x$, for example, $F(-1)<0$.
answered Aug 12 at 7:53
Siong Thye Goh
78.7k134997
add a comment |Â
up vote
2
down vote
accepted
Yes, you are right, it fails the condition that $F(infty)=1$, hence it can't be a CDF.
Also, $F$ takes negative values at negative $x$, for example, $F(-1)<0$.
answered Aug 12 at 7:53
Siong Thye Goh
78.7k134997
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, you are right, it fails the condition that $F(infty)=1$, hence it can't be a CDF.
Also, $F$ takes negative values at negative $x$, for example, $F(-1)<0$.
answered Aug 12 at 7:53
Siong Thye Goh
78.7k134997
Yes, you are right, it fails the condition that $F(infty)=1$, hence it can't be a CDF.
Also, $F$ takes negative values at negative $x$, for example, $F(-1)<0$.
answered Aug 12 at 7:53
Siong Thye Goh
78.7k134997
answered Aug 12 at 7:53
Siong Thye Goh
78.7k134997
answered Aug 12 at 7:53
Siong Thye Goh
78.7k134997
answered Aug 12 at 7:53
Siong Thye Goh
78.7k134997
78.7k134997
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%2f2880073%2fis-fx-frac1-pi-tan-1x-inftyx-infty-a-distribution-function%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
Sure you were not asked about $F(x)=frac12+frac1pitan^-1x$ instead?
– Did
Aug 12 at 15:18
Yes, the question was asked about that function only which I presented here. Of course, the function that you have revealed is a distribution function of the Cauchy Distribution.
– Bhargob
Aug 12 at 18:17