How do I determine the domain of marginal density functions?
$begingroup$
I have the following density function.
$$f(x,y)=frac{1}{32}*(10-3x^2-y)quad,-1<x<1; 0<y<2$$
When calculating the marginal density for $x$, $f_x$, I get
$$int_{0}^2frac{1}{32}*(10-3x^2-y) dy \
= [10y-3x^2y-frac{y^2}{2}]^2_0 \
Rightarrow f_x = frac{9}{16}-frac{3x^2}{16}$$
I am wondering however over which domain this function is defined.
Where do I start thinking/Why?
probability density-function marginal-probability
asked Dec 21 '18 at 7:14
thebillythebilly
566
$endgroup$
add a comment |
$begingroup$
I have the following density function.
$$f(x,y)=frac{1}{32}*(10-3x^2-y)quad,-1<x<1; 0<y<2$$
When calculating the marginal density for $x$, $f_x$, I get
$$int_{0}^2frac{1}{32}*(10-3x^2-y) dy \
= [10y-3x^2y-frac{y^2}{2}]^2_0 \
Rightarrow f_x = frac{9}{16}-frac{3x^2}{16}$$
I am wondering however over which domain this function is defined.
Where do I start thinking/Why?
probability density-function marginal-probability
asked Dec 21 '18 at 7:14
thebillythebilly
566
$endgroup$
add a comment |
$begingroup$
I have the following density function.
$$f(x,y)=frac{1}{32}*(10-3x^2-y)quad,-1<x<1; 0<y<2$$
When calculating the marginal density for $x$, $f_x$, I get
$$int_{0}^2frac{1}{32}*(10-3x^2-y) dy \
= [10y-3x^2y-frac{y^2}{2}]^2_0 \
Rightarrow f_x = frac{9}{16}-frac{3x^2}{16}$$
I am wondering however over which domain this function is defined.
Where do I start thinking/Why?
probability density-function marginal-probability
asked Dec 21 '18 at 7:14
thebillythebilly
566
$endgroup$
I have the following density function.
$$f(x,y)=frac{1}{32}*(10-3x^2-y)quad,-1<x<1; 0<y<2$$
When calculating the marginal density for $x$, $f_x$, I get
$$int_{0}^2frac{1}{32}*(10-3x^2-y) dy \
= [10y-3x^2y-frac{y^2}{2}]^2_0 \
Rightarrow f_x = frac{9}{16}-frac{3x^2}{16}$$
I am wondering however over which domain this function is defined.
Where do I start thinking/Why?
probability density-function marginal-probability
probability density-function marginal-probability
asked Dec 21 '18 at 7:14
thebillythebilly
566
asked Dec 21 '18 at 7:14
thebillythebilly
566
asked Dec 21 '18 at 7:14
thebillythebilly
566
asked Dec 21 '18 at 7:14
thebillythebilly
566
asked Dec 21 '18 at 7:14
thebillythebilly
566
566
add a comment |
add a comment |
1 Answer
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oldest
votes
$begingroup$
It is common to think of PDF's as measurable non-negative functions $mathbb R^ntomathbb R$ and in that context the domain of a PDF is just $mathbb R^n$.
Quite often there is a situation that a random variable $X$ only takes values on e.g. some interval $(a,b)$ and in that situation on forehand we can choose for a corresponding PDF that takes value $0$ outside $(a,b)$.
In your case I would say that the domain of $f_X$ is $mathbb R$ and that it is prescribed by $xmapstofrac9{16}-frac{3x^2}{16}$ for $xin(-1,1)$ and by $xmapsto0$ otherwise.
Observe that for a fixed $xnotin(-1,1)$ indeed you get: $$f_X(x)=int_{-infty}^{infty} f(x,y)dy=0$$ if $f(x,y)$ is prescribed by $(x,y)mapsto0$ for $xnotin(-1,1)$.
answered Dec 21 '18 at 10:00
drhabdrhab
102k545136
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
It is common to think of PDF's as measurable non-negative functions $mathbb R^ntomathbb R$ and in that context the domain of a PDF is just $mathbb R^n$.
