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Is Spearman correlation never greater than Pearson correlation?
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If Spearman correlation is Pearson correlation, but on ranks of the data - does it mean that absolute Spearman will always be lower or equal to absolute Pearson correlation, but never greater? Is it possible to have variables that show greater absolute Spearman than absolute Pearson?
Example:
# Spearman == Pearson
# We're correlating ranked variables from the beginning
foo <- 1:1e3
bar <- 1:1e3
sapply(c("pearson", "spearman"), function(x) abs(cor(foo, bar, method = x)))
# Spearman < Pearson
foo <- rnorm(1e3)
bar <- rnorm(1e3)
sapply(c("pearson", "spearman"), function(x) abs(cor(foo, bar, method = x)))
correlation pearson-r spearman-rho
asked 2 days ago
PoGibas
6118
add a comment |Â
up vote
2
down vote
favorite
If Spearman correlation is Pearson correlation, but on ranks of the data - does it mean that absolute Spearman will always be lower or equal to absolute Pearson correlation, but never greater? Is it possible to have variables that show greater absolute Spearman than absolute Pearson?
Example:
# Spearman == Pearson
# We're correlating ranked variables from the beginning
foo <- 1:1e3
bar <- 1:1e3
sapply(c("pearson", "spearman"), function(x) abs(cor(foo, bar, method = x)))
# Spearman < Pearson
foo <- rnorm(1e3)
bar <- rnorm(1e3)
sapply(c("pearson", "spearman"), function(x) abs(cor(foo, bar, method = x)))
correlation pearson-r spearman-rho
asked 2 days ago
PoGibas
6118
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If Spearman correlation is Pearson correlation, but on ranks of the data - does it mean that absolute Spearman will always be lower or equal to absolute Pearson correlation, but never greater? Is it possible to have variables that show greater absolute Spearman than absolute Pearson?
Example:
# Spearman == Pearson
# We're correlating ranked variables from the beginning
foo <- 1:1e3
bar <- 1:1e3
sapply(c("pearson", "spearman"), function(x) abs(cor(foo, bar, method = x)))
# Spearman < Pearson
foo <- rnorm(1e3)
bar <- rnorm(1e3)
sapply(c("pearson", "spearman"), function(x) abs(cor(foo, bar, method = x)))
correlation pearson-r spearman-rho
asked 2 days ago
PoGibas
6118
If Spearman correlation is Pearson correlation, but on ranks of the data - does it mean that absolute Spearman will always be lower or equal to absolute Pearson correlation, but never greater? Is it possible to have variables that show greater absolute Spearman than absolute Pearson?
Example:
# Spearman == Pearson
# We're correlating ranked variables from the beginning
foo <- 1:1e3
bar <- 1:1e3
sapply(c("pearson", "spearman"), function(x) abs(cor(foo, bar, method = x)))
# Spearman < Pearson
foo <- rnorm(1e3)
bar <- rnorm(1e3)
sapply(c("pearson", "spearman"), function(x) abs(cor(foo, bar, method = x)))
correlation pearson-r spearman-rho
correlation pearson-r spearman-rho
asked 2 days ago
PoGibas
6118
asked 2 days ago
PoGibas
6118
asked 2 days ago
PoGibas
6118
asked 2 days ago
PoGibas
6118
asked 2 days ago
PoGibas
6118
6118
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
10
down vote
accepted
Simple example in which Spearman correlation is greater than Pearson correlation:
x = 1:10; y = x^2
cor(x,y, meth = "p")
[1] 0.9745586
cor(x,y, meth = "s")
[1] 1
answered yesterday
BruceET
2,940418
You're saying - non linear will give greater correlation in Spearman?
– PoGibas
yesterday
3
Yes, as long as the nonlinear function $y = f(x)$ is increasing. (Or, if you're interested in absolute values, decreasing also works.) Spearman only looks at the ranks (relative orders).
– BruceET
yesterday
add a comment |Â
up vote
2
down vote
Using your own example we can see there is no particular bias for two independent normal distributions:
library(tidyverse)
d <- data_frame(
x = replicate(1e4, rnorm(1e3), FALSE),
y = replicate(1e4, rnorm(1e3), FALSE),
pearson = map2_dbl(x, y, cor, method = 'pearson'),
spearman = map2_dbl(x, y, cor, method = 'spearman'),
p_min_s = pearson - spearman
)
qplot(d$p_min_s, xlab = c('pearson - spearman'))
answered yesterday
Axeman
1659
add a comment |Â
up vote
1
down vote
Several Pearson & Spearman correlation values are compared here.
answered yesterday
gunes
3285
New contributor
gunes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.
This link may perish at some point, which would make this answer less than useful.
– Axeman
yesterday
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
Simple example in which Spearman correlation is greater than Pearson correlation:
x = 1:10; y = x^2
cor(x,y, meth = "p")
[1] 0.9745586
cor(x,y, meth = "s")
[1] 1
answered yesterday
BruceET
2,940418
You're saying - non linear will give greater correlation in Spearman?
– PoGibas
yesterday
3
Yes, as long as the nonlinear function $y = f(x)$ is increasing. (Or, if you're interested in absolute values, decreasing also works.) Spearman only looks at the ranks (relative orders).
– BruceET
yesterday
add a comment |Â
up vote
10
down vote
accepted
Simple example in which Spearman correlation is greater than Pearson correlation:
x = 1:10; y = x^2
cor(x,y, meth = "p")
[1] 0.9745586
cor(x,y, meth = "s")
[1] 1
answered yesterday
BruceET
2,940418
You're saying - non linear will give greater correlation in Spearman?
