Interpreting an agreement function
$begingroup$
Consider the following function:
$$
P = frac{1}{n(n - 1)} sum_{j=1}^k n_{j} (n_{j} - 1)
$$
where for $n = sum_{j=1}^k n_{j}$.
Intuitively, this function measure concentration of values the vector $(n_1, ..., n_k)$. Take the edge cases:
Values concentrated: $exists j, n_j = n$ (in other words $forall i neq j, n_i = 0$) $Rightarrow P = 1.0$
Least concentration (uniformly distribution): $n_1 = n_2 = ... = n_k Rightarrow P = 0.0$
The formula is relatively simple-looking, but it's not obvious to interpret. I am looking for better/simpler ways to explain this formula to people who are not mathematically sophisticated (say in my psychology department, who know the basics of statistics like mean and variance). Would appreciate if you provide me with any suggestions on this.
probability combinatorics intuition education
edited Nov 28 '18 at 21:00
greedoid
38.7k114797
asked Nov 28 '18 at 17:57
DanielDaniel
1,1311021
$endgroup$
add a comment |
$begingroup$
Consider the following function:
$$
P = frac{1}{n(n - 1)} sum_{j=1}^k n_{j} (n_{j} - 1)
$$
where for $n = sum_{j=1}^k n_{j}$.
Intuitively, this function measure concentration of values the vector $(n_1, ..., n_k)$. Take the edge cases:
Values concentrated: $exists j, n_j = n$ (in other words $forall i neq j, n_i = 0$) $Rightarrow P = 1.0$
Least concentration (uniformly distribution): $n_1 = n_2 = ... = n_k Rightarrow P = 0.0$
The formula is relatively simple-looking, but it's not obvious to interpret. I am looking for better/simpler ways to explain this formula to people who are not mathematically sophisticated (say in my psychology department, who know the basics of statistics like mean and variance). Would appreciate if you provide me with any suggestions on this.
probability combinatorics intuition education
edited Nov 28 '18 at 21:00
greedoid
38.7k114797
asked Nov 28 '18 at 17:57
DanielDaniel
1,1311021
$endgroup$
add a comment |
$begingroup$
Consider the following function:
$$
P = frac{1}{n(n - 1)} sum_{j=1}^k n_{j} (n_{j} - 1)
$$
where for $n = sum_{j=1}^k n_{j}$.
Intuitively, this function measure concentration of values the vector $(n_1, ..., n_k)$. Take the edge cases:
Values concentrated: $exists j, n_j = n$ (in other words $forall i neq j, n_i = 0$) $Rightarrow P = 1.0$
Least concentration (uniformly distribution): $n_1 = n_2 = ... = n_k Rightarrow P = 0.0$
The formula is relatively simple-looking, but it's not obvious to interpret. I am looking for better/simpler ways to explain this formula to people who are not mathematically sophisticated (say in my psychology department, who know the basics of statistics like mean and variance). Would appreciate if you provide me with any suggestions on this.
probability combinatorics intuition education
edited Nov 28 '18 at 21:00
greedoid
38.7k114797
asked Nov 28 '18 at 17:57
DanielDaniel
1,1311021
$endgroup$
Consider the following function:
$$
P = frac{1}{n(n - 1)} sum_{j=1}^k n_{j} (n_{j} - 1)
$$
where for $n = sum_{j=1}^k n_{j}$.
Intuitively, this function measure concentration of values the vector $(n_1, ..., n_k)$. Take the edge cases:
Values concentrated: $exists j, n_j = n$ (in other words $forall i neq j, n_i = 0$) $Rightarrow P = 1.0$
Least concentration (uniformly distribution): $n_1 = n_2 = ... = n_k Rightarrow P = 0.0$
The formula is relatively simple-looking, but it's not obvious to interpret. I am looking for better/simpler ways to explain this formula to people who are not mathematically sophisticated (say in my psychology department, who know the basics of statistics like mean and variance). Would appreciate if you provide me with any suggestions on this.
probability combinatorics intuition education
probability combinatorics intuition education
edited Nov 28 '18 at 21:00
greedoid
38.7k114797
asked Nov 28 '18 at 17:57
DanielDaniel
1,1311021
edited Nov 28 '18 at 21:00
greedoid
38.7k114797
asked Nov 28 '18 at 17:57
DanielDaniel
1,1311021
edited Nov 28 '18 at 21:00
greedoid
38.7k114797
edited Nov 28 '18 at 21:00
greedoid
38.7k114797
edited Nov 28 '18 at 21:00
greedoid
38.7k114797
38.7k114797
asked Nov 28 '18 at 17:57
DanielDaniel
1,1311021
asked Nov 28 '18 at 17:57
DanielDaniel
1,1311021
asked Nov 28 '18 at 17:57
DanielDaniel
1,1311021
1,1311021
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Supppose we have sets $A_1,...A_k$ which are pairvise disjunct and for each $j$ let $n_j = |A_j|$.
