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If $int_-infty^k f(x)dx = .95$ and $f(x)$ is positive even, then $int_-k^k f(x)dx = .10$?
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If $int_-infty^k f(x)dx = .95$ and $f(x)$ is positive even, then $int_-k^k f(x)dx = .10$?
Does this seem correct? Or is there not enough information to make the last conclusion?
Say for all real $x$, $f(x) > 0$, $f(x) = f(-x)$, and $int_-infty^infty f(x)dx = 1$. If $int_-infty^k f(x)dx = .95$, then $int_-k^k f(x)dx = .10?$.
Can I assume $k=.05$?
integration
edited Jul 26 at 2:23
Parcly Taxel
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asked Jul 26 at 2:23
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up vote
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favorite
If $int_-infty^k f(x)dx = .95$ and $f(x)$ is positive even, then $int_-k^k f(x)dx = .10$?
Does this seem correct? Or is there not enough information to make the last conclusion?
Say for all real $x$, $f(x) > 0$, $f(x) = f(-x)$, and $int_-infty^infty f(x)dx = 1$. If $int_-infty^k f(x)dx = .95$, then $int_-k^k f(x)dx = .10?$.
Can I assume $k=.05$?
integration
edited Jul 26 at 2:23
Parcly Taxel
33.5k136588
asked Jul 26 at 2:23
Vera
11
This does not hold in general, at least for general functions $f(x)$. You seem to have a few errors: $k$ is the bound of your integral, and not what the integral evaluates to, so writing $k = 0.05$ is incorrect. Furthermore, you must have some underlying assumptions about $f(x)$. Is $f(x)$ an odd function (i.e. symmetric about the $y$-axis)? If not, then there is no way to know what $int_-k^k f(x) dx$ evaluates to. If $f(x)$ is symmetric, then you can use that $int_k^infty f(x) dx = 0.05,$ use symmetry to evaluate $int_-infty^-k f(x) dx$, and find $int_-k^k$ from there.
– JavaMan
Jul 26 at 2:34
And of course, above, I mean to say if $f(x) is even, not odd!
– JavaMan
Jul 26 at 2:50
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $int_-infty^k f(x)dx = .95$ and $f(x)$ is positive even, then $int_-k^k f(x)dx = .10$?
Does this seem correct? Or is there not enough information to make the last conclusion?
Say for all real $x$, $f(x) > 0$, $f(x) = f(-x)$, and $int_-infty^infty f(x)dx = 1$. If $int_-infty^k f(x)dx = .95$, then $int_-k^k f(x)dx = .10?$.
Can I assume $k=.05$?
integration
edited Jul 26 at 2:23
Parcly Taxel
33.5k136588
asked Jul 26 at 2:23
Vera
11
If $int_-infty^k f(x)dx = .95$ and $f(x)$ is positive even, then $int_-k^k f(x)dx = .10$?
Does this seem correct? Or is there not enough information to make the last conclusion?
Say for all real $x$, $f(x) > 0$, $f(x) = f(-x)$, and $int_-infty^infty f(x)dx = 1$. If $int_-infty^k f(x)dx = .95$, then $int_-k^k f(x)dx = .10?$.
Can I assume $k=.05$?
integration
edited Jul 26 at 2:23
Parcly Taxel
33.5k136588
asked Jul 26 at 2:23
Vera
11
edited Jul 26 at 2:23
Parcly Taxel
33.5k136588
edited Jul 26 at 2:23
Parcly Taxel
33.5k136588
edited Jul 26 at 2:23
Parcly Taxel
33.5k136588
33.5k136588
asked Jul 26 at 2:23
Vera
11
asked Jul 26 at 2:23
Vera
11
asked Jul 26 at 2:23
Vera
11
11
This does not hold in general, at least for general functions $f(x)$. You seem to have a few errors: $k$ is the bound of your integral, and not what the integral evaluates to, so writing $k = 0.05$ is incorrect. Furthermore, you must have some underlying assumptions about $f(x)$. Is $f(x)$ an odd function (i.e. symmetric about the $y$-axis)? If not, then there is no way to know what $int_-k^k f(x) dx$ evaluates to. If $f(x)$ is symmetric, then you can use that $int_k^infty f(x) dx = 0.05,$ use symmetry to evaluate $int_-infty^-k f(x) dx$, and find $int_-k^k$ from there.
