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What do you call a statistical mean that is calculated from upper and lower extremes in any given dataset?

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4

$begingroup$

What do you call a statistical mean that is calculated from upper and lower extremes in any given dataset?

For example, if you have a set:

{ -2, 0 , 8, 9, 1, 50, -2, 6}

The upper extreme of this set is 50 and lower extreme is -2. So, average of the extremes would be (-2 + 50 / 2) = 48/2 = 24

Is there a term for this kind of statistical mean?

mathematical-statistics mean average

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asked 8 hours ago

blackbeardblackbeard

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$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It's the "midrange".
  $endgroup$
  – jbowman
  8 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

What do you call a statistical mean that is calculated from upper and lower extremes in any given dataset?

For example, if you have a set:

{ -2, 0 , 8, 9, 1, 50, -2, 6}

The upper extreme of this set is 50 and lower extreme is -2. So, average of the extremes would be (-2 + 50 / 2) = 48/2 = 24

Is there a term for this kind of statistical mean?

mathematical-statistics mean average

share|cite|improve this question

asked 8 hours ago

blackbeardblackbeard

2322 bronze badges

New contributor

blackbeard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It's the "midrange".
  $endgroup$
  – jbowman
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

What do you call a statistical mean that is calculated from upper and lower extremes in any given dataset?

For example, if you have a set:

{ -2, 0 , 8, 9, 1, 50, -2, 6}

The upper extreme of this set is 50 and lower extreme is -2. So, average of the extremes would be (-2 + 50 / 2) = 48/2 = 24

Is there a term for this kind of statistical mean?

mathematical-statistics mean average

share|cite|improve this question

asked 8 hours ago

blackbeardblackbeard

2322 bronze badges

New contributor

blackbeard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

{ -2, 0 , 8, 9, 1, 50, -2, 6}

share|cite|improve this question

asked 8 hours ago

blackbeardblackbeard

2322 bronze badges

New contributor

blackbeard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 8 hours ago

blackbeardblackbeard

2322 bronze badges

New contributor

blackbeard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 8 hours ago

blackbeardblackbeard

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New contributor

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asked 8 hours ago

blackbeardblackbeard

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1 Answer
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4

$begingroup$

It's called the midrange and while it's not the most widely used statistic in the world it does have some relevance to the uniform distribution.

Let's introduce the order statistic notation: if have $n$ i.i.d. random variables $X_1, ..., X_n$, then the notation $X_{(i)}$ is used to refer to the $i$-th largest of the set ${X_1, ..., X_n}$. Thus we have:

$$ X_{(1)} ≤ X_{(2)} ≤···≤ X_{(n)} tag{1} $$

Where $X_{(1)}$ is the minimum and $X_{(n)}$ is the maximum element. Then range and midrange are defined as:

$$ begin{align}
R & = X_{(n)} - X_{(1)} tag{2} \
A & = frac{X_{(1)} + X_{(n)}}{2} tag{3} \
end{align}
$$

These formulas are taken from CRC Standard Probability and Statistics Tables and Formulae, section 4.6.6.

If $X_i$ is assumed to have a uniform distribution $X_i sim U(alpha, beta)$, where $alpha$ and $beta$ are the lower and upper bounds respectively, then we can give the MLE estimates in terms of these formulas:

$$
begin{align}
hat{alpha} & = X_{(1)} tag{4} \
hat{beta} & = X_{(n)} tag{5}
end{align}
$$

The mean of the resulting distribution is the same as the midrange:

$$
begin{align}
mu & = A = frac{X_{(1)} + X_{(n)}}{2} tag{6} \
end{align}
$$

This is probably the only use for this particular statistic.

share|cite|improve this answer

answered 7 hours ago

olooneyolooney

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add a comment | 

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4

$begingroup$

It's called the midrange and while it's not the most widely used statistic in the world it does have some relevance to the uniform distribution.

Let's introduce the order statistic notation: if have $n$ i.i.d. random variables $X_1, ..., X_n$, then the notation $X_{(i)}$ is used to refer to the $i$-th largest of the set ${X_1, ..., X_n}$. Thus we have:

$$ X_{(1)} ≤ X_{(2)} ≤···≤ X_{(n)} tag{1} $$

Where $X_{(1)}$ is the minimum and $X_{(n)}$ is the maximum element. Then range and midrange are defined as:

$$ begin{align}
R & = X_{(n)} - X_{(1)} tag{2} \
A & = frac{X_{(1)} + X_{(n)}}{2} tag{3} \
end{align}
$$

These formulas are taken from CRC Standard Probability and Statistics Tables and Formulae, section 4.6.6.

