Correct terminology for the joint probability density function?
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Suppose I have a joint PDF of two random variables given as:
$$f_{X,Y}(x,y).$$
Both of the random variables vary from $0$ to $infty$.
I integrate variable $Y$ to some arbitrary values, i.e.,
$$f_X(x,0leq yleq 5).$$
Now I know that $$f_{X}( x,0leq y leq infty)$$ can be termed as marginal PDF of $X$. But how should I define $f_X(x,0leq yleq 5)$? Is it the marginal PDF of $X$ when $Y$ lies in 0 to 5. Does $f_X(x,0leq yleq 5)$ qualify to be called as PDF as $$f_X(0leq x leq infty ,0leq yleq 5)neq1$$
probability notation terminology
asked yesterday
Ankit
1769
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show 1 more comment
up vote
0
down vote
favorite
Suppose I have a joint PDF of two random variables given as:
$$f_{X,Y}(x,y).$$
Both of the random variables vary from $0$ to $infty$.
I integrate variable $Y$ to some arbitrary values, i.e.,
$$f_X(x,0leq yleq 5).$$
Now I know that $$f_{X}( x,0leq y leq infty)$$ can be termed as marginal PDF of $X$. But how should I define $f_X(x,0leq yleq 5)$? Is it the marginal PDF of $X$ when $Y$ lies in 0 to 5. Does $f_X(x,0leq yleq 5)$ qualify to be called as PDF as $$f_X(0leq x leq infty ,0leq yleq 5)neq1$$
probability notation terminology
asked yesterday
Ankit
1769
$f_X(x,.<y<.)$ is confusing notation. Marginal of $X$ is just $f_X(x)=int f_{X,Y}(x,y),dy$
– StubbornAtom
yesterday
Looks like you are asking about the conditional density of $Xmid a<Y<b$ for some $a,b$, which is given by $$f_{Xmid a<Y<b}(x)=int_a^bfrac{f_{X,Y}(x,y)}{P(a<Y<b)},dy$$
– StubbornAtom
yesterday
Suppose it is a bivariate normal distribution f(x,y) and I integrate y, not from $-infty$ to $infty$, but from $0$ to $5$. What should be the terminology, should that be now just PDF of $x$ when $yin{0,5}$?
– Ankit
yesterday
Not $yin{0,5}$, but $yin(0,5)$. What you are asking for is the conditional distribution of $X$ given that $0<Y<5$. See my last comment.
– StubbornAtom
yesterday
So now the bivariate PDF is conditional distribution of X given that 0<Y<5. It is not PDF of X conditional on 0<Y<5, because then it needs to have integration over range of X as 1 , which is not the case. Am I correct?
– Ankit
yesterday
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose I have a joint PDF of two random variables given as:
$$f_{X,Y}(x,y).$$
Both of the random variables vary from $0$ to $infty$.
I integrate variable $Y$ to some arbitrary values, i.e.,
$$f_X(x,0leq yleq 5).$$
Now I know that $$f_{X}( x,0leq y leq infty)$$ can be termed as marginal PDF of $X$. But how should I define $f_X(x,0leq yleq 5)$? Is it the marginal PDF of $X$ when $Y$ lies in 0 to 5. Does $f_X(x,0leq yleq 5)$ qualify to be called as PDF as $$f_X(0leq x leq infty ,0leq yleq 5)neq1$$
probability notation terminology
asked yesterday
Ankit
1769
Suppose I have a joint PDF of two random variables given as:
$$f_{X,Y}(x,y).$$
Both of the random variables vary from $0$ to $infty$.
I integrate variable $Y$ to some arbitrary values, i.e.,
$$f_X(x,0leq yleq 5).$$
Now I know that $$f_{X}( x,0leq y leq infty)$$ can be termed as marginal PDF of $X$. But how should I define $f_X(x,0leq yleq 5)$? Is it the marginal PDF of $X$ when $Y$ lies in 0 to 5. Does $f_X(x,0leq yleq 5)$ qualify to be called as PDF as $$f_X(0leq x leq infty ,0leq yleq 5)neq1$$
probability notation terminology
probability notation terminology
asked yesterday
Ankit
1769
asked yesterday
Ankit
1769
asked yesterday
Ankit
1769
asked yesterday
Ankit
1769
asked yesterday
Ankit
1769
1769
$f_X(x,.<y<.)$ is confusing notation. Marginal of $X$ is just $f_X(x)=int f_{X,Y}(x,y),dy$
– StubbornAtom
yesterday
Looks like you are asking about the conditional density of $Xmid a<Y<b$ for some $a,b$, which is given by $$f_{Xmid a<Y<b}(x)=int_a^bfrac{f_{X,Y}(x,y)}{P(a<Y<b)},dy$$
– StubbornAtom
yesterday
Suppose it is a bivariate normal distribution f(x,y) and I integrate y, not from $-infty$ to $infty$, but from $0$ to $5$. What should be the terminology, should that be now just PDF of $x$ when $yin{0,5}$?
