Why use indicator variables in p.d.f.s?
$begingroup$
I am slightly confused about the use of indicator variables in probability density functions.
For example, consider the density of $(X,Y)$ uniformly distributed on the unit disc. This density can be written as
$$f_{X,Y}(x,y)= frac{1}{pi} mathbb{1} left { (x,y) mid x^2 + y^2 leq 1 right }.$$
However, I am not really seeing how this makes sense. I understand indicator variables when solving problems with probability, but why not just write the density without it? What is it exactly saying?
Thanks in advance.
probability density-function
asked Dec 23 '18 at 9:11
MathIsLife12MathIsLife12
613211
$endgroup$
add a comment |
$begingroup$
I am slightly confused about the use of indicator variables in probability density functions.
For example, consider the density of $(X,Y)$ uniformly distributed on the unit disc. This density can be written as
$$f_{X,Y}(x,y)= frac{1}{pi} mathbb{1} left { (x,y) mid x^2 + y^2 leq 1 right }.$$
However, I am not really seeing how this makes sense. I understand indicator variables when solving problems with probability, but why not just write the density without it? What is it exactly saying?
Thanks in advance.
probability density-function
asked Dec 23 '18 at 9:11
MathIsLife12MathIsLife12
613211
$endgroup$
add a comment |
$begingroup$
I am slightly confused about the use of indicator variables in probability density functions.
For example, consider the density of $(X,Y)$ uniformly distributed on the unit disc. This density can be written as
$$f_{X,Y}(x,y)= frac{1}{pi} mathbb{1} left { (x,y) mid x^2 + y^2 leq 1 right }.$$
However, I am not really seeing how this makes sense. I understand indicator variables when solving problems with probability, but why not just write the density without it? What is it exactly saying?
Thanks in advance.
probability density-function
asked Dec 23 '18 at 9:11
MathIsLife12MathIsLife12
613211
$endgroup$
I am slightly confused about the use of indicator variables in probability density functions.
For example, consider the density of $(X,Y)$ uniformly distributed on the unit disc. This density can be written as
$$f_{X,Y}(x,y)= frac{1}{pi} mathbb{1} left { (x,y) mid x^2 + y^2 leq 1 right }.$$
However, I am not really seeing how this makes sense. I understand indicator variables when solving problems with probability, but why not just write the density without it? What is it exactly saying?
Thanks in advance.
probability density-function
probability density-function
asked Dec 23 '18 at 9:11
MathIsLife12MathIsLife12
613211
asked Dec 23 '18 at 9:11
MathIsLife12MathIsLife12
613211
asked Dec 23 '18 at 9:11
MathIsLife12MathIsLife12
613211
asked Dec 23 '18 at 9:11
MathIsLife12MathIsLife12
613211
asked Dec 23 '18 at 9:11
MathIsLife12MathIsLife12
613211
613211
add a comment |
add a comment |
1 Answer
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$begingroup$
Note that $f_{X,Y}:mathbb R^2tomathbb R$.
It says exactly that $f_{X,Y}(x,y)$ will take value $frac1{pi}$ if $x^2+y^2leq1$ and will take value $0$ otherwise.
So an excellent presentation of a PDF which is formally for a fixed nonnegative integer $n$ a nonnegative measurable function $mathbb R^ntomathbb R$ that gives value $1$ by integration wrt Lebesgue-measure.
In notation it can be handsome to write things like: $mathbb EX=int f_X(x)xdx$ without bothering on borders.
If it is not your taste then of course you can also choose for discerning cases: $x^2+y^2leq1$ and "otherwise" without any mentioning of indicatorfunction.
answered Dec 23 '18 at 9:21
drhabdrhab
103k545136
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$begingroup$
Note that $f_{X,Y}:mathbb R^2tomathbb R$.
It says exactly that $f_{X,Y}(x,y)$ will take value $frac1{pi}$ if $x^2+y^2leq1$ and will take value $0$ otherwise.
So an excellent presentation of a PDF which is formally for a fixed nonnegative integer $n$ a nonnegative measurable function $mathbb R^ntomathbb R$ that gives value $1$ by integration wrt Lebesgue-measure.
In notation it can be handsome to write things like: $mathbb EX=int f_X(x)xdx$ without bothering on borders.
If it is not your taste then of course you can also choose for discerning cases: $x^2+y^2leq1$ and "otherwise" without any mentioning of indicatorfunction.
answered Dec 23 '18 at 9:21
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Note that $f_{X,Y}:mathbb R^2tomathbb R$.
It says exactly that $f_{X,Y}(x,y)$ will take value $frac1{pi}$ if $x^2+y^2leq1$ and will take value $0$ otherwise.
So an excellent presentation of a PDF which is formally for a fixed nonnegative integer $n$ a nonnegative measurable function $mathbb R^ntomathbb R$ that gives value $1$ by integration wrt Lebesgue-measure.
In notation it can be handsome to write things like: $mathbb EX=int f_X(x)xdx$ without bothering on borders.
If it is not your taste then of course you can also choose for discerning cases: $x^2+y^2leq1$ and "otherwise" without any mentioning of indicatorfunction.
answered Dec 23 '18 at 9:21
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Note that $f_{X,Y}:mathbb R^2tomathbb R$.
It says exactly that $f_{X,Y}(x,y)$ will take value $frac1{pi}$ if $x^2+y^2leq1$ and will take value $0$ otherwise.
So an excellent presentation of a PDF which is formally for a fixed nonnegative integer $n$ a nonnegative measurable function $mathbb R^ntomathbb R$ that gives value $1$ by integration wrt Lebesgue-measure.
In notation it can be handsome to write things like: $mathbb EX=int f_X(x)xdx$ without bothering on borders.
If it is not your taste then of course you can also choose for discerning cases: $x^2+y^2leq1$ and "otherwise" without any mentioning of indicatorfunction.
answered Dec 23 '18 at 9:21
drhabdrhab
103k545136
$endgroup$
Note that $f_{X,Y}:mathbb R^2tomathbb R$.
It says exactly that $f_{X,Y}(x,y)$ will take value $frac1{pi}$ if $x^2+y^2leq1$ and will take value $0$ otherwise.
So an excellent presentation of a PDF which is formally for a fixed nonnegative integer $n$ a nonnegative measurable function $mathbb R^ntomathbb R$ that gives value $1$ by integration wrt Lebesgue-measure.
In notation it can be handsome to write things like: $mathbb EX=int f_X(x)xdx$ without bothering on borders.
If it is not your taste then of course you can also choose for discerning cases: $x^2+y^2leq1$ and "otherwise" without any mentioning of indicatorfunction.
answered Dec 23 '18 at 9:21
drhabdrhab
103k545136
edited Dec 23 '18 at 9:28
answered Dec 23 '18 at 9:21
drhabdrhab
103k545136
answered Dec 23 '18 at 9:21
drhabdrhab
103k545136
answered Dec 23 '18 at 9:21
drhabdrhab
103k545136
103k545136
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