Quantile function questionbinomial quantile functionIs there any probability distribution that can fit based on 3 quantiles?Quantile(X + constant) = Quantile(X) + constant?In biweight midvariance, why would the median absolute deviation (MAD) be multiplied by the 0.75 standard normal quantile?basic Quantile provesQuantile function of the r th order statistics of i.n.n.i.d. random variablesEstimating Quantile Function Using Pseudo SampleQuantile function for binomial distribution?Inverse distribution function (quantile function)Taylor series expansion of quantile function
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Quantile function question
binomial quantile functionIs there any probability distribution that can fit based on 3 quantiles?Quantile(X + constant) = Quantile(X) + constant?In biweight midvariance, why would the median absolute deviation (MAD) be multiplied by the 0.75 standard normal quantile?basic Quantile provesQuantile function of the r th order statistics of i.n.n.i.d. random variablesEstimating Quantile Function Using Pseudo SampleQuantile function for binomial distribution?Inverse distribution function (quantile function)Taylor series expansion of quantile function
$begingroup$
is there a way how to get quantile function from random variable $Y$ defined as: $$ Y = begincases
0 text ... with probability 0.25\
f(x) text ... with probability 0.75
endcases $$
where $f(x)$ is continuous probability distribution function?
Thank you for your time.
quantile
edited 7 hours ago
Cettt
1,888622
asked Mar 11 at 8:26
Matus D.Matus D.
185
$endgroup$
add a comment |
$begingroup$
is there a way how to get quantile function from random variable $Y$ defined as: $$ Y = begincases
0 text ... with probability 0.25\
f(x) text ... with probability 0.75
endcases $$
where $f(x)$ is continuous probability distribution function?
Thank you for your time.
quantile
edited 7 hours ago
Cettt
1,888622
asked Mar 11 at 8:26
Matus D.Matus D.
185
$endgroup$
add a comment |
$begingroup$
is there a way how to get quantile function from random variable $Y$ defined as: $$ Y = begincases
0 text ... with probability 0.25\
f(x) text ... with probability 0.75
endcases $$
where $f(x)$ is continuous probability distribution function?
Thank you for your time.
quantile
edited 7 hours ago
Cettt
1,888622
asked Mar 11 at 8:26
Matus D.Matus D.
185
$endgroup$
is there a way how to get quantile function from random variable $Y$ defined as: $$ Y = begincases
0 text ... with probability 0.25\
f(x) text ... with probability 0.75
endcases $$
where $f(x)$ is continuous probability distribution function?
Thank you for your time.
quantile
quantile
edited 7 hours ago
Cettt
1,888622
asked Mar 11 at 8:26
Matus D.Matus D.
185
edited 7 hours ago
Cettt
1,888622
asked Mar 11 at 8:26
Matus D.Matus D.
185
edited 7 hours ago
Cettt
1,888622
edited 7 hours ago
Cettt
1,888622
edited 7 hours ago
Cettt
1,888622
1,888622
asked Mar 11 at 8:26
Matus D.Matus D.
185
asked Mar 11 at 8:26
Matus D.Matus D.
185
asked Mar 11 at 8:26
Matus D.Matus D.
185
185
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $F_Y^-1$ denote the quantile function of $Y$. Then
$$
forall uin(0,1),quad F_Y^-1(u)=left{beginarrayrcl0&textrmif&ulefrac14\f(x)&textrmif&u>frac14endarrayright.
$$
If you want the details:
Let $F_Y$ denote the cumulative distribution function of $Y$. Then
$F_Y(y)=0$ if $y<0$, $F_Y(y)=frac14$ if $0le y<f(x)$ and $F_Y(y)=1$ if $yge f(x)$. You can then deduce the quantile function from its definition:
$$
F_Y^-1(u)=infyinmathbb Rmid F_Y(y)ge u.
$$
answered Mar 11 at 8:40
WillWill
1213
New contributor
Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I'm sorry I don't understand. Who is $x$? Who is $y$? Are $y$ and $Y$ the same for you? Do you mean that $Y=0$ with probability 0.25 and $f(X)$ with probability 0.75 where $X$ is another random variable? The more you specify, the more I can understand your question and give the answer you're looking for
$endgroup$
– Will
Mar 11 at 19:41
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $F_Y^-1$ denote the quantile function of $Y$. Then
$$
forall uin(0,1),quad F_Y^-1(u)=left{beginarrayrcl0&textrmif&ulefrac14\f(x)&textrmif&u>frac14endarrayright.
$$
If you want the details:
Let $F_Y$ denote the cumulative distribution function of $Y$. Then
$F_Y(y)=0$ if $y<0$, $F_Y(y)=frac14$ if $0le y<f(x)$ and $F_Y(y)=1$ if $yge f(x)$. You can then deduce the quantile function from its definition:
$$
F_Y^-1(u)=infyinmathbb Rmid F_Y(y)ge u.
