Showing that the ratio of two standard independent normals is a Cauchy using Characteristic Functions
$begingroup$
Question
Let $X$ and $Y$ be independent standard normals. Use characteristic functions to find the distribution of $X/Y$.
My attempt
We will attempt to show that $Ee^{itX/Y}=e^{-|t|}$ (the c.f. of a Cauchy random variable) from which the claim will follow. To this end, note that
$$
Ee^{itX/Y}=intfrac{1}{sqrt{2pi}}e^{-y^2/2}int e^{itx/y}frac{1}{sqrt{2pi}}e^{-x^2/2}, dx, dy=int frac{1}{sqrt{2pi}}e^{-y^2/2}expleft(-frac{t^2}{2y^2}right), dy
$$
where we used the fact that $Ee^{it X}=exp(-0.5t^2)$. We can write it as
$$
Ee^{itX/Y}=int frac{1}{sqrt{2pi}}expleft(-frac{1}{2}left[frac{t^2}{y^2}+y^2right]right), dy.
$$
But at this point I don't know how to evaluate the integral. I tried completing the square in the exponent but I couldn't make progress from there.
Any help is appreciated.
probability probability-theory statistics characteristic-functions
asked Dec 9 '18 at 19:18
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Question
Let $X$ and $Y$ be independent standard normals. Use characteristic functions to find the distribution of $X/Y$.
My attempt
We will attempt to show that $Ee^{itX/Y}=e^{-|t|}$ (the c.f. of a Cauchy random variable) from which the claim will follow. To this end, note that
$$
Ee^{itX/Y}=intfrac{1}{sqrt{2pi}}e^{-y^2/2}int e^{itx/y}frac{1}{sqrt{2pi}}e^{-x^2/2}, dx, dy=int frac{1}{sqrt{2pi}}e^{-y^2/2}expleft(-frac{t^2}{2y^2}right), dy
$$
where we used the fact that $Ee^{it X}=exp(-0.5t^2)$. We can write it as
$$
Ee^{itX/Y}=int frac{1}{sqrt{2pi}}expleft(-frac{1}{2}left[frac{t^2}{y^2}+y^2right]right), dy.
$$
But at this point I don't know how to evaluate the integral. I tried completing the square in the exponent but I couldn't make progress from there.
Any help is appreciated.
probability probability-theory statistics characteristic-functions
asked Dec 9 '18 at 19:18
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Question
Let $X$ and $Y$ be independent standard normals. Use characteristic functions to find the distribution of $X/Y$.
My attempt
We will attempt to show that $Ee^{itX/Y}=e^{-|t|}$ (the c.f. of a Cauchy random variable) from which the claim will follow. To this end, note that
$$
Ee^{itX/Y}=intfrac{1}{sqrt{2pi}}e^{-y^2/2}int e^{itx/y}frac{1}{sqrt{2pi}}e^{-x^2/2}, dx, dy=int frac{1}{sqrt{2pi}}e^{-y^2/2}expleft(-frac{t^2}{2y^2}right), dy
$$
where we used the fact that $Ee^{it X}=exp(-0.5t^2)$. We can write it as
$$
Ee^{itX/Y}=int frac{1}{sqrt{2pi}}expleft(-frac{1}{2}left[frac{t^2}{y^2}+y^2right]right), dy.
$$
But at this point I don't know how to evaluate the integral. I tried completing the square in the exponent but I couldn't make progress from there.
Any help is appreciated.
probability probability-theory statistics characteristic-functions
asked Dec 9 '18 at 19:18
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
Question
Let $X$ and $Y$ be independent standard normals. Use characteristic functions to find the distribution of $X/Y$.
My attempt
We will attempt to show that $Ee^{itX/Y}=e^{-|t|}$ (the c.f. of a Cauchy random variable) from which the claim will follow. To this end, note that
$$
Ee^{itX/Y}=intfrac{1}{sqrt{2pi}}e^{-y^2/2}int e^{itx/y}frac{1}{sqrt{2pi}}e^{-x^2/2}, dx, dy=int frac{1}{sqrt{2pi}}e^{-y^2/2}expleft(-frac{t^2}{2y^2}right), dy
$$
where we used the fact that $Ee^{it X}=exp(-0.5t^2)$. We can write it as
$$
Ee^{itX/Y}=int frac{1}{sqrt{2pi}}expleft(-frac{1}{2}left[frac{t^2}{y^2}+y^2right]right), dy.
$$
But at this point I don't know how to evaluate the integral. I tried completing the square in the exponent but I couldn't make progress from there.
