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Claim about Empirical Quantile Function: $hatF_n^-1left(fracin+1right) = X_(i)$
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The lecturer in a class I'm taking defined the empirical quantile function for a sample of $n$ random variables $X_i_i = 1^n$ as follows:
$$
hatF_n^-1(p) = left{beginaligned
&X_(np) &&, np in mathbbN\
&X_(lfloor np+1 rfloor) &&, np notin mathbbN,
endaligned
right.$$
where $X_(i)$ represents the $i^textth$ order statistic of the sample.
Based on this definition, I'm trying to understand the following claim:
$$
hatF_n^-1left(fracin+1right) = X_(i).
$$
My progress to this point:
- I was able to show that: $n in mathbbN$ and $i in [0,n] cap mathbbN implies n+1 nmid ni$. Hence, $fracnin+1 notin mathbbN$ in this situation.
- What then remains to show is that $i leq fracnin+1 + 1 < i+1$. The second inquality is clear, since $fracnn+1 < 1$.
- Thus, showing this boils down to showing: $i leq fracnn+1cdot i + 1$ when $n in mathbbN$, $i in [1,n]$.
statistics inequality
asked 2 days ago
zxmkn
1686
add a comment |Â
up vote
-1
down vote
favorite
The lecturer in a class I'm taking defined the empirical quantile function for a sample of $n$ random variables $X_i_i = 1^n$ as follows:
$$
hatF_n^-1(p) = left{beginaligned
&X_(np) &&, np in mathbbN\
&X_(lfloor np+1 rfloor) &&, np notin mathbbN,
endaligned
right.$$
where $X_(i)$ represents the $i^textth$ order statistic of the sample.
Based on this definition, I'm trying to understand the following claim:
$$
hatF_n^-1left(fracin+1right) = X_(i).
$$
My progress to this point:
- I was able to show that: $n in mathbbN$ and $i in [0,n] cap mathbbN implies n+1 nmid ni$. Hence, $fracnin+1 notin mathbbN$ in this situation.
- What then remains to show is that $i leq fracnin+1 + 1 < i+1$. The second inquality is clear, since $fracnn+1 < 1$.
- Thus, showing this boils down to showing: $i leq fracnn+1cdot i + 1$ when $n in mathbbN$, $i in [1,n]$.
statistics inequality
asked 2 days ago
zxmkn
1686
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
The lecturer in a class I'm taking defined the empirical quantile function for a sample of $n$ random variables $X_i_i = 1^n$ as follows:
$$
hatF_n^-1(p) = left{beginaligned
&X_(np) &&, np in mathbbN\
&X_(lfloor np+1 rfloor) &&, np notin mathbbN,
endaligned
right.$$
where $X_(i)$ represents the $i^textth$ order statistic of the sample.
Based on this definition, I'm trying to understand the following claim:
$$
hatF_n^-1left(fracin+1right) = X_(i).
$$
My progress to this point:
- I was able to show that: $n in mathbbN$ and $i in [0,n] cap mathbbN implies n+1 nmid ni$. Hence, $fracnin+1 notin mathbbN$ in this situation.
- What then remains to show is that $i leq fracnin+1 + 1 < i+1$. The second inquality is clear, since $fracnn+1 < 1$.
- Thus, showing this boils down to showing: $i leq fracnn+1cdot i + 1$ when $n in mathbbN$, $i in [1,n]$.
statistics inequality
asked 2 days ago
zxmkn
1686
The lecturer in a class I'm taking defined the empirical quantile function for a sample of $n$ random variables $X_i_i = 1^n$ as follows:
$$
hatF_n^-1(p) = left{beginaligned
&X_(np) &&, np in mathbbN\
&X_(lfloor np+1 rfloor) &&, np notin mathbbN,
endaligned
right.$$
where $X_(i)$ represents the $i^textth$ order statistic of the sample.
Based on this definition, I'm trying to understand the following claim:
$$
hatF_n^-1left(fracin+1right) = X_(i).
$$
My progress to this point:
- I was able to show that: $n in mathbbN$ and $i in [0,n] cap mathbbN implies n+1 nmid ni$. Hence, $fracnin+1 notin mathbbN$ in this situation.
- What then remains to show is that $i leq fracnin+1 + 1 < i+1$. The second inquality is clear, since $fracnn+1 < 1$.