– JavaMan
Jul 26 at 2:34
And of course, above, I mean to say if $f(x) is even, not odd!
– JavaMan
Jul 26 at 2:50
add a comment |Â
This does not hold in general, at least for general functions $f(x)$. You seem to have a few errors: $k$ is the bound of your integral, and not what the integral evaluates to, so writing $k = 0.05$ is incorrect. Furthermore, you must have some underlying assumptions about $f(x)$. Is $f(x)$ an odd function (i.e. symmetric about the $y$-axis)? If not, then there is no way to know what $int_-k^k f(x) dx$ evaluates to. If $f(x)$ is symmetric, then you can use that $int_k^infty f(x) dx = 0.05,$ use symmetry to evaluate $int_-infty^-k f(x) dx$, and find $int_-k^k$ from there.
– JavaMan
Jul 26 at 2:34
And of course, above, I mean to say if $f(x) is even, not odd!
– JavaMan
Jul 26 at 2:50
This does not hold in general, at least for general functions $f(x)$. You seem to have a few errors: $k$ is the bound of your integral, and not what the integral evaluates to, so writing $k = 0.05$ is incorrect. Furthermore, you must have some underlying assumptions about $f(x)$. Is $f(x)$ an odd function (i.e. symmetric about the $y$-axis)? If not, then there is no way to know what $int_-k^k f(x) dx$ evaluates to. If $f(x)$ is symmetric, then you can use that $int_k^infty f(x) dx = 0.05,$ use symmetry to evaluate $int_-infty^-k f(x) dx$, and find $int_-k^k$ from there.
– JavaMan
Jul 26 at 2:34
This does not hold in general, at least for general functions $f(x)$. You seem to have a few errors: $k$ is the bound of your integral, and not what the integral evaluates to, so writing $k = 0.05$ is incorrect. Furthermore, you must have some underlying assumptions about $f(x)$. Is $f(x)$ an odd function (i.e. symmetric about the $y$-axis)? If not, then there is no way to know what $int_-k^k f(x) dx$ evaluates to. If $f(x)$ is symmetric, then you can use that $int_k^infty f(x) dx = 0.05,$ use symmetry to evaluate $int_-infty^-k f(x) dx$, and find $int_-k^k$ from there.
– JavaMan
Jul 26 at 2:34
And of course, above, I mean to say if $f(x) is even, not odd!
– JavaMan
Jul 26 at 2:50
And of course, above, I mean to say if $f(x) is even, not odd!
– JavaMan
Jul 26 at 2:50
add a comment |Â
3 Answers
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up vote
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and $int_-infty^infty f(x)dx = 1$
It is not clear from the question whether this is a given condition or not. Assuming it is, note that:
$$
beginalign
1 &= int_-infty^infty f(x)dx \
&= int_-infty^k f(x)dx + int_-k^infty f(x)dx - int_-k^k f(x)dx \
&= 2 cdot int_-infty^k f(x)dx - int_-k^k f(x)dx \
&= 2 cdot 0.95 - int_-k^k f(x)dx
endalign
$$
[ EDIT ] Â For a visualization of the above:
$$
underbracerlapoverbracephantomfrac00-inftyquadquadquadquadquad -kquadquadquad 0quadquadquad k^large=0.95 overline-inftyquadquadquadquadquad underbrace-kquadquadquad 0quadquadquad kquadquadquadquadquad +infty;_large=0.95 ; textbecause ;f; textis even_large=1 \
$$
answered Jul 26 at 2:37
dxiv
53.9k64796
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up vote
0
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I interpret the question as $f$ is an even function which is a densify function, then we have
$$int_-infty^0 f(x) , dx = 0.5$$
and since $$int_-infty^k f(x) , dx= 0.95$$
and $f$ is positive everywhere, we have $k>0$.
