Mixing
Proving Cauchy random variables
Trying to prove that if a random variable $T$ has Cauchy distribution
with probability density function: $$f(x)= frac{1}{pi(1+x^2)}$$
then $X = frac{1}{T}$
and $Y = frac{2T}{1-T^2}$ are also Cauchy random variables.
Now, I realize that this specific form of PDF is called a Standard Cauchy distribution and if I am correct then $T$ is of the form:
$$ frac{1}{e^{|t|}} $$
My main problem here is that I don't understand if i should use the characteristic function of $T$ to prove that $X$ and $Y$ are Cauchy or use it's PDF.
I apologize if this question is too simple, just this problem got me confused with the Cauchy distribution and I want to get the right picture of how to approach this type of questions, any hint greatly appreciated.
probability probability-distributions density-function
asked Nov 20 '18 at 18:57
JKM
6415
add a comment |
Trying to prove that if a random variable $T$ has Cauchy distribution
with probability density function: $$f(x)= frac{1}{pi(1+x^2)}$$
then $X = frac{1}{T}$
and $Y = frac{2T}{1-T^2}$ are also Cauchy random variables.
Now, I realize that this specific form of PDF is called a Standard Cauchy distribution and if I am correct then $T$ is of the form:
$$ frac{1}{e^{|t|}} $$
My main problem here is that I don't understand if i should use the characteristic function of $T$ to prove that $X$ and $Y$ are Cauchy or use it's PDF.
I apologize if this question is too simple, just this problem got me confused with the Cauchy distribution and I want to get the right picture of how to approach this type of questions, any hint greatly appreciated.
probability probability-distributions density-function
asked Nov 20 '18 at 18:57
JKM
6415
2
The change of variables formula might help computing the PDF of $f(X)$, given the PDF of $X$...
– Federico
Nov 20 '18 at 19:02
2
Easier than Federico's suggestion: Cauchy distribution = tan(random angle). Then consider some transformations of the random angle (multiply by something, subtract from something...)
– zhoraster
Nov 20 '18 at 19:05
1
The first question is answered here. The second one is answered here, elaborating the argument of the answer below and the comment above.
– StubbornAtom
Nov 20 '18 at 19:48
add a comment |
Trying to prove that if a random variable $T$ has Cauchy distribution
with probability density function: $$f(x)= frac{1}{pi(1+x^2)}$$
then $X = frac{1}{T}$
and $Y = frac{2T}{1-T^2}$ are also Cauchy random variables.
Now, I realize that this specific form of PDF is called a Standard Cauchy distribution and if I am correct then $T$ is of the form:
$$ frac{1}{e^{|t|}} $$
My main problem here is that I don't understand if i should use the characteristic function of $T$ to prove that $X$ and $Y$ are Cauchy or use it's PDF.
I apologize if this question is too simple, just this problem got me confused with the Cauchy distribution and I want to get the right picture of how to approach this type of questions, any hint greatly appreciated.
probability probability-distributions density-function
asked Nov 20 '18 at 18:57
JKM
6415
Trying to prove that if a random variable $T$ has Cauchy distribution
with probability density function: $$f(x)= frac{1}{pi(1+x^2)}$$
then $X = frac{1}{T}$
and $Y = frac{2T}{1-T^2}$ are also Cauchy random variables.
Now, I realize that this specific form of PDF is called a Standard Cauchy distribution and if I am correct then $T$ is of the form:
$$ frac{1}{e^{|t|}} $$
My main problem here is that I don't understand if i should use the characteristic function of $T$ to prove that $X$ and $Y$ are Cauchy or use it's PDF.
I apologize if this question is too simple, just this problem got me confused with the Cauchy distribution and I want to get the right picture of how to approach this type of questions, any hint greatly appreciated.
probability probability-distributions density-function
probability probability-distributions density-function
asked Nov 20 '18 at 18:57
JKM
6415
asked Nov 20 '18 at 18:57
JKM
6415
asked Nov 20 '18 at 18:57
JKM
6415
asked Nov 20 '18 at 18:57
JKM
6415
asked Nov 20 '18 at 18:57
JKM
6415
6415
2
The change of variables formula might help computing the PDF of $f(X)$, given the PDF of $X$...
– Federico
Nov 20 '18 at 19:02
2
Easier than Federico's suggestion: Cauchy distribution = tan(random angle). Then consider some transformations of the random angle (multiply by something, subtract from something...)
– zhoraster
Nov 20 '18 at 19:05
1
The first question is answered here. The second one is answered here, elaborating the argument of the answer below and the comment above.
– StubbornAtom
Nov 20 '18 at 19:48
add a comment |
2
The change of variables formula might help computing the PDF of $f(X)$, given the PDF of $X$...
– Federico
Nov 20 '18 at 19:02
2
Easier than Federico's suggestion: Cauchy distribution = tan(random angle). Then consider some transformations of the random angle (multiply by something, subtract from something...)