- Thus, showing this boils down to showing: $i leq fracnn+1cdot i + 1$ when $n in mathbbN$, $i in [1,n]$.
statistics inequality
asked 2 days ago
zxmkn
1686
asked 2 days ago
zxmkn
1686
asked 2 days ago
zxmkn
1686
asked 2 days ago
zxmkn
1686
1686
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
To prove the third bulletpoint, suppose for purpose of establishing a contradiction that for $k in 1,...,n$:
beginequation
beginsplit
k & > fracnkn+1 + 1\
fracn+1n+1k & > fracnkn+1 + fracn+1n+1\
nk + k & > nk + n + 1\
k & > n + 1
endsplit
endequation
Which by assumption is not true.
answered yesterday
Ryan Warnick
1,21867
add a comment |Â
up vote
1
down vote
Maybe an example in R will help you visualize the ECDF function.
Here is a sorted normal sample rounded to two places:
x = sort( round(rnorm(10, 20, 3),2) ); x
[1] 16.38 18.07 18.23 18.37 19.31 20.29 20.55 20.68 22.20 26.29
The fifth order statistic is 19.31:
x[5]
[1] 19.31
The x-values of the ECDF are the order statistics. Its y-values are shown below:
F = (1:10)/11; F
[1] 0.09090909 0.18181818 0.27272727 0.36363636 0.45454545 0.54545455
[7] 0.63636364 0.72727273 0.81818182 0.90909091
The fifth y-value of the ECDF is $5/11.$
F[5]; 5/11
[1] 0.4545455
[1] 0.4545455
Here is a plot of this ECDF, in which the x- and y-values mentioned above
are emphasized:
plot(x, F, type="s", lwd=2, ylim=c(0,1))
abline(h=0:1, col="green2")
points(x, F, pch=19)
abline(v = x[5], col="red", lty="dotted")
abline(h = F[5], col="blue", lty="dotted")
Some authors say that the ECDF function consists only of the heavy dots,
and some say that the horizontal lines are also part of the function.
(If included in a plot, the vertical lines at 'jump points' or 'knots' are just to help the eye
follow the function; sometimes they are dotted lines.)
You should know that different texts and statistical software programs (R among them) define the ECDF in various slightly (but fundamentally) different ways. Here is a plot of the ECDF from R.
plot(ecdf(x))
answered 2 days ago
BruceET
33k61439
Nice visualization. It looks like R uses $hatF(x) = frac1n sum_i=1^n mathbb1_ X_i leq x $ instead of $hatF(x) = frac1n+1 sum_i=1^n mathbb1_ X_i leq x $. Good to know that different texts treat this differently. Intuition is always important, but I was really trying to prove the claim that the give definition leads to $hatF_n^-1(fracin+1) = X_(i)$.
– zxmkn
yesterday
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
To prove the third bulletpoint, suppose for purpose of establishing a contradiction that for $k in 1,...,n$:
beginequation
beginsplit
k & > fracnkn+1 + 1\
fracn+1n+1k & > fracnkn+1 + fracn+1n+1\
nk + k & > nk + n + 1\
k & > n + 1
endsplit
endequation
Which by assumption is not true.
answered yesterday
Ryan Warnick
1,21867
add a comment |Â
up vote
1
down vote
accepted
To prove the third bulletpoint, suppose for purpose of establishing a contradiction that for $k in 1,...,n$:
beginequation
beginsplit
k & > fracnkn+1 + 1\
fracn+1n+1k & > fracnkn+1 + fracn+1n+1\
nk + k & > nk + n + 1\
k & > n + 1
endsplit
endequation
Which by assumption is not true.
answered yesterday
Ryan Warnick
1,21867
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To prove the third bulletpoint, suppose for purpose of establishing a contradiction that for $k in 1,...,n$:
beginequation
beginsplit
k & > fracnkn+1 + 1\
fracn+1n+1k & > fracnkn+1 + fracn+1n+1\
nk + k & > nk + n + 1\
k & > n + 1
endsplit
endequation
Which by assumption is not true.
answered yesterday
Ryan Warnick
1,21867
To prove the third bulletpoint, suppose for purpose of establishing a contradiction that for $k in 1,...,n$:
beginequation
beginsplit
k & > fracnkn+1 + 1\
fracn+1n+1k & > fracnkn+1 + fracn+1n+1\
nk + k & > nk + n + 1\
k & > n + 1
endsplit
endequation
Which by assumption is not true.
answered yesterday
Ryan Warnick
1,21867
answered yesterday
Ryan Warnick
1,21867
answered yesterday
Ryan Warnick
1,21867
answered yesterday
Ryan Warnick
1,21867
1,21867
add a comment |Â
add a comment |Â
up vote
1
down vote
Maybe an example in R will help you visualize the ECDF function.