We choose randomly two elements from $A= A_1cup A_2cup...cup A_k$. Then the probability that they are from the same set is the expresion you wrote.
answered Nov 28 '18 at 19:59
greedoidgreedoid
38.7k114797
$endgroup$
add a comment |
$begingroup$
You can write your $P$ in terms of $C$, and vice versa, where $C=sum_{i=1}^k (n_i-n/k)^2/(n/k)$ is the conventional chi-square statistic for testing if the vector of $n_i$ values comes from the all-categories-equally-likely flat multinomial model. So $P$ tells you no more or less than $C$ does, but your audience might be familiar with the chi-square statistic as commonly used to measure unevenness of category counts.
answered Nov 28 '18 at 19:52
kimchi loverkimchi lover
9,69631128
$endgroup$
add a comment |
$begingroup$
Probability to take the same element $j$ two times:
$$
p_j=left(frac{n_j}{n}right)left(frac{n_j-1}{n-1}right)=frac{n_j(n_j-1)}{n(n-1)}
$$
Now we do not care about $j$ anymore and therefore marginalize:
$$
p=sum_j p_j=frac{1}{n(n-1)}sum_j n_j(n_j-1)
$$
which is the probability to take the same element two times (independently of $j$)
answered Nov 28 '18 at 21:51
Picaud VincentPicaud Vincent
1,33439
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Supppose we have sets $A_1,...A_k$ which are pairvise disjunct and for each $j$ let $n_j = |A_j|$.
We choose randomly two elements from $A= A_1cup A_2cup...cup A_k$. Then the probability that they are from the same set is the expresion you wrote.
answered Nov 28 '18 at 19:59
greedoidgreedoid
38.7k114797
$endgroup$
add a comment |
$begingroup$
Supppose we have sets $A_1,...A_k$ which are pairvise disjunct and for each $j$ let $n_j = |A_j|$.
We choose randomly two elements from $A= A_1cup A_2cup...cup A_k$. Then the probability that they are from the same set is the expresion you wrote.
answered Nov 28 '18 at 19:59
greedoidgreedoid
38.7k114797
$endgroup$
add a comment |
$begingroup$
Supppose we have sets $A_1,...A_k$ which are pairvise disjunct and for each $j$ let $n_j = |A_j|$.
We choose randomly two elements from $A= A_1cup A_2cup...cup A_k$. Then the probability that they are from the same set is the expresion you wrote.
answered Nov 28 '18 at 19:59
greedoidgreedoid
38.7k114797
$endgroup$
Supppose we have sets $A_1,...A_k$ which are pairvise disjunct and for each $j$ let $n_j = |A_j|$.
We choose randomly two elements from $A= A_1cup A_2cup...cup A_k$. Then the probability that they are from the same set is the expresion you wrote.
answered Nov 28 '18 at 19:59
greedoidgreedoid
38.7k114797
answered Nov 28 '18 at 19:59
greedoidgreedoid
38.7k114797
answered Nov 28 '18 at 19:59
greedoidgreedoid
38.7k114797
answered Nov 28 '18 at 19:59
greedoidgreedoid
38.7k114797
38.7k114797
add a comment |
add a comment |
$begingroup$
You can write your $P$ in terms of $C$, and vice versa, where $C=sum_{i=1}^k (n_i-n/k)^2/(n/k)$ is the conventional chi-square statistic for testing if the vector of $n_i$ values comes from the all-categories-equally-likely flat multinomial model. So $P$ tells you no more or less than $C$ does, but your audience might be familiar with the chi-square statistic as commonly used to measure unevenness of category counts.