– Ankit
yesterday
Not $yin{0,5}$, but $yin(0,5)$. What you are asking for is the conditional distribution of $X$ given that $0<Y<5$. See my last comment.
– StubbornAtom
yesterday
So now the bivariate PDF is conditional distribution of X given that 0<Y<5. It is not PDF of X conditional on 0<Y<5, because then it needs to have integration over range of X as 1 , which is not the case. Am I correct?
– Ankit
yesterday
|
show 1 more comment
$f_X(x,.<y<.)$ is confusing notation. Marginal of $X$ is just $f_X(x)=int f_{X,Y}(x,y),dy$
– StubbornAtom
yesterday
Looks like you are asking about the conditional density of $Xmid a<Y<b$ for some $a,b$, which is given by $$f_{Xmid a<Y<b}(x)=int_a^bfrac{f_{X,Y}(x,y)}{P(a<Y<b)},dy$$
– StubbornAtom
yesterday
Suppose it is a bivariate normal distribution f(x,y) and I integrate y, not from $-infty$ to $infty$, but from $0$ to $5$. What should be the terminology, should that be now just PDF of $x$ when $yin{0,5}$?
– Ankit
yesterday
Not $yin{0,5}$, but $yin(0,5)$. What you are asking for is the conditional distribution of $X$ given that $0<Y<5$. See my last comment.
– StubbornAtom
yesterday
So now the bivariate PDF is conditional distribution of X given that 0<Y<5. It is not PDF of X conditional on 0<Y<5, because then it needs to have integration over range of X as 1 , which is not the case. Am I correct?
– Ankit
yesterday
$f_X(x,.<y<.)$ is confusing notation. Marginal of $X$ is just $f_X(x)=int f_{X,Y}(x,y),dy$
– StubbornAtom
yesterday
$f_X(x,.<y<.)$ is confusing notation. Marginal of $X$ is just $f_X(x)=int f_{X,Y}(x,y),dy$
– StubbornAtom
yesterday
Looks like you are asking about the conditional density of $Xmid a<Y<b$ for some $a,b$, which is given by $$f_{Xmid a<Y<b}(x)=int_a^bfrac{f_{X,Y}(x,y)}{P(a<Y<b)},dy$$
– StubbornAtom
yesterday
Looks like you are asking about the conditional density of $Xmid a<Y<b$ for some $a,b$, which is given by $$f_{Xmid a<Y<b}(x)=int_a^bfrac{f_{X,Y}(x,y)}{P(a<Y<b)},dy$$
– StubbornAtom
yesterday
Suppose it is a bivariate normal distribution f(x,y) and I integrate y, not from $-infty$ to $infty$, but from $0$ to $5$. What should be the terminology, should that be now just PDF of $x$ when $yin{0,5}$?
– Ankit
yesterday
Suppose it is a bivariate normal distribution f(x,y) and I integrate y, not from $-infty$ to $infty$, but from $0$ to $5$. What should be the terminology, should that be now just PDF of $x$ when $yin{0,5}$?
– Ankit
yesterday
Not $yin{0,5}$, but $yin(0,5)$. What you are asking for is the conditional distribution of $X$ given that $0<Y<5$. See my last comment.
– StubbornAtom
yesterday
Not $yin{0,5}$, but $yin(0,5)$. What you are asking for is the conditional distribution of $X$ given that $0<Y<5$. See my last comment.
– StubbornAtom
yesterday
So now the bivariate PDF is conditional distribution of X given that 0<Y<5. It is not PDF of X conditional on 0<Y<5, because then it needs to have integration over range of X as 1 , which is not the case. Am I correct?
– Ankit
yesterday
So now the bivariate PDF is conditional distribution of X given that 0<Y<5. It is not PDF of X conditional on 0<Y<5, because then it needs to have integration over range of X as 1 , which is not the case. Am I correct?
– Ankit
yesterday
|
show 1 more comment
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$f_X(x,.<y<.)$ is confusing notation. Marginal of $X$ is just $f_X(x)=int f_{X,Y}(x,y),dy$
– StubbornAtom
yesterday
Looks like you are asking about the conditional density of $Xmid a<Y<b$ for some $a,b$, which is given by $$f_{Xmid a<Y<b}(x)=int_a^bfrac{f_{X,Y}(x,y)}{P(a<Y<b)},dy$$
– StubbornAtom
yesterday
Suppose it is a bivariate normal distribution f(x,y) and I integrate y, not from $-infty$ to $infty$, but from $0$ to $5$. What should be the terminology, should that be now just PDF of $x$ when $yin{0,5}$?
– Ankit
yesterday
Not $yin{0,5}$, but $yin(0,5)$. What you are asking for is the conditional distribution of $X$ given that $0<Y<5$. See my last comment.
– StubbornAtom
yesterday
So now the bivariate PDF is conditional distribution of X given that 0<Y<5. It is not PDF of X conditional on 0<Y<5, because then it needs to have integration over range of X as 1 , which is not the case. Am I correct?
– Ankit
yesterday