Skewness and kurtosis of given array
I have given array: [1, 3, 5, 6]. How to calculate skewness and kurtosis of it? There is lots of formulas and I don't know which of them to apply? Also there is different calculators which calculate it differently
example 1: 1.180049,
example 2: 1.57339844872
Please explain how to calculate it properly.
probability statistics
asked Dec 3 '18 at 20:20
TeodorKolev
207212
add a comment |
I have given array: [1, 3, 5, 6]. How to calculate skewness and kurtosis of it? There is lots of formulas and I don't know which of them to apply? Also there is different calculators which calculate it differently
example 1: 1.180049,
example 2: 1.57339844872
Please explain how to calculate it properly.
probability statistics
asked Dec 3 '18 at 20:20
TeodorKolev
207212
add a comment |
I have given array: [1, 3, 5, 6]. How to calculate skewness and kurtosis of it? There is lots of formulas and I don't know which of them to apply? Also there is different calculators which calculate it differently
example 1: 1.180049,
example 2: 1.57339844872
Please explain how to calculate it properly.
probability statistics
asked Dec 3 '18 at 20:20
TeodorKolev
207212
I have given array: [1, 3, 5, 6]. How to calculate skewness and kurtosis of it? There is lots of formulas and I don't know which of them to apply? Also there is different calculators which calculate it differently
example 1: 1.180049,
example 2: 1.57339844872
Please explain how to calculate it properly.
probability statistics
probability statistics
asked Dec 3 '18 at 20:20
TeodorKolev
207212
asked Dec 3 '18 at 20:20
TeodorKolev
207212
asked Dec 3 '18 at 20:20
TeodorKolev
207212
asked Dec 3 '18 at 20:20
TeodorKolev
207212
asked Dec 3 '18 at 20:20
TeodorKolev
207212
207212
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
These two values are the respective third and fourth central statistical moments.
AFAIK they're commonly defined to be
$$
begin{equation*}
begin{split}
skewness &= N^{-1} sum_{k=1}^N left(frac{x_k - mu}{sigma}right)^3 \
kurtosis &= N^{-1} sum_{k=1}^N left(frac{x_k - mu}{sigma}right)^4 \
end{split}
end{equation*}
$$
for data $x = {x_1,ldots,x_N}$.
Also cf the definitions for skewness and kurtosis.
So for the calculation of the values, you start by calculating the mean value over the array $a$ of length $N$:
$$ mu = N^{-1} sum_{k=1}^N a_k $$
After that, one needs to compute the standard deviation $sigma$:
$$ sigma^2 = N^{-1} sum_{k=1}^N left(a_k - muright)^2 $$
Note that some definitions require $sigma$ to be divided by $N-1$ instead of $N$.
Having both $mu$ and $sigma$, you can compute the skewness and kurtosis of the array.
Now we apply this to the array $a = [1,3,5,6]$:
$$
begin{equation*}
begin{split}
mu &= frac{1}{4} left(1+3+5+6right) = 3.75 \
sigma &= sqrt{frac{1}{4} left((1-3.75)^2+(3-3.75)^2+(5-3.75)^2+(6-3.75)^2right)} = 1.92 \
skewness &= frac{1}{4}left((frac{1-3.75}{1.92})^3+(frac{3-3.75}{1.92})^3+(frac{5-3.75}{1.92})^3+(frac{6-3.75}{1.92})^3right) = -0.278 \
kurtosis &= frac{1}{4}left((frac{1-3.75}{1.92})^4+(frac{3-3.75}{1.92})^4+(frac{5-3.75}{1.92})^4+(frac{6-3.75}{1.92})^4right) = 1.574 \
end{split}
end{equation*}
$$
Simple verification using Python's numpy:
>>> np.mean([1,3,5,6])
3.75
>>> np.std([1,3,5,6])
1.920286436967152
>>> 0.25 * sum(map(lambda x: ((float(x)-3.75)/1.92)**3, [1,3,5,6]))
-0.278155008951823
>>> 0.25 * sum(map(lambda x: ((float(x)-3.75)/1.92)**4, [1,3,5,6]))
1.5743375744348693
answered Dec 3 '18 at 20:23
Thomas Lang
1624
This answer gives no explanation. I do not understand that formulas. Can you please show how to calculate with numbers of given array?