Any help is appreciated.
probability probability-theory statistics characteristic-functions
probability probability-theory statistics characteristic-functions
asked Dec 9 '18 at 19:18
Foobaz JohnFoobaz John
22.1k41352
asked Dec 9 '18 at 19:18
Foobaz JohnFoobaz John
22.1k41352
asked Dec 9 '18 at 19:18
Foobaz JohnFoobaz John
22.1k41352
asked Dec 9 '18 at 19:18
Foobaz JohnFoobaz John
22.1k41352
asked Dec 9 '18 at 19:18
Foobaz JohnFoobaz John
22.1k41352
22.1k41352
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Set $phi(t):= mathbb{E}exp(itX/Y)$. Since $phi(0)=1$ and $phi$ is even, it suffices to show that $phi(t)=e^{-t}$ for all $t>0$.
Fix $t>0$. Following your calculations we have
$$phi(t) = sqrt{frac{2}{pi}} int_{(0,infty)} exp left(- frac{1}{2} left[ frac{t^2}{y^2}+y^2 right] right) , dy tag{1}$$
Performing a change of variables, $z:=t/y$, we find
$$phi(t) = t sqrt{frac{2}{pi} } int_{(0,infty)}frac{1}{z^2} exp left(- frac{1}{2} left[ z^2 + frac{t^2}{z^2} right] right) , dz stackrel{(1)}{=} - phi'(t).$$
This shows that $phi$ solves the ODE $$phi' = - phi quad text{on $(0,infty)$}.$$
Since $phi(0)=1$ we conclude that $phi(t)=e^{-t}$ for $t geq 0$.
answered Dec 9 '18 at 20:34
sazsaz
80.6k860125
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add a comment |
$begingroup$
Switch to polar coordinates. It is slightly tidier to evaluate $E(e^{iY/X})$, which is expressible by symmetry as
$$
E(e^{iY/X})=2int_{x=0}^inftyint_{y=-infty}^infty
e^{ity/x}frac1{2pi}e^{-(x^2+y^2)/2},dy,dx.
$$
In polar coordinates this equals
$$
2int_{theta=-pi/2}^{pi/2}int_{r=0}^infty e^{ittan theta}frac1{2pi}e^{-r^2/2}r,dr,dtheta=int_{theta=-pi/2}^{pi/2}frac1{pi}e^{ittantheta}dtheta.
$$
Apply the substitution $u=tantheta$, $du=sec^2theta, dtheta=(1+u^2)dtheta$ to obtain
$$
int_{u=-infty}^inftyfrac1pifrac{e^{itu}}{1+u^2}du.
$$
At this point you can quit, because you've just written down the characteristic function of a Cauchy density.
answered Dec 12 '18 at 23:43
grand_chatgrand_chat
20.3k11326
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
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$begingroup$
Set $phi(t):= mathbb{E}exp(itX/Y)$. Since $phi(0)=1$ and $phi$ is even, it suffices to show that $phi(t)=e^{-t}$ for all $t>0$.
Fix $t>0$. Following your calculations we have
$$phi(t) = sqrt{frac{2}{pi}} int_{(0,infty)} exp left(- frac{1}{2} left[ frac{t^2}{y^2}+y^2 right] right) , dy tag{1}$$
Performing a change of variables, $z:=t/y$, we find
$$phi(t) = t sqrt{frac{2}{pi} } int_{(0,infty)}frac{1}{z^2} exp left(- frac{1}{2} left[ z^2 + frac{t^2}{z^2} right] right) , dz stackrel{(1)}{=} - phi'(t).$$
This shows that $phi$ solves the ODE $$phi' = - phi quad text{on $(0,infty)$}.$$
Since $phi(0)=1$ we conclude that $phi(t)=e^{-t}$ for $t geq 0$.
answered Dec 9 '18 at 20:34
sazsaz
80.6k860125
$endgroup$
add a comment |
$begingroup$
Set $phi(t):= mathbb{E}exp(itX/Y)$. Since $phi(0)=1$ and $phi$ is even, it suffices to show that $phi(t)=e^{-t}$ for all $t>0$.
