Meaning of the words “evaluate joint density function (transformed) at the point $(x,y)$”
$begingroup$
You are given that $X$ and $Y$ both have the same uniform distribution on $[0, 1]$, and are
independent.
$U = X + Y$
and $V =
{X over X + Y}$
.
Find the joint probability density function of
$(U, V)$ evaluated at the point $left ({1over 2}, {1over 2} right )$.
So far, I have used the fact that $X=UV$ and $Y=U-UV$ to set up the jacobian determinant and determine that the joint distribution of $(U,V)$ is $u$.
However, I am still confused by the meaning of "evaluate at the point $left ({1over 2}, {1over 2} right )$", as I don't think I have seen something like this before, or if I have I am not understanding it's application here. The correct solution happens to be $1 over 2$.
probability statistics
asked Dec 25 '18 at 23:22
agbltagblt
350114
$endgroup$
add a comment |
$begingroup$
You are given that $X$ and $Y$ both have the same uniform distribution on $[0, 1]$, and are
independent.
$U = X + Y$
and $V =
{X over X + Y}$
.
Find the joint probability density function of
$(U, V)$ evaluated at the point $left ({1over 2}, {1over 2} right )$.
So far, I have used the fact that $X=UV$ and $Y=U-UV$ to set up the jacobian determinant and determine that the joint distribution of $(U,V)$ is $u$.
However, I am still confused by the meaning of "evaluate at the point $left ({1over 2}, {1over 2} right )$", as I don't think I have seen something like this before, or if I have I am not understanding it's application here. The correct solution happens to be $1 over 2$.
probability statistics
asked Dec 25 '18 at 23:22
agbltagblt
350114
$endgroup$
1
$begingroup$
If the joint density is $(u,w)mapsto u$ then clearly $(1/2,1/2)mapsto 1/2$.
$endgroup$
– Math1000
Dec 25 '18 at 23:43
1
$begingroup$
Strictly speaking you cannot evaluate a density function at a point. Any function equal almost everywhere to a density function of a random variable is also a density function for the same variable. However, the title talks about evaluating the distribution function, not the density.
$endgroup$
– Kavi Rama Murthy
Dec 26 '18 at 0:02
1
$begingroup$
IMV the question to evaluate the PDF at a point is a "stupid question" (not your fault of course). See the comment of @Kavi for that. Why not just asking to determine the PDF if on forehand it is clear that you cannot escape from doing so? Or formally even better: determine a (not "the") PDF.
$endgroup$
– drhab
Dec 26 '18 at 9:34
add a comment |
$begingroup$
You are given that $X$ and $Y$ both have the same uniform distribution on $[0, 1]$, and are
independent.
$U = X + Y$
and $V =
{X over X + Y}$
.
Find the joint probability density function of
$(U, V)$ evaluated at the point $left ({1over 2}, {1over 2} right )$.
So far, I have used the fact that $X=UV$ and $Y=U-UV$ to set up the jacobian determinant and determine that the joint distribution of $(U,V)$ is $u$.
However, I am still confused by the meaning of "evaluate at the point $left ({1over 2}, {1over 2} right )$", as I don't think I have seen something like this before, or if I have I am not understanding it's application here. The correct solution happens to be $1 over 2$.
probability statistics
asked Dec 25 '18 at 23:22
agbltagblt
350114
$endgroup$
You are given that $X$ and $Y$ both have the same uniform distribution on $[0, 1]$, and are
independent.
$U = X + Y$
and $V =
{X over X + Y}$
.
Find the joint probability density function of
$(U, V)$ evaluated at the point $left ({1over 2}, {1over 2} right )$.
So far, I have used the fact that $X=UV$ and $Y=U-UV$ to set up the jacobian determinant and determine that the joint distribution of $(U,V)$ is $u$.
However, I am still confused by the meaning of "evaluate at the point $left ({1over 2}, {1over 2} right )$", as I don't think I have seen something like this before, or if I have I am not understanding it's application here. The correct solution happens to be $1 over 2$.
probability statistics
probability statistics
asked Dec 25 '18 at 23:22
agbltagblt
350114
asked Dec 25 '18 at 23:22
agbltagblt
350114
edited Dec 26 '18 at 13:31
agblt
asked Dec 25 '18 at 23:22
agbltagblt
350114
asked Dec 25 '18 at 23:22
agbltagblt
350114
asked Dec 25 '18 at 23:22
agbltagblt
350114
350114
1
$begingroup$
If the joint density is $(u,w)mapsto u$ then clearly $(1/2,1/2)mapsto 1/2$.
$endgroup$
– Math1000
Dec 25 '18 at 23:43
1
$begingroup$
Strictly speaking you cannot evaluate a density function at a point. Any function equal almost everywhere to a density function of a random variable is also a density function for the same variable. However, the title talks about evaluating the distribution function, not the density.
$endgroup$
– Kavi Rama Murthy
Dec 26 '18 at 0:02
1
$begingroup$
IMV the question to evaluate the PDF at a point is a "stupid question" (not your fault of course). See the comment of @Kavi for that. Why not just asking to determine the PDF if on forehand it is clear that you cannot escape from doing so? Or formally even better: determine a (not "the") PDF.
$endgroup$
– drhab
Dec 26 '18 at 9:34
add a comment |
1
$begingroup$
If the joint density is $(u,w)mapsto u$ then clearly $(1/2,1/2)mapsto 1/2$.
$endgroup$
– Math1000
Dec 25 '18 at 23:43
1
$begingroup$
Strictly speaking you cannot evaluate a density function at a point. Any function equal almost everywhere to a density function of a random variable is also a density function for the same variable. However, the title talks about evaluating the distribution function, not the density.