– TeodorKolev
Dec 3 '18 at 20:25
Why online calculators calculate it differently?
– TeodorKolev
Dec 3 '18 at 20:41
@TeodorKolev Sorry, forgot the square root. Check again. The differnce might arise from different scaling in the standard deviation.
– Thomas Lang
Dec 3 '18 at 20:44
Yet again, numbers are different than calculators
– TeodorKolev
Dec 3 '18 at 20:46
Why? The second calculate says 1.573... and my result is 1.574, but notice for simplicity I rounded to two digits, as the Python version shows. Thus with the usual definitions, the second calculator is right.
– Thomas Lang
Dec 3 '18 at 20:47
|
show 4 more comments
Your Answer
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1 Answer
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1 Answer
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These two values are the respective third and fourth central statistical moments.
AFAIK they're commonly defined to be
$$
begin{equation*}
begin{split}
skewness &= N^{-1} sum_{k=1}^N left(frac{x_k - mu}{sigma}right)^3 \
kurtosis &= N^{-1} sum_{k=1}^N left(frac{x_k - mu}{sigma}right)^4 \
end{split}
end{equation*}
$$
for data $x = {x_1,ldots,x_N}$.
Also cf the definitions for skewness and kurtosis.
So for the calculation of the values, you start by calculating the mean value over the array $a$ of length $N$:
$$ mu = N^{-1} sum_{k=1}^N a_k $$
After that, one needs to compute the standard deviation $sigma$:
$$ sigma^2 = N^{-1} sum_{k=1}^N left(a_k - muright)^2 $$
Note that some definitions require $sigma$ to be divided by $N-1$ instead of $N$.
Having both $mu$ and $sigma$, you can compute the skewness and kurtosis of the array.
Now we apply this to the array $a = [1,3,5,6]$:
$$
begin{equation*}
begin{split}
mu &= frac{1}{4} left(1+3+5+6right) = 3.75 \
sigma &= sqrt{frac{1}{4} left((1-3.75)^2+(3-3.75)^2+(5-3.75)^2+(6-3.75)^2right)} = 1.92 \
skewness &= frac{1}{4}left((frac{1-3.75}{1.92})^3+(frac{3-3.75}{1.92})^3+(frac{5-3.75}{1.92})^3+(frac{6-3.75}{1.92})^3right) = -0.278 \
kurtosis &= frac{1}{4}left((frac{1-3.75}{1.92})^4+(frac{3-3.75}{1.92})^4+(frac{5-3.75}{1.92})^4+(frac{6-3.75}{1.92})^4right) = 1.574 \
end{split}
end{equation*}
$$
Simple verification using Python's numpy:
>>> np.mean([1,3,5,6])
3.75
>>> np.std([1,3,5,6])
1.920286436967152
>>> 0.25 * sum(map(lambda x: ((float(x)-3.75)/1.92)**3, [1,3,5,6]))
-0.278155008951823
>>> 0.25 * sum(map(lambda x: ((float(x)-3.75)/1.92)**4, [1,3,5,6]))
1.5743375744348693
answered Dec 3 '18 at 20:23
Thomas Lang
1624
This answer gives no explanation. I do not understand that formulas. Can you please show how to calculate with numbers of given array?
– TeodorKolev
Dec 3 '18 at 20:25
Why online calculators calculate it differently?
– TeodorKolev
Dec 3 '18 at 20:41
@TeodorKolev Sorry, forgot the square root. Check again. The differnce might arise from different scaling in the standard deviation.