Quite often there is a situation that a random variable $X$ only takes values on e.g. some interval $(a,b)$ and in that situation on forehand we can choose for a corresponding PDF that takes value $0$ outside $(a,b)$.
In your case I would say that the domain of $f_X$ is $mathbb R$ and that it is prescribed by $xmapstofrac9{16}-frac{3x^2}{16}$ for $xin(-1,1)$ and by $xmapsto0$ otherwise.
Observe that for a fixed $xnotin(-1,1)$ indeed you get: $$f_X(x)=int_{-infty}^{infty} f(x,y)dy=0$$ if $f(x,y)$ is prescribed by $(x,y)mapsto0$ for $xnotin(-1,1)$.
answered Dec 21 '18 at 10:00
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
It is common to think of PDF's as measurable non-negative functions $mathbb R^ntomathbb R$ and in that context the domain of a PDF is just $mathbb R^n$.
Quite often there is a situation that a random variable $X$ only takes values on e.g. some interval $(a,b)$ and in that situation on forehand we can choose for a corresponding PDF that takes value $0$ outside $(a,b)$.
In your case I would say that the domain of $f_X$ is $mathbb R$ and that it is prescribed by $xmapstofrac9{16}-frac{3x^2}{16}$ for $xin(-1,1)$ and by $xmapsto0$ otherwise.
Observe that for a fixed $xnotin(-1,1)$ indeed you get: $$f_X(x)=int_{-infty}^{infty} f(x,y)dy=0$$ if $f(x,y)$ is prescribed by $(x,y)mapsto0$ for $xnotin(-1,1)$.
answered Dec 21 '18 at 10:00
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
It is common to think of PDF's as measurable non-negative functions $mathbb R^ntomathbb R$ and in that context the domain of a PDF is just $mathbb R^n$.
Quite often there is a situation that a random variable $X$ only takes values on e.g. some interval $(a,b)$ and in that situation on forehand we can choose for a corresponding PDF that takes value $0$ outside $(a,b)$.
In your case I would say that the domain of $f_X$ is $mathbb R$ and that it is prescribed by $xmapstofrac9{16}-frac{3x^2}{16}$ for $xin(-1,1)$ and by $xmapsto0$ otherwise.
Observe that for a fixed $xnotin(-1,1)$ indeed you get: $$f_X(x)=int_{-infty}^{infty} f(x,y)dy=0$$ if $f(x,y)$ is prescribed by $(x,y)mapsto0$ for $xnotin(-1,1)$.
answered Dec 21 '18 at 10:00
drhabdrhab
102k545136
$endgroup$
It is common to think of PDF's as measurable non-negative functions $mathbb R^ntomathbb R$ and in that context the domain of a PDF is just $mathbb R^n$.
Quite often there is a situation that a random variable $X$ only takes values on e.g. some interval $(a,b)$ and in that situation on forehand we can choose for a corresponding PDF that takes value $0$ outside $(a,b)$.
In your case I would say that the domain of $f_X$ is $mathbb R$ and that it is prescribed by $xmapstofrac9{16}-frac{3x^2}{16}$ for $xin(-1,1)$ and by $xmapsto0$ otherwise.
Observe that for a fixed $xnotin(-1,1)$ indeed you get: $$f_X(x)=int_{-infty}^{infty} f(x,y)dy=0$$ if $f(x,y)$ is prescribed by $(x,y)mapsto0$ for $xnotin(-1,1)$.
answered Dec 21 '18 at 10:00
drhabdrhab
102k545136
answered Dec 21 '18 at 10:00
drhabdrhab
102k545136
answered Dec 21 '18 at 10:00
drhabdrhab
102k545136
answered Dec 21 '18 at 10:00
drhabdrhab
102k545136
102k545136
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