How do I approach this intersectionality proof below?

D

ddq

New member
Joined
Feb 20, 2023
Messages
5
Is there a way to prove that, if we achieve equal proportions of a certain outcome among 4 intersectional groups formed from Sex and Ethnicity, we won't create inequalities among the single groups?

sex = {male, female}, ethnicity = {white, black}

single groups: [math]P(male|y=1)[/math] [math]P(female|y=1)[/math] [math]P(white|y=1)[/math] [math]P(Black|y=1)[/math]
intersectional groups: [math]P(white,male|y=1)[/math] [math]P(black,male|y=1)[/math] [math]P(white,female|y=1)[/math] [math]P(black,female|y=1)[/math]

My question is how do I show that if: [imath]P(white,male|y=1)[/imath] == [imath]P(black,male|y=1)[/imath] == [imath]P(white,female|y=1)[/imath] == [imath]P(black,female|y=1)[/imath], it is (not)possible to achieve [imath]P(male|y=1)[/imath] == [imath]P(female|y=1[/imath] and [imath]P(white|y=1)[/imath] == [imath]P(black|y=1)[/imath]
 
ddq said:
Is there a way to prove that, if we achieve equal proportions of a certain outcome among 4 intersectional groups formed from Sex and Ethnicity, we won't create inequalities among the single groups? sex = {male, female}, ethnicity = {white, black}
single groups: [math]P(male|y=1)[/math] [math]P(female|y=1)[/math] [math]P(white|y=1)[/math] [math]P(Black|y=1)[/math]
intersectional groups: [math]P(white,male|y=1)[/math] [math]P(black,male|y=1)[/math] [math]P(white,female|y=1)[/math] [math]P(black,female|y=1)[/math]My question is how do I show that if: [imath]P(white,male|y=1)[/imath] == [imath]P(black,male|y=1)[/imath] == [imath]P(white,female|y=1)[/imath] == [imath]P(black,female|y=1)[/imath], it is (not)possible to achieve [imath]P(male|y=1)[/imath] == [imath]P(female|y=1[/imath] and [imath]P(white|y=1)[/imath] == [imath]P(black|y=1)[/imath]
Click to expand...
To ddq, Do you realize that that this is an English Language mathematics-help website?
Well we are those things. So why did you post the above utter gibberish ?
 
Is there a way to prove that, if we achieve equal proportions of a certain outcome among 4 intersectional groups formed from Sex and Ethnicity, is it possible to achieve equality on the sex and ethnicity groups?

sex = {male, female},

ethnicity = {white, black}

[math]y= label[/math]
[math]\hat{y} = predicted-label[/math]
single groups:
[math]P(\hat{y}=1|y=1, male)[/math] [math]P(\hat{y}=1|y=1, male)[/math] [math]P(\hat{y}=1|y=1, white)[/math] [math]P(\hat{y}=1|y=1, black)[/math]

intersectional groups:
[math]P(\hat{y}=1|y=1, white, male)[/math] [math]P(\hat{y}=1|y=1, black, male)[/math] [math]P(\hat{y}=1|y=1, white, female)[/math] [math]P(\hat{y}=1|y=1, black, female)[/math]

My question is how do I show that if: [math]P(\hat{y}=1|y=1, white, male) = P(\hat{y}=1|y=1, black, male) = P(\hat{y}=1|y=1, white, female) = P(\hat{y}=1|y=1, black, female)[/math] it is (not)possible to have: [math]P(\hat{y}=1|y=1, male) = P(\hat{y}=1|y=1, female)[/math] and [math]P(\hat{y}=1|y=1, white) = P(\hat{y}=1|y=1, black)[/math]
 
ddq said:
Is there a way to prove that, if we achieve equal proportions of a certain outcome among 4 intersectional groups formed from Sex and Ethnicity, we won't create inequalities among the single groups?

sex = {male, female}, ethnicity = {white, black}

single groups: [math]P(male|y=1)[/math] [math]P(female|y=1)[/math] [math]P(white|y=1)[/math] [math]P(Black|y=1)[/math]
intersectional groups: [math]P(white,male|y=1)[/math] [math]P(black,male|y=1)[/math] [math]P(white,female|y=1)[/math] [math]P(black,female|y=1)[/math]

My question is how do I show that if: [imath]P(white,male|y=1)[/imath] == [imath]P(black,male|y=1)[/imath] == [imath]P(white,female|y=1)[/imath] == [imath]P(black,female|y=1)[/imath], it is (not)possible to achieve [imath]P(male|y=1)[/imath] == [imath]P(female|y=1[/imath] and [imath]P(white|y=1)[/imath] == [imath]P(black|y=1)[/imath]
Click to expand...
An update to my question for clarity.

