If $f_X$ is a pdf of the r.v. $X$ what would be $g(s)=int_{mathbb R}f_X(x)delta (s-x)dx$?
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Let $X$ a real r.v. and $f_X$ it's pdf. What would be $$g(s)=int_{mathbb R}f_X(x)delta (s-x),mathrm d x,$$
where $delta $ is the Dirac distribution ? It look to be a pdf of some r.v. but which one ?
The problem is exactly as follow : Let $X$ and $Y$ two $mathcal N(0,1)$ r.v. with join pdf $f_{X,Y}$ (they are not supposed independent). Let $s=|x-y|$. Consider $$g(s)=int_{mathbb R^2}f_{X,Y}(x,y)delta (s-|x-y|),mathrm d x,mathrm d y.$$
What is exactely $g(s)$ ?
probability
asked Dec 12 '18 at 11:19
user623855user623855
1507
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add a comment |
$begingroup$
Let $X$ a real r.v. and $f_X$ it's pdf. What would be $$g(s)=int_{mathbb R}f_X(x)delta (s-x),mathrm d x,$$
where $delta $ is the Dirac distribution ? It look to be a pdf of some r.v. but which one ?
The problem is exactly as follow : Let $X$ and $Y$ two $mathcal N(0,1)$ r.v. with join pdf $f_{X,Y}$ (they are not supposed independent). Let $s=|x-y|$. Consider $$g(s)=int_{mathbb R^2}f_{X,Y}(x,y)delta (s-|x-y|),mathrm d x,mathrm d y.$$
What is exactely $g(s)$ ?
probability
asked Dec 12 '18 at 11:19
user623855user623855
1507
$endgroup$
add a comment |
$begingroup$
Let $X$ a real r.v. and $f_X$ it's pdf. What would be $$g(s)=int_{mathbb R}f_X(x)delta (s-x),mathrm d x,$$
where $delta $ is the Dirac distribution ? It look to be a pdf of some r.v. but which one ?
The problem is exactly as follow : Let $X$ and $Y$ two $mathcal N(0,1)$ r.v. with join pdf $f_{X,Y}$ (they are not supposed independent). Let $s=|x-y|$. Consider $$g(s)=int_{mathbb R^2}f_{X,Y}(x,y)delta (s-|x-y|),mathrm d x,mathrm d y.$$
What is exactely $g(s)$ ?
probability
asked Dec 12 '18 at 11:19
user623855user623855
1507
$endgroup$
Let $X$ a real r.v. and $f_X$ it's pdf. What would be $$g(s)=int_{mathbb R}f_X(x)delta (s-x),mathrm d x,$$
where $delta $ is the Dirac distribution ? It look to be a pdf of some r.v. but which one ?
The problem is exactly as follow : Let $X$ and $Y$ two $mathcal N(0,1)$ r.v. with join pdf $f_{X,Y}$ (they are not supposed independent). Let $s=|x-y|$. Consider $$g(s)=int_{mathbb R^2}f_{X,Y}(x,y)delta (s-|x-y|),mathrm d x,mathrm d y.$$
What is exactely $g(s)$ ?
probability
probability
asked Dec 12 '18 at 11:19
user623855user623855
1507
asked Dec 12 '18 at 11:19
user623855user623855
1507
asked Dec 12 '18 at 11:19
user623855user623855
1507
asked Dec 12 '18 at 11:19
user623855user623855
1507
asked Dec 12 '18 at 11:19
user623855user623855
1507
1507
add a comment |
add a comment |
1 Answer
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Let $phi$ denote the $mathcal{N}(0,,1)$ pdf. Since $X-Ysimmathcal{N}(0,,2)$ has pdf $tfrac{1}{sqrt{2}}phi(tfrac{x}{sqrt{2}})$, $S:=|X-Y|$ has pdf $g(s)=sqrt{2}phi(tfrac{s}{sqrt{2}})=frac{1}{sqrt{pi}}exp-tfrac{s^2}{4}$ on $Bbb R^+$.
answered Dec 12 '18 at 11:36
J.G.J.G.