$$
answered Mar 11 at 8:40
WillWill
1213
New contributor
Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I'm sorry I don't understand. Who is $x$? Who is $y$? Are $y$ and $Y$ the same for you? Do you mean that $Y=0$ with probability 0.25 and $f(X)$ with probability 0.75 where $X$ is another random variable? The more you specify, the more I can understand your question and give the answer you're looking for
$endgroup$
– Will
Mar 11 at 19:41
add a comment |
$begingroup$
Let $F_Y^-1$ denote the quantile function of $Y$. Then
$$
forall uin(0,1),quad F_Y^-1(u)=left{beginarrayrcl0&textrmif&ulefrac14\f(x)&textrmif&u>frac14endarrayright.
$$
If you want the details:
Let $F_Y$ denote the cumulative distribution function of $Y$. Then
$F_Y(y)=0$ if $y<0$, $F_Y(y)=frac14$ if $0le y<f(x)$ and $F_Y(y)=1$ if $yge f(x)$. You can then deduce the quantile function from its definition:
$$
F_Y^-1(u)=infyinmathbb Rmid F_Y(y)ge u.
$$
answered Mar 11 at 8:40
WillWill
1213
New contributor
Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I'm sorry I don't understand. Who is $x$? Who is $y$? Are $y$ and $Y$ the same for you? Do you mean that $Y=0$ with probability 0.25 and $f(X)$ with probability 0.75 where $X$ is another random variable? The more you specify, the more I can understand your question and give the answer you're looking for
$endgroup$
– Will
Mar 11 at 19:41
add a comment |
$begingroup$
Let $F_Y^-1$ denote the quantile function of $Y$. Then
$$
forall uin(0,1),quad F_Y^-1(u)=left{beginarrayrcl0&textrmif&ulefrac14\f(x)&textrmif&u>frac14endarrayright.
$$
If you want the details:
Let $F_Y$ denote the cumulative distribution function of $Y$. Then
$F_Y(y)=0$ if $y<0$, $F_Y(y)=frac14$ if $0le y<f(x)$ and $F_Y(y)=1$ if $yge f(x)$. You can then deduce the quantile function from its definition:
$$
F_Y^-1(u)=infyinmathbb Rmid F_Y(y)ge u.
$$
answered Mar 11 at 8:40
WillWill
1213
New contributor
Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $F_Y^-1$ denote the quantile function of $Y$. Then
$$
forall uin(0,1),quad F_Y^-1(u)=left{beginarrayrcl0&textrmif&ulefrac14\f(x)&textrmif&u>frac14endarrayright.
$$
If you want the details:
Let $F_Y$ denote the cumulative distribution function of $Y$. Then
$F_Y(y)=0$ if $y<0$, $F_Y(y)=frac14$ if $0le y<f(x)$ and $F_Y(y)=1$ if $yge f(x)$. You can then deduce the quantile function from its definition:
$$
F_Y^-1(u)=infyinmathbb Rmid F_Y(y)ge u.
$$
answered Mar 11 at 8:40
WillWill
1213
New contributor
Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 11 at 8:40
WillWill
1213
New contributor
Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 11 at 8:40
WillWill
1213
answered Mar 11 at 8:40
WillWill
1213
1213
New contributor
Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I'm sorry I don't understand. Who is $x$? Who is $y$? Are $y$ and $Y$ the same for you? Do you mean that $Y=0$ with probability 0.25 and $f(X)$ with probability 0.75 where $X$ is another random variable? The more you specify, the more I can understand your question and give the answer you're looking for
$endgroup$
– Will
Mar 11 at 19:41
add a comment |
$begingroup$
I'm sorry I don't understand. Who is $x$? Who is $y$? Are $y$ and $Y$ the same for you? Do you mean that $Y=0$ with probability 0.25 and $f(X)$ with probability 0.75 where $X$ is another random variable? The more you specify, the more I can understand your question and give the answer you're looking for
$endgroup$
– Will
Mar 11 at 19:41
$begingroup$
I'm sorry I don't understand. Who is $x$? Who is $y$? Are $y$ and $Y$ the same for you? Do you mean that $Y=0$ with probability 0.25 and $f(X)$ with probability 0.75 where $X$ is another random variable? The more you specify, the more I can understand your question and give the answer you're looking for
$endgroup$
– Will
Mar 11 at 19:41
$begingroup$
I'm sorry I don't understand. Who is $x$? Who is $y$? Are $y$ and $Y$ the same for you? Do you mean that $Y=0$ with probability 0.25 and $f(X)$ with probability 0.75 where $X$ is another random variable? The more you specify, the more I can understand your question and give the answer you're looking for
$endgroup$
– Will
Mar 11 at 19:41
add a comment |
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