How do I approach this intersectionality proof below?

[ddq]

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: $$P(male|y=1)$$ $$P(female|y=1)$$ $$P(white|y=1)$$ $$P(Black|y=1)$$
intersectional groups: $$P(white,male|y=1)$$ $$P(black,male|y=1)$$ $$P(white,female|y=1)$$ $$P(black,female|y=1)$$

My question is how do I show that if: $P(white,male|y=1)$ == $P(black,male|y=1)$ == $P(white,female|y=1)$ == $P(black,female|y=1)$, it is (not)possible to achieve $P(male|y=1)$ == $P(female|y=1$ and $P(white|y=1)$ == $P(black|y=1)$

 

[pka]

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: $$P(male|y=1)$$ $$P(female|y=1)$$ $$P(white|y=1)$$ $$P(Black|y=1)$$
intersectional groups: $$P(white,male|y=1)$$ $$P(black,male|y=1)$$ $$P(white,female|y=1)$$ $$P(black,female|y=1)$$My question is how do I show that if: $P(white,male|y=1)$ == $P(black,male|y=1)$ == $P(white,female|y=1)$ == $P(black,female|y=1)$, it is (not)possible to achieve $P(male|y=1)$ == $P(female|y=1$ and $P(white|y=1)$ == $P(black|y=1)$

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 ?

 

[Steven G]

Maybe it would be helpful if you define y?

 

[Dr.Peterson]

We aren't alone in bafflement, even after an edit:

How do I approach this intersectionality proof below?

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?...

math.stackexchange.com

 

[ddq]

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}

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

intersectional groups:
$$P(\hat{y}=1|y=1, white, male)$$ $$P(\hat{y}=1|y=1, black, male)$$ $$P(\hat{y}=1|y=1, white, female)$$ $$P(\hat{y}=1|y=1, black, female)$$

My question is how do I show that if: $$P(\hat{y}=1|y=1, white, male) = P(\hat{y}=1|y=1, black, male) = P(\hat{y}=1|y=1, white, female) = P(\hat{y}=1|y=1, black, female)$$ it is (not)possible to have: $$P(\hat{y}=1|y=1, male) = P(\hat{y}=1|y=1, female)$$ and $$P(\hat{y}=1|y=1, white) = P(\hat{y}=1|y=1, black)$$

 

[ddq]

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: $$P(male|y=1)$$ $$P(female|y=1)$$ $$P(white|y=1)$$ $$P(Black|y=1)$$
intersectional groups: $$P(white,male|y=1)$$ $$P(black,male|y=1)$$ $$P(white,female|y=1)$$ $$P(black,female|y=1)$$

My question is how do I show that if: $P(white,male|y=1)$ == $P(black,male|y=1)$ == $P(white,female|y=1)$ == $P(black,female|y=1)$, it is (not)possible to achieve $P(male|y=1)$ == $P(female|y=1$ and $P(white|y=1)$ == $P(black|y=1)$

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}

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

intersectional groups:
$$P(\hat{y}=1|y=1, white, male)$$ $$P(\hat{y}=1|y=1, black, male)$$ $$P(\hat{y}=1|y=1, white, female)$$ $$P(\hat{y}=1|y=1, black, female)$$

My question is how do I show that if: $$P(\hat{y}=1|y=1, white, male) = P(\hat{y}=1|y=1, black, male) = P(\hat{y}=1|y=1, white, female) = P(\hat{y}=1|y=1, black, female)$$ it is (not)possible to have: $$P(\hat{y}=1|y=1, male) = P(\hat{y}=1|y=1, female)$$ and $$P(\hat{y}=1|y=1, white) = P(\hat{y}=1|y=1, black)$$

 

[Cubist]

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: $$P(male|y=1)$$ $$P(female|y=1)$$ $$P(white|y=1)$$ $$P(Black|y=1)$$
intersectional groups: $$P(white,male|y=1)$$ $$P(black,male|y=1)$$ $$P(white,female|y=1)$$ $$P(black,female|y=1)$$

My question is how do I show that if: $P(white,male|y=1)$ == $P(black,male|y=1)$ == $P(white,female|y=1)$ == $P(black,female|y=1)$, it is (not)possible to achieve $P(male|y=1)$ == $P(female|y=1$ and $P(white|y=1)$ == $P(black|y=1)$

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?

 

[ddq]

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?

Just an idea born out of my own thoughts. Please I have updated the question. Please lemme know if it is clear enough.

 

[ddq]

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}

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

intersectional groups:
$$P(\hat{y}=1|y=1, white, male)$$ $$P(\hat{y}=1|y=1, black, male)$$ $$P(\hat{y}=1|y=1, white, female)$$ $$P(\hat{y}=1|y=1, black, female)$$

My question is how do I show that if: $$P(\hat{y}=1|y=1, white, male) = P(\hat{y}=1|y=1, black, male) = P(\hat{y}=1|y=1, white, female) = P(\hat{y}=1|y=1, black, female)$$ it is (not)possible to have: $$P(\hat{y}=1|y=1, male) = P(\hat{y}=1|y=1, female)$$ and $$P(\hat{y}=1|y=1, white) = P(\hat{y}=1|y=1, black)$$

@Cubist, please see the update here.