Fix $t>0$. Following your calculations we have
$$phi(t) = sqrt{frac{2}{pi}} int_{(0,infty)} exp left(- frac{1}{2} left[ frac{t^2}{y^2}+y^2 right] right) , dy tag{1}$$
Performing a change of variables, $z:=t/y$, we find
$$phi(t) = t sqrt{frac{2}{pi} } int_{(0,infty)}frac{1}{z^2} exp left(- frac{1}{2} left[ z^2 + frac{t^2}{z^2} right] right) , dz stackrel{(1)}{=} - phi'(t).$$
This shows that $phi$ solves the ODE $$phi' = - phi quad text{on $(0,infty)$}.$$
Since $phi(0)=1$ we conclude that $phi(t)=e^{-t}$ for $t geq 0$.
answered Dec 9 '18 at 20:34
sazsaz
80.6k860125
$endgroup$
add a comment |
$begingroup$
Set $phi(t):= mathbb{E}exp(itX/Y)$. Since $phi(0)=1$ and $phi$ is even, it suffices to show that $phi(t)=e^{-t}$ for all $t>0$.
Fix $t>0$. Following your calculations we have
$$phi(t) = sqrt{frac{2}{pi}} int_{(0,infty)} exp left(- frac{1}{2} left[ frac{t^2}{y^2}+y^2 right] right) , dy tag{1}$$
Performing a change of variables, $z:=t/y$, we find
$$phi(t) = t sqrt{frac{2}{pi} } int_{(0,infty)}frac{1}{z^2} exp left(- frac{1}{2} left[ z^2 + frac{t^2}{z^2} right] right) , dz stackrel{(1)}{=} - phi'(t).$$
This shows that $phi$ solves the ODE $$phi' = - phi quad text{on $(0,infty)$}.$$
Since $phi(0)=1$ we conclude that $phi(t)=e^{-t}$ for $t geq 0$.
answered Dec 9 '18 at 20:34
sazsaz
80.6k860125
$endgroup$
Set $phi(t):= mathbb{E}exp(itX/Y)$. Since $phi(0)=1$ and $phi$ is even, it suffices to show that $phi(t)=e^{-t}$ for all $t>0$.
Fix $t>0$. Following your calculations we have
$$phi(t) = sqrt{frac{2}{pi}} int_{(0,infty)} exp left(- frac{1}{2} left[ frac{t^2}{y^2}+y^2 right] right) , dy tag{1}$$
Performing a change of variables, $z:=t/y$, we find
$$phi(t) = t sqrt{frac{2}{pi} } int_{(0,infty)}frac{1}{z^2} exp left(- frac{1}{2} left[ z^2 + frac{t^2}{z^2} right] right) , dz stackrel{(1)}{=} - phi'(t).$$
This shows that $phi$ solves the ODE $$phi' = - phi quad text{on $(0,infty)$}.$$
Since $phi(0)=1$ we conclude that $phi(t)=e^{-t}$ for $t geq 0$.
answered Dec 9 '18 at 20:34
sazsaz
80.6k860125
edited Dec 9 '18 at 20:49
answered Dec 9 '18 at 20:34
sazsaz
80.6k860125
answered Dec 9 '18 at 20:34
sazsaz
80.6k860125
answered Dec 9 '18 at 20:34
sazsaz
80.6k860125
80.6k860125
add a comment |
add a comment |
$begingroup$
Switch to polar coordinates. It is slightly tidier to evaluate $E(e^{iY/X})$, which is expressible by symmetry as
$$
E(e^{iY/X})=2int_{x=0}^inftyint_{y=-infty}^infty
e^{ity/x}frac1{2pi}e^{-(x^2+y^2)/2},dy,dx.