– Thomas Lang
Dec 3 '18 at 20:44
Yet again, numbers are different than calculators
– TeodorKolev
Dec 3 '18 at 20:46
Why? The second calculate says 1.573... and my result is 1.574, but notice for simplicity I rounded to two digits, as the Python version shows. Thus with the usual definitions, the second calculator is right.
– Thomas Lang
Dec 3 '18 at 20:47
|
show 4 more comments
These two values are the respective third and fourth central statistical moments.
AFAIK they're commonly defined to be
$$
begin{equation*}
begin{split}
skewness &= N^{-1} sum_{k=1}^N left(frac{x_k - mu}{sigma}right)^3 \
kurtosis &= N^{-1} sum_{k=1}^N left(frac{x_k - mu}{sigma}right)^4 \
end{split}
end{equation*}
$$
for data $x = {x_1,ldots,x_N}$.
Also cf the definitions for skewness and kurtosis.
So for the calculation of the values, you start by calculating the mean value over the array $a$ of length $N$:
$$ mu = N^{-1} sum_{k=1}^N a_k $$
After that, one needs to compute the standard deviation $sigma$:
$$ sigma^2 = N^{-1} sum_{k=1}^N left(a_k - muright)^2 $$
Note that some definitions require $sigma$ to be divided by $N-1$ instead of $N$.
Having both $mu$ and $sigma$, you can compute the skewness and kurtosis of the array.
Now we apply this to the array $a = [1,3,5,6]$:
$$
begin{equation*}
begin{split}
mu &= frac{1}{4} left(1+3+5+6right) = 3.75 \
sigma &= sqrt{frac{1}{4} left((1-3.75)^2+(3-3.75)^2+(5-3.75)^2+(6-3.75)^2right)} = 1.92 \
skewness &= frac{1}{4}left((frac{1-3.75}{1.92})^3+(frac{3-3.75}{1.92})^3+(frac{5-3.75}{1.92})^3+(frac{6-3.75}{1.92})^3right) = -0.278 \
kurtosis &= frac{1}{4}left((frac{1-3.75}{1.92})^4+(frac{3-3.75}{1.92})^4+(frac{5-3.75}{1.92})^4+(frac{6-3.75}{1.92})^4right) = 1.574 \
end{split}
end{equation*}
$$
Simple verification using Python's numpy:
>>> np.mean([1,3,5,6])
3.75
>>> np.std([1,3,5,6])
1.920286436967152
>>> 0.25 * sum(map(lambda x: ((float(x)-3.75)/1.92)**3, [1,3,5,6]))
-0.278155008951823
>>> 0.25 * sum(map(lambda x: ((float(x)-3.75)/1.92)**4, [1,3,5,6]))
1.5743375744348693
answered Dec 3 '18 at 20:23
Thomas Lang
1624
This answer gives no explanation. I do not understand that formulas. Can you please show how to calculate with numbers of given array?
– TeodorKolev
Dec 3 '18 at 20:25
Why online calculators calculate it differently?
– TeodorKolev
Dec 3 '18 at 20:41
@TeodorKolev Sorry, forgot the square root. Check again. The differnce might arise from different scaling in the standard deviation.
– Thomas Lang
Dec 3 '18 at 20:44
Yet again, numbers are different than calculators
– TeodorKolev
Dec 3 '18 at 20:46
Why? The second calculate says 1.573... and my result is 1.574, but notice for simplicity I rounded to two digits, as the Python version shows. Thus with the usual definitions, the second calculator is right.
– Thomas Lang
Dec 3 '18 at 20:47
|
show 4 more comments
These two values are the respective third and fourth central statistical moments.
AFAIK they're commonly defined to be
$$
begin{equation*}
begin{split}
skewness &= N^{-1} sum_{k=1}^N left(frac{x_k - mu}{sigma}right)^3 \
kurtosis &= N^{-1} sum_{k=1}^N left(frac{x_k - mu}{sigma}right)^4 \
end{split}
end{equation*}
$$
for data $x = {x_1,ldots,x_N}$.