If $X_i$ is assumed to have a uniform distribution $X_i sim U(alpha, beta)$, where $alpha$ and $beta$ are the lower and upper bounds respectively, then we can give the MLE estimates in terms of these formulas:

$$
begin{align}
hat{alpha} & = X_{(1)} tag{4} \
hat{beta} & = X_{(n)} tag{5}
end{align}
$$

The mean of the resulting distribution is the same as the midrange:

$$
begin{align}
mu & = A = frac{X_{(1)} + X_{(n)}}{2} tag{6} \
end{align}
$$

This is probably the only use for this particular statistic.

share|cite|improve this answer

answered 7 hours ago

olooneyolooney

2,05888 silver badges1919 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

It's called the midrange and while it's not the most widely used statistic in the world it does have some relevance to the uniform distribution.

Let's introduce the order statistic notation: if have $n$ i.i.d. random variables $X_1, ..., X_n$, then the notation $X_{(i)}$ is used to refer to the $i$-th largest of the set ${X_1, ..., X_n}$. Thus we have:

$$ X_{(1)} ≤ X_{(2)} ≤···≤ X_{(n)} tag{1} $$

Where $X_{(1)}$ is the minimum and $X_{(n)}$ is the maximum element. Then range and midrange are defined as:

$$ begin{align}
R & = X_{(n)} - X_{(1)} tag{2} \
A & = frac{X_{(1)} + X_{(n)}}{2} tag{3} \
end{align}
$$

These formulas are taken from CRC Standard Probability and Statistics Tables and Formulae, section 4.6.6.

If $X_i$ is assumed to have a uniform distribution $X_i sim U(alpha, beta)$, where $alpha$ and $beta$ are the lower and upper bounds respectively, then we can give the MLE estimates in terms of these formulas:

$$
begin{align}
hat{alpha} & = X_{(1)} tag{4} \
hat{beta} & = X_{(n)} tag{5}
end{align}
$$

The mean of the resulting distribution is the same as the midrange:

$$
begin{align}
mu & = A = frac{X_{(1)} + X_{(n)}}{2} tag{6} \
end{align}
$$

This is probably the only use for this particular statistic.

share|cite|improve this answer

answered 7 hours ago

olooneyolooney

2,05888 silver badges1919 bronze badges

$endgroup$

add a comment | 

$begingroup$

It's called the midrange and while it's not the most widely used statistic in the world it does have some relevance to the uniform distribution.

Let's introduce the order statistic notation: if have $n$ i.i.d. random variables $X_1, ..., X_n$, then the notation $X_{(i)}$ is used to refer to the $i$-th largest of the set ${X_1, ..., X_n}$. Thus we have:

$$ X_{(1)} ≤ X_{(2)} ≤···≤ X_{(n)} tag{1} $$

Where $X_{(1)}$ is the minimum and $X_{(n)}$ is the maximum element. Then range and midrange are defined as:

$$ begin{align}
R & = X_{(n)} - X_{(1)} tag{2} \
A & = frac{X_{(1)} + X_{(n)}}{2} tag{3} \
end{align}
$$

These formulas are taken from CRC Standard Probability and Statistics Tables and Formulae, section 4.6.6.

If $X_i$ is assumed to have a uniform distribution $X_i sim U(alpha, beta)$, where $alpha$ and $beta$ are the lower and upper bounds respectively, then we can give the MLE estimates in terms of these formulas:

$$
begin{align}
hat{alpha} & = X_{(1)} tag{4} \
hat{beta} & = X_{(n)} tag{5}
end{align}
$$

The mean of the resulting distribution is the same as the midrange:

$$
begin{align}
mu & = A = frac{X_{(1)} + X_{(n)}}{2} tag{6} \
end{align}
$$

This is probably the only use for this particular statistic.

share|cite|improve this answer

answered 7 hours ago

olooneyolooney

2,05888 silver badges1919 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

olooneyolooney

2,05888 silver badges1919 bronze badges

answered 7 hours ago

olooneyolooney

2,05888 silver badges1919 bronze badges

answered 7 hours ago

olooneyolooney

2,05888 silver badges1919 bronze badges