– PoGibas
yesterday
3
Yes, as long as the nonlinear function $y = f(x)$ is increasing. (Or, if you're interested in absolute values, decreasing also works.) Spearman only looks at the ranks (relative orders).
– BruceET
yesterday
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
Simple example in which Spearman correlation is greater than Pearson correlation:
x = 1:10; y = x^2
cor(x,y, meth = "p")
[1] 0.9745586
cor(x,y, meth = "s")
[1] 1
answered yesterday
BruceET
2,940418
Simple example in which Spearman correlation is greater than Pearson correlation:
x = 1:10; y = x^2
cor(x,y, meth = "p")
[1] 0.9745586
cor(x,y, meth = "s")
[1] 1
answered yesterday
BruceET
2,940418
answered yesterday
BruceET
2,940418
answered yesterday
BruceET
2,940418
answered yesterday
BruceET
2,940418
2,940418
You're saying - non linear will give greater correlation in Spearman?
– PoGibas
yesterday
3
Yes, as long as the nonlinear function $y = f(x)$ is increasing. (Or, if you're interested in absolute values, decreasing also works.) Spearman only looks at the ranks (relative orders).
– BruceET
yesterday
add a comment |Â
You're saying - non linear will give greater correlation in Spearman?
– PoGibas
yesterday
3
Yes, as long as the nonlinear function $y = f(x)$ is increasing. (Or, if you're interested in absolute values, decreasing also works.) Spearman only looks at the ranks (relative orders).
– BruceET
yesterday
You're saying - non linear will give greater correlation in Spearman?
– PoGibas
yesterday
You're saying - non linear will give greater correlation in Spearman?
– PoGibas
yesterday
3
3
Yes, as long as the nonlinear function $y = f(x)$ is increasing. (Or, if you're interested in absolute values, decreasing also works.) Spearman only looks at the ranks (relative orders).
– BruceET
yesterday
Yes, as long as the nonlinear function $y = f(x)$ is increasing. (Or, if you're interested in absolute values, decreasing also works.) Spearman only looks at the ranks (relative orders).
– BruceET
yesterday
add a comment |Â
up vote
2
down vote
Using your own example we can see there is no particular bias for two independent normal distributions:
library(tidyverse)
d <- data_frame(
x = replicate(1e4, rnorm(1e3), FALSE),
y = replicate(1e4, rnorm(1e3), FALSE),
pearson = map2_dbl(x, y, cor, method = 'pearson'),
spearman = map2_dbl(x, y, cor, method = 'spearman'),
p_min_s = pearson - spearman
)
qplot(d$p_min_s, xlab = c('pearson - spearman'))
answered yesterday
Axeman
1659
add a comment |Â
up vote
2
down vote
Using your own example we can see there is no particular bias for two independent normal distributions:
library(tidyverse)
d <- data_frame(
x = replicate(1e4, rnorm(1e3), FALSE),
y = replicate(1e4, rnorm(1e3), FALSE),
pearson = map2_dbl(x, y, cor, method = 'pearson'),
spearman = map2_dbl(x, y, cor, method = 'spearman'),
p_min_s = pearson - spearman
)
qplot(d$p_min_s, xlab = c('pearson - spearman'))
answered yesterday
Axeman
1659
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Using your own example we can see there is no particular bias for two independent normal distributions:
library(tidyverse)
d <- data_frame(
x = replicate(1e4, rnorm(1e3), FALSE),
y = replicate(1e4, rnorm(1e3), FALSE),
pearson = map2_dbl(x, y, cor, method = 'pearson'),
spearman = map2_dbl(x, y, cor, method = 'spearman'),
p_min_s = pearson - spearman
)
qplot(d$p_min_s, xlab = c('pearson - spearman'))
answered yesterday
Axeman
1659
Using your own example we can see there is no particular bias for two independent normal distributions:
library(tidyverse)
d <- data_frame(
x = replicate(1e4, rnorm(1e3), FALSE),
y = replicate(1e4, rnorm(1e3), FALSE),
pearson = map2_dbl(x, y, cor, method = 'pearson'),
spearman = map2_dbl(x, y, cor, method = 'spearman'),
p_min_s = pearson - spearman
)
qplot(d$p_min_s, xlab = c('pearson - spearman'))
answered yesterday
Axeman
1659
answered yesterday
Axeman
1659
answered yesterday
Axeman
1659
answered yesterday
Axeman
1659
1659
add a comment |Â
add a comment |Â
up vote
1
down vote
Several Pearson & Spearman correlation values are compared here.
answered yesterday
gunes
3285
New contributor
gunes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.
This link may perish at some point, which would make this answer less than useful.
– Axeman
yesterday
add a comment |Â
up vote
1
down vote
Several Pearson & Spearman correlation values are compared here.
answered yesterday
gunes
3285
New contributor
gunes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.
This link may perish at some point, which would make this answer less than useful.
– Axeman
yesterday
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Several Pearson & Spearman correlation values are compared here.
answered yesterday
gunes
3285
New contributor
gunes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Several Pearson & Spearman correlation values are compared here.
answered yesterday
gunes
3285
New contributor
gunes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
gunes
3285
New contributor
gunes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
gunes
3285
answered yesterday
gunes
3285
3285
New contributor
gunes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
gunes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
gunes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.
We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.
This link may perish at some point, which would make this answer less than useful.
– Axeman
yesterday
add a comment |Â
This link may perish at some point, which would make this answer less than useful.
– Axeman
yesterday
This link may perish at some point, which would make this answer less than useful.
– Axeman
yesterday
This link may perish at some point, which would make this answer less than useful.
– Axeman
yesterday
add a comment |Â
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