$$int_-infty^-k f(x) , dx+int_-k^k f(x) , dx + int_k^infty f(x) , dx= 1$$
$$int_-infty^-k f(x) , dx+int_-k^k f(x) , dx = 0.95$$
subtracting the difference,
$$int_k^infty f(x) , dx = 0.05$$
Let $y = -x$, then $$-int_-k^-infty f(-y) , dy=0.05$$
using the information that the function is even,
$$int_-infty^-kf(x) , dx =0.05$$
Hence $$int_-k^k f(x) , dx =0.9$$
$k$ need not be $0.05$, for example, consider the standard normal distribution.
Remark: If it is not a density function and we know that $int_-infty^infty f(x) , dx = M>0$, we can repeat the procedure to deduce the desired quantity.
answered Jul 26 at 2:42
Siong Thye Goh
77.1k134794
add a comment |Â
up vote
0
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so if $int_-infty^inftyf(x)dx=1$
then $int_0^inftyf(x)dx+int_-infty^0f(x)dx=1$
and it has been said that $f(x)=f(-x)$ so let $x=-u,,fracdxdu=-1$
so $int_-infty^0f(x)dx=-int_infty^0f(-u)du=int_0^inftyf(-x)dx$
again since $f(x)=f(-x)$
$int_0^inftyf(-x)dx=int_0^inftyf(x)dx$
so $2int_0^inftyf(x)dx=1$ therefore $int_0^inftyf(x)dx=0.5$ which is what you said.
now if $int_-infty^kf(x)dx=0.95$ then $int_-infty^0f(x)dx+int_0^kf(x)dx=0.95$ so $int_0^kf(x)dx=0.45$
and because $f(x)=f(-x),,int_-k^kf(x)dx=2int_0^kf(x)dx=0.9$
It seems to hold for all even functions provided they can be integated in the domain $(-infty,infty)$
answered Jul 27 at 11:47
Henry Lee
49210
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
and $int_-infty^infty f(x)dx = 1$
It is not clear from the question whether this is a given condition or not. Assuming it is, note that:
$$
beginalign
1 &= int_-infty^infty f(x)dx \
&= int_-infty^k f(x)dx + int_-k^infty f(x)dx - int_-k^k f(x)dx \
&= 2 cdot int_-infty^k f(x)dx - int_-k^k f(x)dx \
&= 2 cdot 0.95 - int_-k^k f(x)dx
endalign
$$
[ EDIT ] Â For a visualization of the above:
$$
underbracerlapoverbracephantomfrac00-inftyquadquadquadquadquad -kquadquadquad 0quadquadquad k^large=0.95 overline-inftyquadquadquadquadquad underbrace-kquadquadquad 0quadquadquad kquadquadquadquadquad +infty;_large=0.95 ; textbecause ;f; textis even_large=1 \
$$
answered Jul 26 at 2:37
dxiv
53.9k64796
add a comment |Â
up vote
1
down vote
and $int_-infty^infty f(x)dx = 1$
It is not clear from the question whether this is a given condition or not. Assuming it is, note that:
$$
beginalign
1 &= int_-infty^infty f(x)dx \
&= int_-infty^k f(x)dx + int_-k^infty f(x)dx - int_-k^k f(x)dx \
&= 2 cdot int_-infty^k f(x)dx - int_-k^k f(x)dx \
&= 2 cdot 0.95 - int_-k^k f(x)dx
endalign
$$
[ EDIT ] Â For a visualization of the above:
$$
underbracerlapoverbracephantomfrac00-inftyquadquadquadquadquad -kquadquadquad 0quadquadquad k^large=0.95 overline-inftyquadquadquadquadquad underbrace-kquadquadquad 0quadquadquad kquadquadquadquadquad +infty;_large=0.95 ; textbecause ;f; textis even_large=1 \
$$
answered Jul 26 at 2:37
dxiv
53.9k64796
add a comment |Â
up vote
1
down vote
up vote
1
down vote
and $int_-infty^infty f(x)dx = 1$
It is not clear from the question whether this is a given condition or not. Assuming it is, note that:
