3 Corrected Math typo based on the reference provided

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$$$\sigma\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$$$\sigma\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.
n <- 10000 # number of samples
x <- array(list(), n)

set.seed(2014)

for(i in 1:n){
x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample
}

dn <- rep(0, n) # mad
sn <- rep(0, n) # sd

for(i in 1:1000){
sn[i] <- sqrt(var(x[[i]])*999/1000) # sd
}

mean(dn) # **0.07979068 check out**
mean(sn) # **0.09995901 check out**

var(dn)/var(sn) # **0.6371817**


As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.
n <- 10000 # number of samples
x <- array(list(), n)

set.seed(2014)

for(i in 1:n){
x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample
}

dn <- rep(0, n) # mad
sn <- rep(0, n) # sd

for(i in 1:1000){
sn[i] <- sqrt(var(x[[i]])*999/1000) # sd
}

mean(dn) # **0.07979068 check out**
mean(sn) # **0.09995901 check out**

var(dn)/var(sn) # **0.6371817**


As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sigma\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sigma\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.
n <- 10000 # number of samples
x <- array(list(), n)

set.seed(2014)

for(i in 1:n){
x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample
}

dn <- rep(0, n) # mad
sn <- rep(0, n) # sd

for(i in 1:1000){
sn[i] <- sqrt(var(x[[i]])*999/1000) # sd
}

mean(dn) # **0.07979068 check out**
mean(sn) # **0.09995901 check out**

var(dn)/var(sn) # **0.6371817**


As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

2 fixed code format

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.

n <- 10000 # number of samples x <- array(list(), n)

set.seed(2014)

for(i in 1:n){ x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample }

dn <- rep(0, n) # mad sn <- rep(0, n) # sd

for(i in 1:1000){ dn[i] <- mean(abs(x[[i]]-mean(x[[i]]))) # mad sn[i] <- sqrt(var(x[[i]])*999/1000) # sd }

mean(dn) # 0.07979068 check out mean(sn) # 0.09995901 check out

var(dn)/var(sn) # 0.6371817

I did some simulation to test all the above statements. Here are the codes and outcome.
n <- 10000 # number of samples
x <- array(list(), n)

set.seed(2014)

for(i in 1:n){
x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample
}

dn <- rep(0, n) # mad
sn <- rep(0, n) # sd

for(i in 1:1000){
sn[i] <- sqrt(var(x[[i]])*999/1000) # sd
}

mean(dn) # **0.07979068 check out**
mean(sn) # **0.09995901 check out**

var(dn)/var(sn) # **0.6371817**


As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.

n <- 10000 # number of samples x <- array(list(), n)

set.seed(2014)

for(i in 1:n){ x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample }

dn <- rep(0, n) # mad sn <- rep(0, n) # sd

for(i in 1:1000){ dn[i] <- mean(abs(x[[i]]-mean(x[[i]]))) # mad sn[i] <- sqrt(var(x[[i]])*999/1000) # sd }

mean(dn) # 0.07979068 check out mean(sn) # 0.09995901 check out

var(dn)/var(sn) # 0.6371817

As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.
n <- 10000 # number of samples
x <- array(list(), n)

set.seed(2014)

for(i in 1:n){
x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample
}

dn <- rep(0, n) # mad
sn <- rep(0, n) # sd

for(i in 1:1000){
sn[i] <- sqrt(var(x[[i]])*999/1000) # sd
}

mean(dn) # **0.07979068 check out**
mean(sn) # **0.09995901 check out**

var(dn)/var(sn) # **0.6371817**


As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!

1

# Comparison between MAD and SD

I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $$(X_n)$$ be a sequence of random variables and define two measures of spread as follows.

1. Mean Absolute Deviation: $$d_n := \frac{1}{n}\sum|x_i-\bar x|$$.
2. Standard Deviation: $$s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$$.

Then he mentioned that Fisher claimed that for identically distributed normal observations $$s_n$$ is about 12% more efficient than $$d_n$$. In addition, $$s_n$$ converges to $$\sigma$$ while $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$. I have several questions about these statements.

1. How to prove that $$s_n$$ is 12% more efficient, please? As least where to find the proof, please?
2. How to prove that $$d_n$$ converges to $$\sqrt{2/\pi}\doteq 0.8\sigma$$, please? Again at least where to find the proof, please?
3. I did some simulation to test all the above statements. Here are the codes and outcome.

n <- 10000 # number of samples x <- array(list(), n)

set.seed(2014)

for(i in 1:n){ x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample }

dn <- rep(0, n) # mad sn <- rep(0, n) # sd

for(i in 1:1000){ dn[i] <- mean(abs(x[[i]]-mean(x[[i]]))) # mad sn[i] <- sqrt(var(x[[i]])*999/1000) # sd }

mean(dn) # 0.07979068 check out mean(sn) # 0.09995901 check out

var(dn)/var(sn) # 0.6371817

As the above simulation shows, the 12% efficiency of $$s_n$$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!