Finding the transform function of two histograms [closed]
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May you please help me how can I find the transform function of the following histograms?
Thank you
transformation
edited Dec 10 '18 at 19:30
gt6989b
34.2k22455
asked Dec 10 '18 at 19:24
AliAli
53
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closed as unclear what you're asking by Federico, amWhy, Xander Henderson, Matt Samuel, mlc Dec 10 '18 at 22:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
May you please help me how can I find the transform function of the following histograms?
Thank you
transformation
edited Dec 10 '18 at 19:30
gt6989b
34.2k22455
asked Dec 10 '18 at 19:24
AliAli
53
$endgroup$
closed as unclear what you're asking by Federico, amWhy, Xander Henderson, Matt Samuel, mlc Dec 10 '18 at 22:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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not clear what you are asking. You have some $r = P(z)$ and you want to have $q = f(r) = f(P(z))$ to have the distribution on the right?
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– gt6989b
Dec 10 '18 at 19:31
add a comment |
$begingroup$
May you please help me how can I find the transform function of the following histograms?
Thank you
transformation
edited Dec 10 '18 at 19:30
gt6989b
34.2k22455
asked Dec 10 '18 at 19:24
AliAli
53
$endgroup$
May you please help me how can I find the transform function of the following histograms?
Thank you
transformation
transformation
edited Dec 10 '18 at 19:30
gt6989b
34.2k22455
asked Dec 10 '18 at 19:24
AliAli
53
edited Dec 10 '18 at 19:30
gt6989b
34.2k22455
asked Dec 10 '18 at 19:24
AliAli
53
edited Dec 10 '18 at 19:30
gt6989b
34.2k22455
edited Dec 10 '18 at 19:30
gt6989b
34.2k22455
edited Dec 10 '18 at 19:30
gt6989b
34.2k22455
34.2k22455
asked Dec 10 '18 at 19:24
AliAli
53
asked Dec 10 '18 at 19:24
AliAli
53
asked Dec 10 '18 at 19:24
AliAli
53
53
closed as unclear what you're asking by Federico, amWhy, Xander Henderson, Matt Samuel, mlc Dec 10 '18 at 22:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Federico, amWhy, Xander Henderson, Matt Samuel, mlc Dec 10 '18 at 22:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
not clear what you are asking. You have some $r = P(z)$ and you want to have $q = f(r) = f(P(z))$ to have the distribution on the right?
$endgroup$
– gt6989b
Dec 10 '18 at 19:31
add a comment |
$begingroup$
not clear what you are asking. You have some $r = P(z)$ and you want to have $q = f(r) = f(P(z))$ to have the distribution on the right?
$endgroup$
– gt6989b
Dec 10 '18 at 19:31
$begingroup$
not clear what you are asking. You have some $r = P(z)$ and you want to have $q = f(r) = f(P(z))$ to have the distribution on the right?
$endgroup$
– gt6989b
Dec 10 '18 at 19:31
$begingroup$
not clear what you are asking. You have some $r = P(z)$ and you want to have $q = f(r) = f(P(z))$ to have the distribution on the right?
$endgroup$
– gt6989b
Dec 10 '18 at 19:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It appears that you are want to transformation the random variable $Z$ to random variable $R$. You are given the pdfs for the two random variables, labelled $P_z(z)$ and $P_r(r)$.