answered Nov 28 '18 at 19:52
kimchi loverkimchi lover
9,69631128
$endgroup$
add a comment |
$begingroup$
You can write your $P$ in terms of $C$, and vice versa, where $C=sum_{i=1}^k (n_i-n/k)^2/(n/k)$ is the conventional chi-square statistic for testing if the vector of $n_i$ values comes from the all-categories-equally-likely flat multinomial model. So $P$ tells you no more or less than $C$ does, but your audience might be familiar with the chi-square statistic as commonly used to measure unevenness of category counts.
answered Nov 28 '18 at 19:52
kimchi loverkimchi lover
9,69631128
$endgroup$
add a comment |
$begingroup$
You can write your $P$ in terms of $C$, and vice versa, where $C=sum_{i=1}^k (n_i-n/k)^2/(n/k)$ is the conventional chi-square statistic for testing if the vector of $n_i$ values comes from the all-categories-equally-likely flat multinomial model. So $P$ tells you no more or less than $C$ does, but your audience might be familiar with the chi-square statistic as commonly used to measure unevenness of category counts.
answered Nov 28 '18 at 19:52
kimchi loverkimchi lover
9,69631128
$endgroup$
You can write your $P$ in terms of $C$, and vice versa, where $C=sum_{i=1}^k (n_i-n/k)^2/(n/k)$ is the conventional chi-square statistic for testing if the vector of $n_i$ values comes from the all-categories-equally-likely flat multinomial model. So $P$ tells you no more or less than $C$ does, but your audience might be familiar with the chi-square statistic as commonly used to measure unevenness of category counts.
answered Nov 28 '18 at 19:52
kimchi loverkimchi lover
9,69631128
answered Nov 28 '18 at 19:52
kimchi loverkimchi lover
9,69631128
answered Nov 28 '18 at 19:52
kimchi loverkimchi lover
9,69631128
answered Nov 28 '18 at 19:52
kimchi loverkimchi lover
9,69631128
9,69631128
add a comment |
add a comment |
$begingroup$
Probability to take the same element $j$ two times:
$$
p_j=left(frac{n_j}{n}right)left(frac{n_j-1}{n-1}right)=frac{n_j(n_j-1)}{n(n-1)}
$$
Now we do not care about $j$ anymore and therefore marginalize:
$$
p=sum_j p_j=frac{1}{n(n-1)}sum_j n_j(n_j-1)
$$
which is the probability to take the same element two times (independently of $j$)
answered Nov 28 '18 at 21:51
Picaud VincentPicaud Vincent
1,33439
$endgroup$
add a comment |
$begingroup$
Probability to take the same element $j$ two times:
$$
p_j=left(frac{n_j}{n}right)left(frac{n_j-1}{n-1}right)=frac{n_j(n_j-1)}{n(n-1)}
$$
Now we do not care about $j$ anymore and therefore marginalize:
$$
p=sum_j p_j=frac{1}{n(n-1)}sum_j n_j(n_j-1)
$$
which is the probability to take the same element two times (independently of $j$)
answered Nov 28 '18 at 21:51
Picaud VincentPicaud Vincent
1,33439
$endgroup$
add a comment |
$begingroup$
Probability to take the same element $j$ two times:
$$
p_j=left(frac{n_j}{n}right)left(frac{n_j-1}{n-1}right)=frac{n_j(n_j-1)}{n(n-1)}
$$
Now we do not care about $j$ anymore and therefore marginalize:
$$
p=sum_j p_j=frac{1}{n(n-1)}sum_j n_j(n_j-1)
$$
which is the probability to take the same element two times (independently of $j$)
answered Nov 28 '18 at 21:51
Picaud VincentPicaud Vincent
1,33439
$endgroup$
Probability to take the same element $j$ two times:
$$
p_j=left(frac{n_j}{n}right)left(frac{n_j-1}{n-1}right)=frac{n_j(n_j-1)}{n(n-1)}
$$
Now we do not care about $j$ anymore and therefore marginalize:
$$
p=sum_j p_j=frac{1}{n(n-1)}sum_j n_j(n_j-1)
$$
which is the probability to take the same element two times (independently of $j$)
answered Nov 28 '18 at 21:51
Picaud VincentPicaud Vincent
1,33439
edited Dec 16 '18 at 20:06
answered Nov 28 '18 at 21:51
Picaud VincentPicaud Vincent
1,33439
answered Nov 28 '18 at 21:51
Picaud VincentPicaud Vincent
1,33439
answered Nov 28 '18 at 21:51
Picaud VincentPicaud Vincent
1,33439
1,33439
add a comment |
add a comment |
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