Explain in simple words the difference between the Causal and Correlated metric types.

Jaelyn Payne
2022-10-15
Answered

Explain in simple words the difference between the Causal and Correlated metric types.

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Miah Scott

Answered 2022-10-16
Author has **19** answers

causations and correlation (connections) can exist simultaneously, relationship doesn't infer causation.

Causation unequivocally applies to situations where activity A causes result B. Then again, connections(correlation) is essentially a relationship. Activity A connects with Action B-however one occasion doesn't be guaranteed to make the other occasion occur.

Relationship(correlation)and causation are frequently confounded in light of the fact that the human psyche likes to observe designs in any event, when they don't exist. We regularly manufactures these examples when two factors have all the earmarks of being so firmly related that one is reliant upon the other. That would suggests a circumstances and logicals results relationships where the reliant occasion is the aftereffect of an autonomous occasion.

Causation unequivocally applies to situations where activity A causes result B. Then again, connections(correlation) is essentially a relationship. Activity A connects with Action B-however one occasion doesn't be guaranteed to make the other occasion occur.

Relationship(correlation)and causation are frequently confounded in light of the fact that the human psyche likes to observe designs in any event, when they don't exist. We regularly manufactures these examples when two factors have all the earmarks of being so firmly related that one is reliant upon the other. That would suggests a circumstances and logicals results relationships where the reliant occasion is the aftereffect of an autonomous occasion.

asked 2022-07-21

Causation doesn't imply correlation and correlation does not imply

causation

Let P := |corr(X,Y) > .5| let Q := exits a relation F: $X\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}Y$

Then the often stated line of correlation does not imply causation is simply! $Q\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P$.

It is also true that causation does not imply correlation. So! $Q\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P$

But $(P\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}Q)\vee (Q\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P)$ is a tautology.

causation

Let P := |corr(X,Y) > .5| let Q := exits a relation F: $X\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}Y$

Then the often stated line of correlation does not imply causation is simply! $Q\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P$.

It is also true that causation does not imply correlation. So! $Q\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P$

But $(P\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}Q)\vee (Q\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P)$ is a tautology.

asked 2022-11-03

Constructing a sample by correlation

Suppose we have two samples with known correlation (should be relatively high). Say both samples have n data points. What if now we still know the correlation factor but one sample only consistent of the first 5 data point.

Could one still construct the remaining data points solely using the correlation with the other sample?

My idea would be to look at the relative differences in the known sample and compensate by the correlation. Could this work? Thanks for any assistance.

Suppose we have two samples with known correlation (should be relatively high). Say both samples have n data points. What if now we still know the correlation factor but one sample only consistent of the first 5 data point.

Could one still construct the remaining data points solely using the correlation with the other sample?

My idea would be to look at the relative differences in the known sample and compensate by the correlation. Could this work? Thanks for any assistance.

asked 2022-10-07

Probability of infection by staphylococcus aureus

Please forgive my innumeracy, but I have a question with which I am hoping someone might be able to help me.

Suppose the following be true. The chance of a prosthetic hip joint becoming infected by staphylococcus aureus is one per cent. The chance of a natural hip joint becoming infected by staphylococcus aureus is 0.1%. In other words (in case I am misusing the word 'chance'), one in one hundred people with prosthetic hip joints will become infected by staphylococcus aureus, whereas only one in one thousand people with natural hip joints will become so infected.

Now suppose that X has a prosthetic hip joint and that X's hip joint becomes infected by staphylococcus aureus.

Given only the information provided here, is it correct to say that X's hip joint probably would not have become infected but for the fact that X has a prosthetic hip joint (instead of a natural hip joint)? In other words (to make it clear what I mean by 'probably'), is it correct to say that there is a greater than 50 per cent chance that X's hip joint would not have become infected but for the fact that X has a prosthetic hip joint? Why or why not?

Please forgive my innumeracy, but I have a question with which I am hoping someone might be able to help me.

Suppose the following be true. The chance of a prosthetic hip joint becoming infected by staphylococcus aureus is one per cent. The chance of a natural hip joint becoming infected by staphylococcus aureus is 0.1%. In other words (in case I am misusing the word 'chance'), one in one hundred people with prosthetic hip joints will become infected by staphylococcus aureus, whereas only one in one thousand people with natural hip joints will become so infected.

Now suppose that X has a prosthetic hip joint and that X's hip joint becomes infected by staphylococcus aureus.

Given only the information provided here, is it correct to say that X's hip joint probably would not have become infected but for the fact that X has a prosthetic hip joint (instead of a natural hip joint)? In other words (to make it clear what I mean by 'probably'), is it correct to say that there is a greater than 50 per cent chance that X's hip joint would not have become infected but for the fact that X has a prosthetic hip joint? Why or why not?

asked 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 $P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$, because the tests are independent.