To find a monotonic transformation that will do this, insist that the cumulative probability matches:
$$eqalign{
mathcal{P}(Z<z) &= mathcal{P}(R<r) \
int_0^z P_z(z'), dz' &= int_0^r P_r(r'),dr' \
int_0^zleft(2-4leftvert z'-frac{1}{2}rightvertright),dz' &=int_0^rleft(2 - 2r'right),dr'}$$
For $z<frac{1}{2}$:
$$eqalign{
int_0^z 4z',dz' &=2r-r^2 \
2z^2 &=2r-r^2 \
r &= 1 - sqrt{1-2z^2}}$$
For $z>frac{1}{2}$:
$$eqalign{
int_0^frac{1}{2}left(2-4leftvert z'-frac{1}{2}rightvertright),dz' +
int_frac{1}{2}^zleft(2-4leftvert z'-frac{1}{2}rightvertright),dz' &=int_0^rleft(2 - 2r'right),dr' \
frac{1}{2}+int_frac{1}{2}^z (4-4z'),dz' &=2r-r^2 \
4z-2z^2-1 &=2r-r^2 \
r &= 1 - sqrt{2z^2-4z+2} \
r &=1-sqrt{2}(1-z)
}$$
An alternative transformation can be found intuitively, by transforming the random variables directly:
$$ R=2leftvert Z-frac{1}{2} rightvert$$
so that,
$$ r=2leftvert z-frac{1}{2} rightvert$$
answered Dec 10 '18 at 22:27
DeanDean
1,00537
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It appears that you are want to transformation the random variable $Z$ to random variable $R$. You are given the pdfs for the two random variables, labelled $P_z(z)$ and $P_r(r)$.
To find a monotonic transformation that will do this, insist that the cumulative probability matches:
$$eqalign{
mathcal{P}(Z<z) &= mathcal{P}(R<r) \
int_0^z P_z(z'), dz' &= int_0^r P_r(r'),dr' \
int_0^zleft(2-4leftvert z'-frac{1}{2}rightvertright),dz' &=int_0^rleft(2 - 2r'right),dr'}$$
For $z<frac{1}{2}$:
$$eqalign{
int_0^z 4z',dz' &=2r-r^2 \
2z^2 &=2r-r^2 \
r &= 1 - sqrt{1-2z^2}}$$
For $z>frac{1}{2}$:
$$eqalign{
int_0^frac{1}{2}left(2-4leftvert z'-frac{1}{2}rightvertright),dz' +
int_frac{1}{2}^zleft(2-4leftvert z'-frac{1}{2}rightvertright),dz' &=int_0^rleft(2 - 2r'right),dr' \
frac{1}{2}+int_frac{1}{2}^z (4-4z'),dz' &=2r-r^2 \
4z-2z^2-1 &=2r-r^2 \
r &= 1 - sqrt{2z^2-4z+2} \
r &=1-sqrt{2}(1-z)
}$$
An alternative transformation can be found intuitively, by transforming the random variables directly:
$$ R=2leftvert Z-frac{1}{2} rightvert$$
so that,
$$ r=2leftvert z-frac{1}{2} rightvert$$
answered Dec 10 '18 at 22:27
DeanDean
1,00537
$endgroup$
add a comment |
$begingroup$
It appears that you are want to transformation the random variable $Z$ to random variable $R$. You are given the pdfs for the two random variables, labelled $P_z(z)$ and $P_r(r)$.
To find a monotonic transformation that will do this, insist that the cumulative probability matches:
$$eqalign{
mathcal{P}(Z<z) &= mathcal{P}(R<r) \
int_0^z P_z(z'), dz' &= int_0^r P_r(r'),dr' \
int_0^zleft(2-4leftvert z'-frac{1}{2}rightvertright),dz' &=int_0^rleft(2 - 2r'right),dr'}$$
For $z<frac{1}{2}$:
$$eqalign{
int_0^z 4z',dz' &=2r-r^2 \
2z^2 &=2r-r^2 \
r &= 1 - sqrt{1-2z^2}}$$
For $z>frac{1}{2}$:
$$eqalign{
int_0^frac{1}{2}left(2-4leftvert z'-frac{1}{2}rightvertright),dz' +
int_frac{1}{2}^zleft(2-4leftvert z'-frac{1}{2}rightvertright),dz' &=int_0^rleft(2 - 2r'right),dr' \
frac{1}{2}+int_frac{1}{2}^z (4-4z'),dz' &=2r-r^2 \
4z-2z^2-1 &=2r-r^2 \
r &= 1 - sqrt{2z^2-4z+2} \
r &=1-sqrt{2}(1-z)
}$$
An alternative transformation can be found intuitively, by transforming the random variables directly:
$$ R=2leftvert Z-frac{1}{2} rightvert$$
so that,
$$ r=2leftvert z-frac{1}{2} rightvert$$
answered Dec 10 '18 at 22:27
DeanDean
1,00537
$endgroup$
add a comment |
$begingroup$
It appears that you are want to transformation the random variable $Z$ to random variable $R$. You are given the pdfs for the two random variables, labelled $P_z(z)$ and $P_r(r)$.