$$
In polar coordinates this equals
$$
2int_{theta=-pi/2}^{pi/2}int_{r=0}^infty e^{ittan theta}frac1{2pi}e^{-r^2/2}r,dr,dtheta=int_{theta=-pi/2}^{pi/2}frac1{pi}e^{ittantheta}dtheta.
$$
Apply the substitution $u=tantheta$, $du=sec^2theta, dtheta=(1+u^2)dtheta$ to obtain
$$
int_{u=-infty}^inftyfrac1pifrac{e^{itu}}{1+u^2}du.
$$
At this point you can quit, because you've just written down the characteristic function of a Cauchy density.
answered Dec 12 '18 at 23:43
grand_chatgrand_chat
20.3k11326
$endgroup$
add a comment |
$begingroup$
Switch to polar coordinates. It is slightly tidier to evaluate $E(e^{iY/X})$, which is expressible by symmetry as
$$
E(e^{iY/X})=2int_{x=0}^inftyint_{y=-infty}^infty
e^{ity/x}frac1{2pi}e^{-(x^2+y^2)/2},dy,dx.
$$
In polar coordinates this equals
$$
2int_{theta=-pi/2}^{pi/2}int_{r=0}^infty e^{ittan theta}frac1{2pi}e^{-r^2/2}r,dr,dtheta=int_{theta=-pi/2}^{pi/2}frac1{pi}e^{ittantheta}dtheta.
$$
Apply the substitution $u=tantheta$, $du=sec^2theta, dtheta=(1+u^2)dtheta$ to obtain
$$
int_{u=-infty}^inftyfrac1pifrac{e^{itu}}{1+u^2}du.
$$
At this point you can quit, because you've just written down the characteristic function of a Cauchy density.
answered Dec 12 '18 at 23:43
grand_chatgrand_chat
20.3k11326
$endgroup$
add a comment |
$begingroup$
Switch to polar coordinates. It is slightly tidier to evaluate $E(e^{iY/X})$, which is expressible by symmetry as
$$
E(e^{iY/X})=2int_{x=0}^inftyint_{y=-infty}^infty
e^{ity/x}frac1{2pi}e^{-(x^2+y^2)/2},dy,dx.
$$
In polar coordinates this equals
$$
2int_{theta=-pi/2}^{pi/2}int_{r=0}^infty e^{ittan theta}frac1{2pi}e^{-r^2/2}r,dr,dtheta=int_{theta=-pi/2}^{pi/2}frac1{pi}e^{ittantheta}dtheta.
$$
Apply the substitution $u=tantheta$, $du=sec^2theta, dtheta=(1+u^2)dtheta$ to obtain
$$
int_{u=-infty}^inftyfrac1pifrac{e^{itu}}{1+u^2}du.
$$
At this point you can quit, because you've just written down the characteristic function of a Cauchy density.
answered Dec 12 '18 at 23:43
grand_chatgrand_chat
20.3k11326
$endgroup$
Switch to polar coordinates. It is slightly tidier to evaluate $E(e^{iY/X})$, which is expressible by symmetry as
$$
E(e^{iY/X})=2int_{x=0}^inftyint_{y=-infty}^infty
e^{ity/x}frac1{2pi}e^{-(x^2+y^2)/2},dy,dx.
$$
In polar coordinates this equals
$$
2int_{theta=-pi/2}^{pi/2}int_{r=0}^infty e^{ittan theta}frac1{2pi}e^{-r^2/2}r,dr,dtheta=int_{theta=-pi/2}^{pi/2}frac1{pi}e^{ittantheta}dtheta.
$$
Apply the substitution $u=tantheta$, $du=sec^2theta, dtheta=(1+u^2)dtheta$ to obtain
$$
int_{u=-infty}^inftyfrac1pifrac{e^{itu}}{1+u^2}du.
$$
At this point you can quit, because you've just written down the characteristic function of a Cauchy density.
answered Dec 12 '18 at 23:43
grand_chatgrand_chat
20.3k11326
answered Dec 12 '18 at 23:43
grand_chatgrand_chat
20.3k11326
answered Dec 12 '18 at 23:43
grand_chatgrand_chat
20.3k11326
answered Dec 12 '18 at 23:43
grand_chatgrand_chat
20.3k11326
20.3k11326
add a comment |
add a comment |
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