Here is a sorted normal sample rounded to two places:
x = sort( round(rnorm(10, 20, 3),2) ); x
[1] 16.38 18.07 18.23 18.37 19.31 20.29 20.55 20.68 22.20 26.29
The fifth order statistic is 19.31:
x[5]
[1] 19.31
The x-values of the ECDF are the order statistics. Its y-values are shown below:
F = (1:10)/11; F
[1] 0.09090909 0.18181818 0.27272727 0.36363636 0.45454545 0.54545455
[7] 0.63636364 0.72727273 0.81818182 0.90909091
The fifth y-value of the ECDF is $5/11.$
F[5]; 5/11
[1] 0.4545455
[1] 0.4545455
Here is a plot of this ECDF, in which the x- and y-values mentioned above
are emphasized:
plot(x, F, type="s", lwd=2, ylim=c(0,1))
abline(h=0:1, col="green2")
points(x, F, pch=19)
abline(v = x[5], col="red", lty="dotted")
abline(h = F[5], col="blue", lty="dotted")
Some authors say that the ECDF function consists only of the heavy dots,
and some say that the horizontal lines are also part of the function.
(If included in a plot, the vertical lines at 'jump points' or 'knots' are just to help the eye
follow the function; sometimes they are dotted lines.)
You should know that different texts and statistical software programs (R among them) define the ECDF in various slightly (but fundamentally) different ways. Here is a plot of the ECDF from R.
plot(ecdf(x))
answered 2 days ago
BruceET
33k61439
Nice visualization. It looks like R uses $hatF(x) = frac1n sum_i=1^n mathbb1_ X_i leq x $ instead of $hatF(x) = frac1n+1 sum_i=1^n mathbb1_ X_i leq x $. Good to know that different texts treat this differently. Intuition is always important, but I was really trying to prove the claim that the give definition leads to $hatF_n^-1(fracin+1) = X_(i)$.
– zxmkn
yesterday
add a comment |Â
up vote
1
down vote
Maybe an example in R will help you visualize the ECDF function.
Here is a sorted normal sample rounded to two places:
x = sort( round(rnorm(10, 20, 3),2) ); x
[1] 16.38 18.07 18.23 18.37 19.31 20.29 20.55 20.68 22.20 26.29
The fifth order statistic is 19.31:
x[5]
[1] 19.31
The x-values of the ECDF are the order statistics. Its y-values are shown below:
F = (1:10)/11; F
[1] 0.09090909 0.18181818 0.27272727 0.36363636 0.45454545 0.54545455
[7] 0.63636364 0.72727273 0.81818182 0.90909091
The fifth y-value of the ECDF is $5/11.$
F[5]; 5/11
[1] 0.4545455
[1] 0.4545455
Here is a plot of this ECDF, in which the x- and y-values mentioned above
are emphasized:
plot(x, F, type="s", lwd=2, ylim=c(0,1))
abline(h=0:1, col="green2")
points(x, F, pch=19)
abline(v = x[5], col="red", lty="dotted")
abline(h = F[5], col="blue", lty="dotted")
Some authors say that the ECDF function consists only of the heavy dots,
and some say that the horizontal lines are also part of the function.
(If included in a plot, the vertical lines at 'jump points' or 'knots' are just to help the eye
follow the function; sometimes they are dotted lines.)
You should know that different texts and statistical software programs (R among them) define the ECDF in various slightly (but fundamentally) different ways. Here is a plot of the ECDF from R.
plot(ecdf(x))
answered 2 days ago
BruceET
33k61439
Nice visualization. It looks like R uses $hatF(x) = frac1n sum_i=1^n mathbb1_ X_i leq x $ instead of $hatF(x) = frac1n+1 sum_i=1^n mathbb1_ X_i leq x $. Good to know that different texts treat this differently. Intuition is always important, but I was really trying to prove the claim that the give definition leads to $hatF_n^-1(fracin+1) = X_(i)$.
– zxmkn
yesterday
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Maybe an example in R will help you visualize the ECDF function.