By law of total probability we know that:

$P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$

Replacing, and assuming conditional independence given D, we have:

$P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$

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:

$P({T}_{1})=0.95\ast 0.01+0.05\ast 0.99\approx 0.059$

$P({T}_{1})=0.95\ast 0.01+0.05\ast 0.99\approx 0.059$

So we have:

$P({T}_{1},{T}_{2})=P({T}_{1}{)}^{2}\approx {0.059}^{2}\approx 0.003481$

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.

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 $P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$, because the tests are independent.

By law of total probability we know that:

$P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$

Replacing, and assuming conditional independence given D, we have:

$P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$

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:

$P({T}_{1})=0.95\ast 0.01+0.05\ast 0.99\approx 0.059$

$P({T}_{1})=0.95\ast 0.01+0.05\ast 0.99\approx 0.059$

So we have:

$P({T}_{1},{T}_{2})=P({T}_{1}{)}^{2}\approx {0.059}^{2}\approx 0.003481$

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.

asked 2022-09-26

Pearson Correlation Coefficient Interpretation

Let X=(1,2,3,...,20). Suppose that $Y=({y}_{1},{y}_{2},...,{y}_{20})$ with ${y}_{i}={x}_{i}^{2}$ and $Z=({z}_{1},{z}_{2},...,{z}_{20})$ with ${z}_{i}={e}^{{x}_{i}}$. Pearson correlation coefficient is defined by formula

$\rho (X,Y)=\frac{\sum _{i=1}^{20}({x}_{i}-\overline{x})({y}_{i}-\overline{y})}{\sqrt{(\sum _{i=1}^{20}({x}_{i}-\overline{x}{)}^{2})(\sum _{i=1}^{20}({y}_{i}-\overline{y}{)}^{2})}}$

If $\rho (X,Y)=1$, we can say that X and Y have a linear correlation. If $0.7\le \rho (X,Y)<1$ then X and Y has a strong linear correlation, if $0.5\le \rho (X,Y)<0.7$ then X and Y has a modest linear correlation, and if $0\le \rho (X,Y)<0.5$ then X and Y has a weak linear correlation. Using this formula, we get $\rho $(X,Y)=0.9 and $\rho $(X,Z)=0.5. However, the relationship between X and Y is actually quadratic but they have the high correlation coefficient that indicate linear correlation.

So, my question is what is "linear correlation" actually between X and Y ? Since $\rho $(X,Z)=0.5 indicate the modest correlation coefficient, what is another intepretation of this value? What is the difference between $\rho $(X,Y) and $\rho $(X,Z), noting that Y and Z is not a linear function of X.

Let X=(1,2,3,...,20). Suppose that $Y=({y}_{1},{y}_{2},...,{y}_{20})$ with ${y}_{i}={x}_{i}^{2}$ and $Z=({z}_{1},{z}_{2},...,{z}_{20})$ with ${z}_{i}={e}^{{x}_{i}}$. Pearson correlation coefficient is defined by formula

$\rho (X,Y)=\frac{\sum _{i=1}^{20}({x}_{i}-\overline{x})({y}_{i}-\overline{y})}{\sqrt{(\sum _{i=1}^{20}({x}_{i}-\overline{x}{)}^{2})(\sum _{i=1}^{20}({y}_{i}-\overline{y}{)}^{2})}}$

If $\rho (X,Y)=1$, we can say that X and Y have a linear correlation. If $0.7\le \rho (X,Y)<1$ then X and Y has a strong linear correlation, if $0.5\le \rho (X,Y)<0.7$ then X and Y has a modest linear correlation, and if $0\le \rho (X,Y)<0.5$ then X and Y has a weak linear correlation. Using this formula, we get $\rho $(X,Y)=0.9 and $\rho $(X,Z)=0.5. However, the relationship between X and Y is actually quadratic but they have the high correlation coefficient that indicate linear correlation.

So, my question is what is "linear correlation" actually between X and Y ? Since $\rho $(X,Z)=0.5 indicate the modest correlation coefficient, what is another intepretation of this value? What is the difference between $\rho $(X,Y) and $\rho $(X,Z), noting that Y and Z is not a linear function of X.

asked 2022-08-13

If we have two non-zero correlated random variables then they are dependent.

Why then do we have the saying "Correlation does not imply Causation". A change in one variable may not cause exactly the same change in another but there is at least some 'causal' link.

Why then do we have the saying "Correlation does not imply Causation". A change in one variable may not cause exactly the same change in another but there is at least some 'causal' link.

asked 2022-07-22

Correlation and causation data

What are some real life examples (data) in which high correlation:

1) implies causation

2) doesn't imply causation.

I know that there is a lot of data out there of weird correlations, like divorce rate and consuption of margarine, but the problem with this data is, that we don't really know if one is caused by the other, because nobody tested it, so we cannot strictly say that they're unrelated.

What are some real life examples (data) in which high correlation:

1) implies causation

2) doesn't imply causation.

I know that there is a lot of data out there of weird correlations, like divorce rate and consuption of margarine, but the problem with this data is, that we don't really know if one is caused by the other, because nobody tested it, so we cannot strictly say that they're unrelated.