$$
beginalign
1 &= int_-infty^infty f(x)dx \
&= int_-infty^k f(x)dx + int_-k^infty f(x)dx - int_-k^k f(x)dx \
&= 2 cdot int_-infty^k f(x)dx - int_-k^k f(x)dx \
&= 2 cdot 0.95 - int_-k^k f(x)dx
endalign
$$
[ EDIT ] Â For a visualization of the above:
$$
underbracerlapoverbracephantomfrac00-inftyquadquadquadquadquad -kquadquadquad 0quadquadquad k^large=0.95 overline-inftyquadquadquadquadquad underbrace-kquadquadquad 0quadquadquad kquadquadquadquadquad +infty;_large=0.95 ; textbecause ;f; textis even_large=1 \
$$
answered Jul 26 at 2:37
dxiv
53.9k64796
and $int_-infty^infty f(x)dx = 1$
It is not clear from the question whether this is a given condition or not. Assuming it is, note that:
$$
beginalign
1 &= int_-infty^infty f(x)dx \
&= int_-infty^k f(x)dx + int_-k^infty f(x)dx - int_-k^k f(x)dx \
&= 2 cdot int_-infty^k f(x)dx - int_-k^k f(x)dx \
&= 2 cdot 0.95 - int_-k^k f(x)dx
endalign
$$
[ EDIT ] Â For a visualization of the above:
$$
underbracerlapoverbracephantomfrac00-inftyquadquadquadquadquad -kquadquadquad 0quadquadquad k^large=0.95 overline-inftyquadquadquadquadquad underbrace-kquadquadquad 0quadquadquad kquadquadquadquadquad +infty;_large=0.95 ; textbecause ;f; textis even_large=1 \
$$
answered Jul 26 at 2:37
dxiv
53.9k64796
edited Jul 26 at 4:16
answered Jul 26 at 2:37
dxiv
53.9k64796
answered Jul 26 at 2:37
dxiv
53.9k64796
answered Jul 26 at 2:37
dxiv
53.9k64796
53.9k64796
add a comment |Â
add a comment |Â
up vote
0
down vote
I interpret the question as $f$ is an even function which is a densify function, then we have
$$int_-infty^0 f(x) , dx = 0.5$$
and since $$int_-infty^k f(x) , dx= 0.95$$
and $f$ is positive everywhere, we have $k>0$.
$$int_-infty^-k f(x) , dx+int_-k^k f(x) , dx + int_k^infty f(x) , dx= 1$$
$$int_-infty^-k f(x) , dx+int_-k^k f(x) , dx = 0.95$$
subtracting the difference,
$$int_k^infty f(x) , dx = 0.05$$
Let $y = -x$, then $$-int_-k^-infty f(-y) , dy=0.05$$
using the information that the function is even,
$$int_-infty^-kf(x) , dx =0.05$$
Hence $$int_-k^k f(x) , dx =0.9$$
$k$ need not be $0.05$, for example, consider the standard normal distribution.
Remark: If it is not a density function and we know that $int_-infty^infty f(x) , dx = M>0$, we can repeat the procedure to deduce the desired quantity.
answered Jul 26 at 2:42
Siong Thye Goh
77.1k134794
add a comment |Â
up vote
0
down vote
I interpret the question as $f$ is an even function which is a densify function, then we have
$$int_-infty^0 f(x) , dx = 0.5$$
and since $$int_-infty^k f(x) , dx= 0.95$$
and $f$ is positive everywhere, we have $k>0$.
$$int_-infty^-k f(x) , dx+int_-k^k f(x) , dx + int_k^infty f(x) , dx= 1$$
$$int_-infty^-k f(x) , dx+int_-k^k f(x) , dx = 0.95$$
subtracting the difference,
$$int_k^infty f(x) , dx = 0.05$$
Let $y = -x$, then $$-int_-k^-infty f(-y) , dy=0.05$$
using the information that the function is even,
$$int_-infty^-kf(x) , dx =0.05$$
Hence $$int_-k^k f(x) , dx =0.9$$
$k$ need not be $0.05$, for example, consider the standard normal distribution.