Here is a sorted normal sample rounded to two places:
x = sort( round(rnorm(10, 20, 3),2) ); x
[1] 16.38 18.07 18.23 18.37 19.31 20.29 20.55 20.68 22.20 26.29
The fifth order statistic is 19.31:
x[5]
[1] 19.31
The x-values of the ECDF are the order statistics. Its y-values are shown below:
F = (1:10)/11; F
[1] 0.09090909 0.18181818 0.27272727 0.36363636 0.45454545 0.54545455
[7] 0.63636364 0.72727273 0.81818182 0.90909091
The fifth y-value of the ECDF is $5/11.$
F[5]; 5/11
[1] 0.4545455
[1] 0.4545455
Here is a plot of this ECDF, in which the x- and y-values mentioned above
are emphasized:
plot(x, F, type="s", lwd=2, ylim=c(0,1))
abline(h=0:1, col="green2")
points(x, F, pch=19)
abline(v = x[5], col="red", lty="dotted")
abline(h = F[5], col="blue", lty="dotted")
Some authors say that the ECDF function consists only of the heavy dots,
and some say that the horizontal lines are also part of the function.
(If included in a plot, the vertical lines at 'jump points' or 'knots' are just to help the eye
follow the function; sometimes they are dotted lines.)
You should know that different texts and statistical software programs (R among them) define the ECDF in various slightly (but fundamentally) different ways. Here is a plot of the ECDF from R.
plot(ecdf(x))
answered 2 days ago
BruceET
33k61439
Maybe an example in R will help you visualize the ECDF function.
Here is a sorted normal sample rounded to two places:
x = sort( round(rnorm(10, 20, 3),2) ); x
[1] 16.38 18.07 18.23 18.37 19.31 20.29 20.55 20.68 22.20 26.29
The fifth order statistic is 19.31:
x[5]
[1] 19.31
The x-values of the ECDF are the order statistics. Its y-values are shown below:
F = (1:10)/11; F
[1] 0.09090909 0.18181818 0.27272727 0.36363636 0.45454545 0.54545455
[7] 0.63636364 0.72727273 0.81818182 0.90909091
The fifth y-value of the ECDF is $5/11.$
F[5]; 5/11
[1] 0.4545455
[1] 0.4545455
Here is a plot of this ECDF, in which the x- and y-values mentioned above
are emphasized:
plot(x, F, type="s", lwd=2, ylim=c(0,1))
abline(h=0:1, col="green2")
points(x, F, pch=19)
abline(v = x[5], col="red", lty="dotted")
abline(h = F[5], col="blue", lty="dotted")
Some authors say that the ECDF function consists only of the heavy dots,
and some say that the horizontal lines are also part of the function.
(If included in a plot, the vertical lines at 'jump points' or 'knots' are just to help the eye
follow the function; sometimes they are dotted lines.)
You should know that different texts and statistical software programs (R among them) define the ECDF in various slightly (but fundamentally) different ways. Here is a plot of the ECDF from R.
plot(ecdf(x))
answered 2 days ago
BruceET
33k61439
edited 2 days ago
answered 2 days ago
BruceET
33k61439
answered 2 days ago
BruceET
33k61439
answered 2 days ago
BruceET
33k61439
33k61439
Nice visualization. It looks like R uses $hatF(x) = frac1n sum_i=1^n mathbb1_ X_i leq x $ instead of $hatF(x) = frac1n+1 sum_i=1^n mathbb1_ X_i leq x $. Good to know that different texts treat this differently. Intuition is always important, but I was really trying to prove the claim that the give definition leads to $hatF_n^-1(fracin+1) = X_(i)$.
– zxmkn
yesterday
add a comment |Â
Nice visualization. It looks like R uses $hatF(x) = frac1n sum_i=1^n mathbb1_ X_i leq x $ instead of $hatF(x) = frac1n+1 sum_i=1^n mathbb1_ X_i leq x $. Good to know that different texts treat this differently. Intuition is always important, but I was really trying to prove the claim that the give definition leads to $hatF_n^-1(fracin+1) = X_(i)$.
– zxmkn
yesterday
Nice visualization. It looks like R uses $hatF(x) = frac1n sum_i=1^n mathbb1_ X_i leq x $ instead of $hatF(x) = frac1n+1 sum_i=1^n mathbb1_ X_i leq x $. Good to know that different texts treat this differently. Intuition is always important, but I was really trying to prove the claim that the give definition leads to $hatF_n^-1(fracin+1) = X_(i)$.
– zxmkn
yesterday
Nice visualization. It looks like R uses $hatF(x) = frac1n sum_i=1^n mathbb1_ X_i leq x $ instead of $hatF(x) = frac1n+1 sum_i=1^n mathbb1_ X_i leq x $. Good to know that different texts treat this differently. Intuition is always important, but I was really trying to prove the claim that the give definition leads to $hatF_n^-1(fracin+1) = X_(i)$.
– zxmkn
yesterday
add a comment |Â
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