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Recent questions in Correlation And Causation
Inferential StatisticsAnswered question
Cael Dickerson Cael Dickerson 2022-11-09

Correlation matrix with same pairwise correlation coefficientQuestion
Correlation matrix. Consider 𝑛 random variables with the same pairwise correlation coefficient ρ n . Find the highest possible value of ρ n for
a) n=3
b) n=4
c) general, n β‰₯2
HINT: Correlation matrix must be positive semi-definite.
My Workings
This is what I infer from "same pairwise coefficients":
( 1 ρ n ρ n ρ n 1 ρ n ρ n ρ n 1 )
Because a correlation matrix is positive semi-definite, all principal minors have to be positive
For n=3, the principal minors calculations yield
∣ H 1 ∣=1
∣ H 2 ∣=1- ρ n 2 β‰₯ 0 β‡’ ρ n ≀ 1
∣ H 3 ∣= ρ n 2 - ρ n β‰₯ 0 β‡’ ρ n β‰₯ 1
∣ H 4 ∣= ( 1 3 +2 ρ n 3 )-(3 ρ n 2 ) β‰₯ 0 . I solved for the roots and found 1
Conclusion: For n=3, max ρ n =1
I did the same method for n=4 and found max ρ n =1 again
My Problem
Result looks too simple and false. Method is tedious
I have no idea how to do the general case. (by induction ?)
Thank you for your help
INDUCTION ATTEMPT
H 0
( 1 ρ 2 ρ 2 1 )
β‡’ βˆ’ 1 2 βˆ’ 1 ≀ ρ 2 ≀ 1
H n : Pn: Suppose that for a nxn correlation matrix An with same pairwise coefficients ρ n , βˆ’ 1 n βˆ’ 1 ≀ ρ n ≀ 1 holds
H n + 1
βˆ’ 1 n βˆ’ 1 ≀ ρ n ≀ 1
⇐⇒ βˆ’ 1 ( n + 1 ) βˆ’ 1 ≀ ρ n + 1 ≀ 1
⇐⇒ βˆ’ 1 n ≀ ρ n + 1 ≀ 1
And because for all βˆ’ 1 n β‰₯ βˆ’ 1 n βˆ’ 1
βˆ’ 1 n βˆ’ 1 ≀ ρ n + 1 ≀ 1

Inferential StatisticsAnswered question
wijii4 wijii4 2022-09-17

Conditional Probability and Independence nonsense in a problem
The statement:
Suppose that a patient tests positive for a disease affecting 1% of the population. For a patient who has the disease, there is a 95% chance of testing positive, and for a patient who doesn't has the disease, there is a 95% chance of testing negative. The patient gets a second, independent, test done, and again tests positive. Find the probability that the patient has the disease.
The problem:
I can solve this problem, but I'm unable to understand what is wrong with the following:
Let T i be the event that the patient tests positive in the i-th test, and let D be the event that the patient has the disease.
The problem says that P ( T 1 , T 2 ) = 0.95 2 βˆ— 0.01 + 0.05 2 βˆ— 0.99 = 0.0115, because the tests are independent.
By law of total probability we know that:
P ( T 1 , T 2 ) = 0.95 2 βˆ— 0.01 + 0.05 2 βˆ— 0.99 = 0.0115
Replacing, and assuming conditional independence given D, we have:
P ( T 1 , T 2 ) = 0.95 2 βˆ— 0.01 + 0.05 2 βˆ— 0.99 = 0.0115
This is the correct result, but now let's consider that:
P ( T 1 , T 2 ) = P ( T 1 ) 2
We know that P ( T 1 , T 2 ) = P ( T 1 ) 2 for all i because of symmetry, so we have P ( T 1 , T 2 ) = P ( T 1 ) 2 . Again, by law of total probability:
P ( T 1 ) = 0.95 βˆ— 0.01 + 0.05 βˆ— 0.99 β‰ˆ 0.059
P ( T 1 ) = 0.95 βˆ— 0.01 + 0.05 βˆ— 0.99 β‰ˆ 0.059
So we have:
P ( T 1 , T 2 ) = P ( T 1 ) 2 β‰ˆ 0.059 2 β‰ˆ 0.003481
The second approach is wrong, but it seems legitimate to me, and I'm unable to find what's wrong.
Thank's for your help, you make self studying easier.

Correlation and causation are two important concepts in mathematics. Correlation is when two variables are related, meaning a change in one variable can cause a change in the other. Causation is when one variable causes a change in the other. To understand the difference between these two concepts, it is important to understand how equations can help answer questions about correlation and causation. Equations can help determine if a change in one variable will lead to a change in the other, and if so, how much. When studying correlation and causation, equations are a useful tool for determining the strength of the relationship between the two variables. If you need help understanding correlation and causation, on our site you can find the answers you need.