Is there a way to prove that, if we achieve equal proportions of a certain outcome among 4 intersectional groups formed from Sex and Ethnicity, is it possible to achieve equality on the sex and ethnicity groups?

sex = {male, female},

ethnicity = {white, black}

[math]y= label[/math]
[math]\hat{y} = predicted-label[/math]
single groups:
[math]P(\hat{y}=1|y=1, male)[/math] [math]P(\hat{y}=1|y=1, male)[/math] [math]P(\hat{y}=1|y=1, white)[/math] [math]P(\hat{y}=1|y=1, black)[/math]

intersectional groups:
[math]P(\hat{y}=1|y=1, white, male)[/math] [math]P(\hat{y}=1|y=1, black, male)[/math] [math]P(\hat{y}=1|y=1, white, female)[/math] [math]P(\hat{y}=1|y=1, black, female)[/math]

My question is how do I show that if: [math]P(\hat{y}=1|y=1, white, male) = P(\hat{y}=1|y=1, black, male) = P(\hat{y}=1|y=1, white, female) = P(\hat{y}=1|y=1, black, female)[/math] it is (not)possible to have: [math]P(\hat{y}=1|y=1, male) = P(\hat{y}=1|y=1, female)[/math] and [math]P(\hat{y}=1|y=1, white) = P(\hat{y}=1|y=1, black)[/math]
 
ddq said:
Is there a way to prove that, if we achieve equal proportions of a certain outcome among 4 intersectional groups formed from Sex and Ethnicity, we won't create inequalities among the single groups?

sex = {male, female}, ethnicity = {white, black}

single groups: [math]P(male|y=1)[/math] [math]P(female|y=1)[/math] [math]P(white|y=1)[/math] [math]P(Black|y=1)[/math]
intersectional groups: [math]P(white,male|y=1)[/math] [math]P(black,male|y=1)[/math] [math]P(white,female|y=1)[/math] [math]P(black,female|y=1)[/math]

My question is how do I show that if: [imath]P(white,male|y=1)[/imath] == [imath]P(black,male|y=1)[/imath] == [imath]P(white,female|y=1)[/imath] == [imath]P(black,female|y=1)[/imath], it is (not)possible to achieve [imath]P(male|y=1)[/imath] == [imath]P(female|y=1[/imath] and [imath]P(white|y=1)[/imath] == [imath]P(black|y=1)[/imath]
Click to expand...


Please give us some context. Did this question arise during a discussion with your friends, or was it given in a class (because the notation isn't clear)? Please be honest and then we can answer you appropriately.

--

This might relate to your question...

We have 4 separate sets of shapes...

Set A are black circles ⚫
Set B are white circles ⚪
Set C are black squares ⬛
Set D are white squares ⬜

Let's use the notation that n(A) is the number of elements in set A.

a = n(A)
b = n(B)
c = n(C)
d = n(D)

Can you answer the following questions:-
How many circles are in the four sets? ⚪ ⚫
How many squares are in the four sets? ⬜ ⬛
How many white shapes are in the four sets? ⚪ ⬜
How many black shapes are in the four sets? ⚫ ⬛

Now suppose that the four sets have the same number of elements "x" (a=x, b=x, c=x, and d=x) then can you answer the above questions again?
 
Cubist said:
Please give us some context. Did this question arise during a discussion with your friends, or was it given in a class (because the notation isn't clear)? Please be honest and then we can answer you appropriately.

--

This might relate to your question...

We have 4 separate sets of shapes...

Set A are black circles ⚫
Set B are white circles ⚪
Set C are black squares ⬛
Set D are white squares ⬜

Let's use the notation that n(A) is the number of elements in set A.

a = n(A)
b = n(B)
c = n(C)
d = n(D)

Can you answer the following questions:-
How many circles are in the four sets? ⚪ ⚫
How many squares are in the four sets? ⬜ ⬛
How many white shapes are in the four sets? ⚪ ⬜
How many black shapes are in the four sets? ⚫ ⬛

Now suppose that the four sets have the same number of elements "x" (a=x, b=x, c=x, and d=x) then can you answer the above questions again?
Click to expand...
Just an idea born out of my own thoughts. Please I have updated the question. Please lemme know if it is clear enough.
 
ddq said:
An update to my question for clarity.

Is there a way to prove that, if we achieve equal proportions of a certain outcome among 4 intersectional groups formed from Sex and Ethnicity, is it possible to achieve equality on the sex and ethnicity groups?

sex = {male, female},

ethnicity = {white, black}

[math]y= label[/math]
[math]\hat{y} = predicted-label[/math]
single groups:
[math]P(\hat{y}=1|y=1, male)[/math] [math]P(\hat{y}=1|y=1, male)[/math] [math]P(\hat{y}=1|y=1, white)[/math] [math]P(\hat{y}=1|y=1, black)[/math]

intersectional groups:
[math]P(\hat{y}=1|y=1, white, male)[/math] [math]P(\hat{y}=1|y=1, black, male)[/math] [math]P(\hat{y}=1|y=1, white, female)[/math] [math]P(\hat{y}=1|y=1, black, female)[/math]

My question is how do I show that if: [math]P(\hat{y}=1|y=1, white, male) = P(\hat{y}=1|y=1, black, male) = P(\hat{y}=1|y=1, white, female) = P(\hat{y}=1|y=1, black, female)[/math] it is (not)possible to have: [math]P(\hat{y}=1|y=1, male) = P(\hat{y}=1|y=1, female)[/math] and [math]P(\hat{y}=1|y=1, white) = P(\hat{y}=1|y=1, black)[/math]
Click to expand...
@Cubist, please see the update here.
 
You must log in or register to reply here.
Top