$endgroup$
– Kavi Rama Murthy
Dec 26 '18 at 0:02
1
$begingroup$
IMV the question to evaluate the PDF at a point is a "stupid question" (not your fault of course). See the comment of @Kavi for that. Why not just asking to determine the PDF if on forehand it is clear that you cannot escape from doing so? Or formally even better: determine a (not "the") PDF.
$endgroup$
– drhab
Dec 26 '18 at 9:34
1
1
$begingroup$
If the joint density is $(u,w)mapsto u$ then clearly $(1/2,1/2)mapsto 1/2$.
$endgroup$
– Math1000
Dec 25 '18 at 23:43
$begingroup$
If the joint density is $(u,w)mapsto u$ then clearly $(1/2,1/2)mapsto 1/2$.
$endgroup$
– Math1000
Dec 25 '18 at 23:43
1
1
$begingroup$
Strictly speaking you cannot evaluate a density function at a point. Any function equal almost everywhere to a density function of a random variable is also a density function for the same variable. However, the title talks about evaluating the distribution function, not the density.
$endgroup$
– Kavi Rama Murthy
Dec 26 '18 at 0:02
$begingroup$
Strictly speaking you cannot evaluate a density function at a point. Any function equal almost everywhere to a density function of a random variable is also a density function for the same variable. However, the title talks about evaluating the distribution function, not the density.
$endgroup$
– Kavi Rama Murthy
Dec 26 '18 at 0:02
1
1
$begingroup$
IMV the question to evaluate the PDF at a point is a "stupid question" (not your fault of course). See the comment of @Kavi for that. Why not just asking to determine the PDF if on forehand it is clear that you cannot escape from doing so? Or formally even better: determine a (not "the") PDF.
$endgroup$
– drhab
Dec 26 '18 at 9:34
$begingroup$
IMV the question to evaluate the PDF at a point is a "stupid question" (not your fault of course). See the comment of @Kavi for that. Why not just asking to determine the PDF if on forehand it is clear that you cannot escape from doing so? Or formally even better: determine a (not "the") PDF.
$endgroup$
– drhab
Dec 26 '18 at 9:34
add a comment |
1 Answer
1
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oldest
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$begingroup$
If $f$ is a function on a set $X$, and $xin X$, the English phrase "evaluate $f$ at $x$" means "compute $f(x)$".
Here if $f$ is the probability density function on $[0,1]times[0, 1]$, (not the distribution, which is a different object) then to evaluate $f$ at $(frac12, frac12)$ just means to compute $f(frac12,frac12)$.
answered Dec 26 '18 at 0:16
Jack MJack M
18.9k33882
$endgroup$
add a comment |
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$begingroup$
If $f$ is a function on a set $X$, and $xin X$, the English phrase "evaluate $f$ at $x$" means "compute $f(x)$".
Here if $f$ is the probability density function on $[0,1]times[0, 1]$, (not the distribution, which is a different object) then to evaluate $f$ at $(frac12, frac12)$ just means to compute $f(frac12,frac12)$.
answered Dec 26 '18 at 0:16
Jack MJack M
18.9k33882
$endgroup$
add a comment |
$begingroup$
If $f$ is a function on a set $X$, and $xin X$, the English phrase "evaluate $f$ at $x$" means "compute $f(x)$".
Here if $f$ is the probability density function on $[0,1]times[0, 1]$, (not the distribution, which is a different object) then to evaluate $f$ at $(frac12, frac12)$ just means to compute $f(frac12,frac12)$.
answered Dec 26 '18 at 0:16
Jack MJack M
18.9k33882
$endgroup$
add a comment |
$begingroup$
If $f$ is a function on a set $X$, and $xin X$, the English phrase "evaluate $f$ at $x$" means "compute $f(x)$".
Here if $f$ is the probability density function on $[0,1]times[0, 1]$, (not the distribution, which is a different object) then to evaluate $f$ at $(frac12, frac12)$ just means to compute $f(frac12,frac12)$.
answered Dec 26 '18 at 0:16
Jack MJack M
18.9k33882
$endgroup$
If $f$ is a function on a set $X$, and $xin X$, the English phrase "evaluate $f$ at $x$" means "compute $f(x)$".
Here if $f$ is the probability density function on $[0,1]times[0, 1]$, (not the distribution, which is a different object) then to evaluate $f$ at $(frac12, frac12)$ just means to compute $f(frac12,frac12)$.
answered Dec 26 '18 at 0:16
Jack MJack M
18.9k33882
answered Dec 26 '18 at 0:16
Jack MJack M
18.9k33882
answered Dec 26 '18 at 0:16
Jack MJack M
18.9k33882
answered Dec 26 '18 at 0:16
Jack MJack M
18.9k33882
18.9k33882
add a comment |
add a comment |
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1
$begingroup$
If the joint density is $(u,w)mapsto u$ then clearly $(1/2,1/2)mapsto 1/2$.
$endgroup$
– Math1000
Dec 25 '18 at 23:43
1
$begingroup$
Strictly speaking you cannot evaluate a density function at a point. Any function equal almost everywhere to a density function of a random variable is also a density function for the same variable. However, the title talks about evaluating the distribution function, not the density.
$endgroup$
– Kavi Rama Murthy
Dec 26 '18 at 0:02
1
$begingroup$
IMV the question to evaluate the PDF at a point is a "stupid question" (not your fault of course). See the comment of @Kavi for that. Why not just asking to determine the PDF if on forehand it is clear that you cannot escape from doing so? Or formally even better: determine a (not "the") PDF.
$endgroup$
– drhab
Dec 26 '18 at 9:34