Remark: If it is not a density function and we know that $int_-infty^infty f(x) , dx = M>0$, we can repeat the procedure to deduce the desired quantity.
answered Jul 26 at 2:42
Siong Thye Goh
77.1k134794
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I interpret the question as $f$ is an even function which is a densify function, then we have
$$int_-infty^0 f(x) , dx = 0.5$$
and since $$int_-infty^k f(x) , dx= 0.95$$
and $f$ is positive everywhere, we have $k>0$.
$$int_-infty^-k f(x) , dx+int_-k^k f(x) , dx + int_k^infty f(x) , dx= 1$$
$$int_-infty^-k f(x) , dx+int_-k^k f(x) , dx = 0.95$$
subtracting the difference,
$$int_k^infty f(x) , dx = 0.05$$
Let $y = -x$, then $$-int_-k^-infty f(-y) , dy=0.05$$
using the information that the function is even,
$$int_-infty^-kf(x) , dx =0.05$$
Hence $$int_-k^k f(x) , dx =0.9$$
$k$ need not be $0.05$, for example, consider the standard normal distribution.
Remark: If it is not a density function and we know that $int_-infty^infty f(x) , dx = M>0$, we can repeat the procedure to deduce the desired quantity.
answered Jul 26 at 2:42
Siong Thye Goh
77.1k134794
I interpret the question as $f$ is an even function which is a densify function, then we have
$$int_-infty^0 f(x) , dx = 0.5$$
and since $$int_-infty^k f(x) , dx= 0.95$$
and $f$ is positive everywhere, we have $k>0$.
$$int_-infty^-k f(x) , dx+int_-k^k f(x) , dx + int_k^infty f(x) , dx= 1$$
$$int_-infty^-k f(x) , dx+int_-k^k f(x) , dx = 0.95$$
subtracting the difference,
$$int_k^infty f(x) , dx = 0.05$$
Let $y = -x$, then $$-int_-k^-infty f(-y) , dy=0.05$$
using the information that the function is even,
$$int_-infty^-kf(x) , dx =0.05$$
Hence $$int_-k^k f(x) , dx =0.9$$
$k$ need not be $0.05$, for example, consider the standard normal distribution.
Remark: If it is not a density function and we know that $int_-infty^infty f(x) , dx = M>0$, we can repeat the procedure to deduce the desired quantity.
answered Jul 26 at 2:42
Siong Thye Goh
77.1k134794
edited Jul 26 at 3:02
answered Jul 26 at 2:42
Siong Thye Goh
77.1k134794
answered Jul 26 at 2:42
Siong Thye Goh
77.1k134794
answered Jul 26 at 2:42
Siong Thye Goh
77.1k134794
77.1k134794
add a comment |Â
add a comment |Â
up vote
0
down vote
so if $int_-infty^inftyf(x)dx=1$
then $int_0^inftyf(x)dx+int_-infty^0f(x)dx=1$
and it has been said that $f(x)=f(-x)$ so let $x=-u,,fracdxdu=-1$
so $int_-infty^0f(x)dx=-int_infty^0f(-u)du=int_0^inftyf(-x)dx$
again since $f(x)=f(-x)$
$int_0^inftyf(-x)dx=int_0^inftyf(x)dx$
so $2int_0^inftyf(x)dx=1$ therefore $int_0^inftyf(x)dx=0.5$ which is what you said.
now if $int_-infty^kf(x)dx=0.95$ then $int_-infty^0f(x)dx+int_0^kf(x)dx=0.95$ so $int_0^kf(x)dx=0.45$
and because $f(x)=f(-x),,int_-k^kf(x)dx=2int_0^kf(x)dx=0.9$
It seems to hold for all even functions provided they can be integated in the domain $(-infty,infty)$
answered Jul 27 at 11:47
Henry Lee
49210
add a comment |Â
up vote
0
down vote
so if $int_-infty^inftyf(x)dx=1$
then $int_0^inftyf(x)dx+int_-infty^0f(x)dx=1$
and it has been said that $f(x)=f(-x)$ so let $x=-u,,fracdxdu=-1$
so $int_-infty^0f(x)dx=-int_infty^0f(-u)du=int_0^inftyf(-x)dx$
again since $f(x)=f(-x)$
$int_0^inftyf(-x)dx=int_0^inftyf(x)dx$
so $2int_0^inftyf(x)dx=1$ therefore $int_0^inftyf(x)dx=0.5$ which is what you said.