Also cf the definitions for skewness and kurtosis.
So for the calculation of the values, you start by calculating the mean value over the array $a$ of length $N$:
$$ mu = N^{-1} sum_{k=1}^N a_k $$
After that, one needs to compute the standard deviation $sigma$:
$$ sigma^2 = N^{-1} sum_{k=1}^N left(a_k - muright)^2 $$
Note that some definitions require $sigma$ to be divided by $N-1$ instead of $N$.
Having both $mu$ and $sigma$, you can compute the skewness and kurtosis of the array.
Now we apply this to the array $a = [1,3,5,6]$:
$$
begin{equation*}
begin{split}
mu &= frac{1}{4} left(1+3+5+6right) = 3.75 \
sigma &= sqrt{frac{1}{4} left((1-3.75)^2+(3-3.75)^2+(5-3.75)^2+(6-3.75)^2right)} = 1.92 \
skewness &= frac{1}{4}left((frac{1-3.75}{1.92})^3+(frac{3-3.75}{1.92})^3+(frac{5-3.75}{1.92})^3+(frac{6-3.75}{1.92})^3right) = -0.278 \
kurtosis &= frac{1}{4}left((frac{1-3.75}{1.92})^4+(frac{3-3.75}{1.92})^4+(frac{5-3.75}{1.92})^4+(frac{6-3.75}{1.92})^4right) = 1.574 \
end{split}
end{equation*}
$$
Simple verification using Python's numpy:
>>> np.mean([1,3,5,6])
3.75
>>> np.std([1,3,5,6])
1.920286436967152
>>> 0.25 * sum(map(lambda x: ((float(x)-3.75)/1.92)**3, [1,3,5,6]))
-0.278155008951823
>>> 0.25 * sum(map(lambda x: ((float(x)-3.75)/1.92)**4, [1,3,5,6]))
1.5743375744348693
answered Dec 3 '18 at 20:23
Thomas Lang
1624
These two values are the respective third and fourth central statistical moments.
AFAIK they're commonly defined to be
$$
begin{equation*}
begin{split}
skewness &= N^{-1} sum_{k=1}^N left(frac{x_k - mu}{sigma}right)^3 \
kurtosis &= N^{-1} sum_{k=1}^N left(frac{x_k - mu}{sigma}right)^4 \
end{split}
end{equation*}
$$
for data $x = {x_1,ldots,x_N}$.
Also cf the definitions for skewness and kurtosis.
So for the calculation of the values, you start by calculating the mean value over the array $a$ of length $N$:
$$ mu = N^{-1} sum_{k=1}^N a_k $$
After that, one needs to compute the standard deviation $sigma$:
$$ sigma^2 = N^{-1} sum_{k=1}^N left(a_k - muright)^2 $$
Note that some definitions require $sigma$ to be divided by $N-1$ instead of $N$.
Having both $mu$ and $sigma$, you can compute the skewness and kurtosis of the array.