To find a monotonic transformation that will do this, insist that the cumulative probability matches:
$$eqalign{
mathcal{P}(Z<z) &= mathcal{P}(R<r) \
int_0^z P_z(z'), dz' &= int_0^r P_r(r'),dr' \
int_0^zleft(2-4leftvert z'-frac{1}{2}rightvertright),dz' &=int_0^rleft(2 - 2r'right),dr'}$$
For $z<frac{1}{2}$:
$$eqalign{
int_0^z 4z',dz' &=2r-r^2 \
2z^2 &=2r-r^2 \
r &= 1 - sqrt{1-2z^2}}$$
For $z>frac{1}{2}$:
$$eqalign{
int_0^frac{1}{2}left(2-4leftvert z'-frac{1}{2}rightvertright),dz' +
int_frac{1}{2}^zleft(2-4leftvert z'-frac{1}{2}rightvertright),dz' &=int_0^rleft(2 - 2r'right),dr' \
frac{1}{2}+int_frac{1}{2}^z (4-4z'),dz' &=2r-r^2 \
4z-2z^2-1 &=2r-r^2 \
r &= 1 - sqrt{2z^2-4z+2} \
r &=1-sqrt{2}(1-z)
}$$
An alternative transformation can be found intuitively, by transforming the random variables directly:
$$ R=2leftvert Z-frac{1}{2} rightvert$$
so that,
$$ r=2leftvert z-frac{1}{2} rightvert$$
answered Dec 10 '18 at 22:27
DeanDean
1,00537
$endgroup$
It appears that you are want to transformation the random variable $Z$ to random variable $R$. You are given the pdfs for the two random variables, labelled $P_z(z)$ and $P_r(r)$.
To find a monotonic transformation that will do this, insist that the cumulative probability matches:
$$eqalign{
mathcal{P}(Z<z) &= mathcal{P}(R<r) \
int_0^z P_z(z'), dz' &= int_0^r P_r(r'),dr' \
int_0^zleft(2-4leftvert z'-frac{1}{2}rightvertright),dz' &=int_0^rleft(2 - 2r'right),dr'}$$
For $z<frac{1}{2}$:
$$eqalign{
int_0^z 4z',dz' &=2r-r^2 \
2z^2 &=2r-r^2 \
r &= 1 - sqrt{1-2z^2}}$$
For $z>frac{1}{2}$:
$$eqalign{
int_0^frac{1}{2}left(2-4leftvert z'-frac{1}{2}rightvertright),dz' +
int_frac{1}{2}^zleft(2-4leftvert z'-frac{1}{2}rightvertright),dz' &=int_0^rleft(2 - 2r'right),dr' \
frac{1}{2}+int_frac{1}{2}^z (4-4z'),dz' &=2r-r^2 \
4z-2z^2-1 &=2r-r^2 \
r &= 1 - sqrt{2z^2-4z+2} \
r &=1-sqrt{2}(1-z)
}$$
An alternative transformation can be found intuitively, by transforming the random variables directly:
$$ R=2leftvert Z-frac{1}{2} rightvert$$
so that,
$$ r=2leftvert z-frac{1}{2} rightvert$$
answered Dec 10 '18 at 22:27
DeanDean
1,00537
answered Dec 10 '18 at 22:27
DeanDean
1,00537
answered Dec 10 '18 at 22:27
DeanDean
1,00537
answered Dec 10 '18 at 22:27
DeanDean
1,00537
1,00537
add a comment |
add a comment |
$begingroup$
not clear what you are asking. You have some $r = P(z)$ and you want to have $q = f(r) = f(P(z))$ to have the distribution on the right?
$endgroup$
– gt6989b
Dec 10 '18 at 19:31