now if $int_-infty^kf(x)dx=0.95$ then $int_-infty^0f(x)dx+int_0^kf(x)dx=0.95$ so $int_0^kf(x)dx=0.45$
and because $f(x)=f(-x),,int_-k^kf(x)dx=2int_0^kf(x)dx=0.9$
It seems to hold for all even functions provided they can be integated in the domain $(-infty,infty)$
answered Jul 27 at 11:47
Henry Lee
49210
add a comment |Â
up vote
0
down vote
up vote
0
down vote
so if $int_-infty^inftyf(x)dx=1$
then $int_0^inftyf(x)dx+int_-infty^0f(x)dx=1$
and it has been said that $f(x)=f(-x)$ so let $x=-u,,fracdxdu=-1$
so $int_-infty^0f(x)dx=-int_infty^0f(-u)du=int_0^inftyf(-x)dx$
again since $f(x)=f(-x)$
$int_0^inftyf(-x)dx=int_0^inftyf(x)dx$
so $2int_0^inftyf(x)dx=1$ therefore $int_0^inftyf(x)dx=0.5$ which is what you said.
now if $int_-infty^kf(x)dx=0.95$ then $int_-infty^0f(x)dx+int_0^kf(x)dx=0.95$ so $int_0^kf(x)dx=0.45$
and because $f(x)=f(-x),,int_-k^kf(x)dx=2int_0^kf(x)dx=0.9$
It seems to hold for all even functions provided they can be integated in the domain $(-infty,infty)$
answered Jul 27 at 11:47
Henry Lee
49210
so if $int_-infty^inftyf(x)dx=1$
then $int_0^inftyf(x)dx+int_-infty^0f(x)dx=1$
and it has been said that $f(x)=f(-x)$ so let $x=-u,,fracdxdu=-1$
so $int_-infty^0f(x)dx=-int_infty^0f(-u)du=int_0^inftyf(-x)dx$
again since $f(x)=f(-x)$
$int_0^inftyf(-x)dx=int_0^inftyf(x)dx$
so $2int_0^inftyf(x)dx=1$ therefore $int_0^inftyf(x)dx=0.5$ which is what you said.
now if $int_-infty^kf(x)dx=0.95$ then $int_-infty^0f(x)dx+int_0^kf(x)dx=0.95$ so $int_0^kf(x)dx=0.45$
and because $f(x)=f(-x),,int_-k^kf(x)dx=2int_0^kf(x)dx=0.9$
It seems to hold for all even functions provided they can be integated in the domain $(-infty,infty)$
answered Jul 27 at 11:47
Henry Lee
49210
answered Jul 27 at 11:47
Henry Lee
49210
answered Jul 27 at 11:47
Henry Lee
49210
answered Jul 27 at 11:47
Henry Lee
49210
49210
add a comment |Â
add a comment |Â
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This does not hold in general, at least for general functions $f(x)$. You seem to have a few errors: $k$ is the bound of your integral, and not what the integral evaluates to, so writing $k = 0.05$ is incorrect. Furthermore, you must have some underlying assumptions about $f(x)$. Is $f(x)$ an odd function (i.e. symmetric about the $y$-axis)? If not, then there is no way to know what $int_-k^k f(x) dx$ evaluates to. If $f(x)$ is symmetric, then you can use that $int_k^infty f(x) dx = 0.05,$ use symmetry to evaluate $int_-infty^-k f(x) dx$, and find $int_-k^k$ from there.
– JavaMan
Jul 26 at 2:34
And of course, above, I mean to say if $f(x) is even, not odd!
– JavaMan
Jul 26 at 2:50