Now we apply this to the array $a = [1,3,5,6]$:
$$
begin{equation*}
begin{split}
mu &= frac{1}{4} left(1+3+5+6right) = 3.75 \
sigma &= sqrt{frac{1}{4} left((1-3.75)^2+(3-3.75)^2+(5-3.75)^2+(6-3.75)^2right)} = 1.92 \
skewness &= frac{1}{4}left((frac{1-3.75}{1.92})^3+(frac{3-3.75}{1.92})^3+(frac{5-3.75}{1.92})^3+(frac{6-3.75}{1.92})^3right) = -0.278 \
kurtosis &= frac{1}{4}left((frac{1-3.75}{1.92})^4+(frac{3-3.75}{1.92})^4+(frac{5-3.75}{1.92})^4+(frac{6-3.75}{1.92})^4right) = 1.574 \
end{split}
end{equation*}
$$
Simple verification using Python's numpy:
>>> np.mean([1,3,5,6])
3.75
>>> np.std([1,3,5,6])
1.920286436967152
>>> 0.25 * sum(map(lambda x: ((float(x)-3.75)/1.92)**3, [1,3,5,6]))
-0.278155008951823
>>> 0.25 * sum(map(lambda x: ((float(x)-3.75)/1.92)**4, [1,3,5,6]))
1.5743375744348693
answered Dec 3 '18 at 20:23
Thomas Lang
1624
edited Dec 3 '18 at 20:46
answered Dec 3 '18 at 20:23
Thomas Lang
1624
answered Dec 3 '18 at 20:23
Thomas Lang
1624
answered Dec 3 '18 at 20:23
Thomas Lang
1624
1624
This answer gives no explanation. I do not understand that formulas. Can you please show how to calculate with numbers of given array?
– TeodorKolev
Dec 3 '18 at 20:25
Why online calculators calculate it differently?
– TeodorKolev
Dec 3 '18 at 20:41
@TeodorKolev Sorry, forgot the square root. Check again. The differnce might arise from different scaling in the standard deviation.
– Thomas Lang
Dec 3 '18 at 20:44
Yet again, numbers are different than calculators
– TeodorKolev
Dec 3 '18 at 20:46
Why? The second calculate says 1.573... and my result is 1.574, but notice for simplicity I rounded to two digits, as the Python version shows. Thus with the usual definitions, the second calculator is right.
– Thomas Lang
Dec 3 '18 at 20:47
|
show 4 more comments
This answer gives no explanation. I do not understand that formulas. Can you please show how to calculate with numbers of given array?
– TeodorKolev
Dec 3 '18 at 20:25
Why online calculators calculate it differently?
– TeodorKolev
Dec 3 '18 at 20:41
@TeodorKolev Sorry, forgot the square root. Check again. The differnce might arise from different scaling in the standard deviation.
– Thomas Lang
Dec 3 '18 at 20:44
Yet again, numbers are different than calculators
– TeodorKolev
Dec 3 '18 at 20:46
Why? The second calculate says 1.573... and my result is 1.574, but notice for simplicity I rounded to two digits, as the Python version shows. Thus with the usual definitions, the second calculator is right.
– Thomas Lang
Dec 3 '18 at 20:47
This answer gives no explanation. I do not understand that formulas. Can you please show how to calculate with numbers of given array?
– TeodorKolev
Dec 3 '18 at 20:25
This answer gives no explanation. I do not understand that formulas. Can you please show how to calculate with numbers of given array?
– TeodorKolev
Dec 3 '18 at 20:25
Why online calculators calculate it differently?
– TeodorKolev
Dec 3 '18 at 20:41
Why online calculators calculate it differently?
– TeodorKolev
Dec 3 '18 at 20:41
@TeodorKolev Sorry, forgot the square root. Check again. The differnce might arise from different scaling in the standard deviation.
– Thomas Lang
Dec 3 '18 at 20:44
@TeodorKolev Sorry, forgot the square root. Check again. The differnce might arise from different scaling in the standard deviation.
– Thomas Lang
Dec 3 '18 at 20:44
Yet again, numbers are different than calculators
– TeodorKolev
Dec 3 '18 at 20:46
Yet again, numbers are different than calculators
– TeodorKolev
Dec 3 '18 at 20:46
Why? The second calculate says 1.573... and my result is 1.574, but notice for simplicity I rounded to two digits, as the Python version shows. Thus with the usual definitions, the second calculator is right.
– Thomas Lang
Dec 3 '18 at 20:47
Why? The second calculate says 1.573... and my result is 1.574, but notice for simplicity I rounded to two digits, as the Python version shows. Thus with the usual definitions, the second calculator is right.
– Thomas Lang
Dec 3 '18 at 20:47
|
show 4 more comments
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