26.4k22541
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$begingroup$
in what it answer to my question ?
$endgroup$
– user623855
Dec 12 '18 at 11:54
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@user623855 See my edit.
$endgroup$
– J.G.
Dec 12 '18 at 11:55
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Can we say in general that $f(X_1,...,X_n)$ has pdf $$g(s)=int_{mathbb R^n}f_{X_1,...,X_n}(x_1,...,x_n)delta (s-f(x_1,...,x_n))dx_1...dx_n ?$$
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– user623855
Dec 12 '18 at 11:58
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@user623855 Yes. PMFs of functions of discrete variables have a similar form.
$endgroup$
– J.G.
Dec 12 '18 at 12:00
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $phi$ denote the $mathcal{N}(0,,1)$ pdf. Since $X-Ysimmathcal{N}(0,,2)$ has pdf $tfrac{1}{sqrt{2}}phi(tfrac{x}{sqrt{2}})$, $S:=|X-Y|$ has pdf $g(s)=sqrt{2}phi(tfrac{s}{sqrt{2}})=frac{1}{sqrt{pi}}exp-tfrac{s^2}{4}$ on $Bbb R^+$.
answered Dec 12 '18 at 11:36
J.G.J.G.
26.4k22541
$endgroup$
$begingroup$
in what it answer to my question ?
$endgroup$
– user623855
Dec 12 '18 at 11:54
$begingroup$
@user623855 See my edit.
$endgroup$
– J.G.
Dec 12 '18 at 11:55
$begingroup$
Can we say in general that $f(X_1,...,X_n)$ has pdf $$g(s)=int_{mathbb R^n}f_{X_1,...,X_n}(x_1,...,x_n)delta (s-f(x_1,...,x_n))dx_1...dx_n ?$$
$endgroup$
– user623855
Dec 12 '18 at 11:58
$begingroup$
@user623855 Yes. PMFs of functions of discrete variables have a similar form.
$endgroup$
– J.G.
Dec 12 '18 at 12:00
add a comment |
$begingroup$
Let $phi$ denote the $mathcal{N}(0,,1)$ pdf. Since $X-Ysimmathcal{N}(0,,2)$ has pdf $tfrac{1}{sqrt{2}}phi(tfrac{x}{sqrt{2}})$, $S:=|X-Y|$ has pdf $g(s)=sqrt{2}phi(tfrac{s}{sqrt{2}})=frac{1}{sqrt{pi}}exp-tfrac{s^2}{4}$ on $Bbb R^+$.
answered Dec 12 '18 at 11:36
J.G.J.G.
26.4k22541
$endgroup$
$begingroup$
in what it answer to my question ?
$endgroup$
– user623855
Dec 12 '18 at 11:54
$begingroup$
@user623855 See my edit.
$endgroup$
– J.G.
Dec 12 '18 at 11:55
$begingroup$
Can we say in general that $f(X_1,...,X_n)$ has pdf $$g(s)=int_{mathbb R^n}f_{X_1,...,X_n}(x_1,...,x_n)delta (s-f(x_1,...,x_n))dx_1...dx_n ?$$
$endgroup$
– user623855
Dec 12 '18 at 11:58
$begingroup$
@user623855 Yes. PMFs of functions of discrete variables have a similar form.
$endgroup$
– J.G.
Dec 12 '18 at 12:00
add a comment |
$begingroup$
Let $phi$ denote the $mathcal{N}(0,,1)$ pdf. Since $X-Ysimmathcal{N}(0,,2)$ has pdf $tfrac{1}{sqrt{2}}phi(tfrac{x}{sqrt{2}})$, $S:=|X-Y|$ has pdf $g(s)=sqrt{2}phi(tfrac{s}{sqrt{2}})=frac{1}{sqrt{pi}}exp-tfrac{s^2}{4}$ on $Bbb R^+$.
answered Dec 12 '18 at 11:36
J.G.J.G.