– zhoraster
Nov 20 '18 at 19:05
1
The first question is answered here. The second one is answered here, elaborating the argument of the answer below and the comment above.
– StubbornAtom
Nov 20 '18 at 19:48
2
2
The change of variables formula might help computing the PDF of $f(X)$, given the PDF of $X$...
– Federico
Nov 20 '18 at 19:02
The change of variables formula might help computing the PDF of $f(X)$, given the PDF of $X$...
– Federico
Nov 20 '18 at 19:02
2
2
Easier than Federico's suggestion: Cauchy distribution = tan(random angle). Then consider some transformations of the random angle (multiply by something, subtract from something...)
– zhoraster
Nov 20 '18 at 19:05
Easier than Federico's suggestion: Cauchy distribution = tan(random angle). Then consider some transformations of the random angle (multiply by something, subtract from something...)
– zhoraster
Nov 20 '18 at 19:05
1
1
The first question is answered here. The second one is answered here, elaborating the argument of the answer below and the comment above.
– StubbornAtom
Nov 20 '18 at 19:48
The first question is answered here. The second one is answered here, elaborating the argument of the answer below and the comment above.
– StubbornAtom
Nov 20 '18 at 19:48
add a comment |
1 Answer
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oldest
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One way of obtaining a Cauchy random variable is as $T=tan U$
where $U$ is a uniform random variable on the interval $(-pi/2,pi/2)$.
Then $X=cot U=tan(pi/2-U)$ and $Y=tan 2U$. Now $pi/2-U$ is uniform
on $(0,pi)$. Thinking of this modulo $pi$, it's essentially the same as
the distribution of $U$. Likewise $2U$ is uniform on $(-pi,pi)$.
Again modulo $pi$, that's essentially the same distribution as $U$.
answered Nov 20 '18 at 19:07
Lord Shark the Unknown
101k958132
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
One way of obtaining a Cauchy random variable is as $T=tan U$
where $U$ is a uniform random variable on the interval $(-pi/2,pi/2)$.
Then $X=cot U=tan(pi/2-U)$ and $Y=tan 2U$. Now $pi/2-U$ is uniform
on $(0,pi)$. Thinking of this modulo $pi$, it's essentially the same as
the distribution of $U$. Likewise $2U$ is uniform on $(-pi,pi)$.
Again modulo $pi$, that's essentially the same distribution as $U$.
answered Nov 20 '18 at 19:07
Lord Shark the Unknown
101k958132
add a comment |
One way of obtaining a Cauchy random variable is as $T=tan U$
where $U$ is a uniform random variable on the interval $(-pi/2,pi/2)$.
Then $X=cot U=tan(pi/2-U)$ and $Y=tan 2U$. Now $pi/2-U$ is uniform
on $(0,pi)$. Thinking of this modulo $pi$, it's essentially the same as
the distribution of $U$. Likewise $2U$ is uniform on $(-pi,pi)$.
Again modulo $pi$, that's essentially the same distribution as $U$.
answered Nov 20 '18 at 19:07
Lord Shark the Unknown
101k958132
add a comment |
One way of obtaining a Cauchy random variable is as $T=tan U$
where $U$ is a uniform random variable on the interval $(-pi/2,pi/2)$.
Then $X=cot U=tan(pi/2-U)$ and $Y=tan 2U$. Now $pi/2-U$ is uniform
on $(0,pi)$. Thinking of this modulo $pi$, it's essentially the same as
the distribution of $U$. Likewise $2U$ is uniform on $(-pi,pi)$.
Again modulo $pi$, that's essentially the same distribution as $U$.
answered Nov 20 '18 at 19:07
Lord Shark the Unknown
101k958132
One way of obtaining a Cauchy random variable is as $T=tan U$
where $U$ is a uniform random variable on the interval $(-pi/2,pi/2)$.
Then $X=cot U=tan(pi/2-U)$ and $Y=tan 2U$. Now $pi/2-U$ is uniform
on $(0,pi)$. Thinking of this modulo $pi$, it's essentially the same as
the distribution of $U$. Likewise $2U$ is uniform on $(-pi,pi)$.
Again modulo $pi$, that's essentially the same distribution as $U$.
answered Nov 20 '18 at 19:07
Lord Shark the Unknown
101k958132
answered Nov 20 '18 at 19:07
Lord Shark the Unknown
101k958132
answered Nov 20 '18 at 19:07
Lord Shark the Unknown
101k958132
answered Nov 20 '18 at 19:07
Lord Shark the Unknown
101k958132
101k958132
add a comment |
add a comment |
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2
The change of variables formula might help computing the PDF of $f(X)$, given the PDF of $X$...
– Federico
Nov 20 '18 at 19:02
2
Easier than Federico's suggestion: Cauchy distribution = tan(random angle). Then consider some transformations of the random angle (multiply by something, subtract from something...)
– zhoraster
Nov 20 '18 at 19:05
1
The first question is answered here. The second one is answered here, elaborating the argument of the answer below and the comment above.
– StubbornAtom
Nov 20 '18 at 19:48