26.4k22541
$endgroup$
Let $phi$ denote the $mathcal{N}(0,,1)$ pdf. Since $X-Ysimmathcal{N}(0,,2)$ has pdf $tfrac{1}{sqrt{2}}phi(tfrac{x}{sqrt{2}})$, $S:=|X-Y|$ has pdf $g(s)=sqrt{2}phi(tfrac{s}{sqrt{2}})=frac{1}{sqrt{pi}}exp-tfrac{s^2}{4}$ on $Bbb R^+$.
answered Dec 12 '18 at 11:36
J.G.J.G.
26.4k22541
edited Dec 12 '18 at 11:55
answered Dec 12 '18 at 11:36
J.G.J.G.
26.4k22541
answered Dec 12 '18 at 11:36
J.G.J.G.
26.4k22541
answered Dec 12 '18 at 11:36
J.G.J.G.
26.4k22541
26.4k22541
$begingroup$
in what it answer to my question ?
$endgroup$
– user623855
Dec 12 '18 at 11:54
$begingroup$
@user623855 See my edit.
$endgroup$
– J.G.
Dec 12 '18 at 11:55
$begingroup$
Can we say in general that $f(X_1,...,X_n)$ has pdf $$g(s)=int_{mathbb R^n}f_{X_1,...,X_n}(x_1,...,x_n)delta (s-f(x_1,...,x_n))dx_1...dx_n ?$$
$endgroup$
– user623855
Dec 12 '18 at 11:58
$begingroup$
@user623855 Yes. PMFs of functions of discrete variables have a similar form.
$endgroup$
– J.G.
Dec 12 '18 at 12:00
add a comment |
$begingroup$
in what it answer to my question ?
$endgroup$
– user623855
Dec 12 '18 at 11:54
$begingroup$
@user623855 See my edit.
$endgroup$
– J.G.
Dec 12 '18 at 11:55
$begingroup$
Can we say in general that $f(X_1,...,X_n)$ has pdf $$g(s)=int_{mathbb R^n}f_{X_1,...,X_n}(x_1,...,x_n)delta (s-f(x_1,...,x_n))dx_1...dx_n ?$$
$endgroup$
– user623855
Dec 12 '18 at 11:58
$begingroup$
@user623855 Yes. PMFs of functions of discrete variables have a similar form.
$endgroup$
– J.G.
Dec 12 '18 at 12:00
$begingroup$
in what it answer to my question ?
$endgroup$
– user623855
Dec 12 '18 at 11:54
$begingroup$
in what it answer to my question ?
$endgroup$
– user623855
Dec 12 '18 at 11:54
$begingroup$
@user623855 See my edit.
$endgroup$
– J.G.
Dec 12 '18 at 11:55
$begingroup$
@user623855 See my edit.
$endgroup$
– J.G.
Dec 12 '18 at 11:55
$begingroup$
Can we say in general that $f(X_1,...,X_n)$ has pdf $$g(s)=int_{mathbb R^n}f_{X_1,...,X_n}(x_1,...,x_n)delta (s-f(x_1,...,x_n))dx_1...dx_n ?$$
$endgroup$
– user623855
Dec 12 '18 at 11:58
$begingroup$
Can we say in general that $f(X_1,...,X_n)$ has pdf $$g(s)=int_{mathbb R^n}f_{X_1,...,X_n}(x_1,...,x_n)delta (s-f(x_1,...,x_n))dx_1...dx_n ?$$
$endgroup$
– user623855
Dec 12 '18 at 11:58
$begingroup$
@user623855 Yes. PMFs of functions of discrete variables have a similar form.
$endgroup$
– J.G.
Dec 12 '18 at 12:00
$begingroup$
@user623855 Yes. PMFs of functions of discrete variables have a similar form.
$endgroup$
– J.G.
Dec 12 '18 at 12:00
add a comment |
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