Recent questions in Correlation And Causation

Inferential StatisticsAnswered question

Laila Murphy 2022-11-20

Correlation bound

Let x and y be two random variables such that:

Corr(x,y) = b, where Corr(x,y) represents correlation between x and y, b is a scalar number in range of [-1, 1]. Let y' be an estimation of y. An example could be y'=y+(rand(0,1)-0.5)*.1, rand(0,1) gives random number between 0, 1. I am adding some noise to the data.

My questions are:

Is there a way where I can bound the correlation between x, y' i.e. Corr(x,y')?I mentioned y' in light of random perturbation, I would like to know what if I don't have that information, where I only know that y' is a estimation of y. Are there any literature that cover it?

Let x and y be two random variables such that:

Corr(x,y) = b, where Corr(x,y) represents correlation between x and y, b is a scalar number in range of [-1, 1]. Let y' be an estimation of y. An example could be y'=y+(rand(0,1)-0.5)*.1, rand(0,1) gives random number between 0, 1. I am adding some noise to the data.

My questions are:

Is there a way where I can bound the correlation between x, y' i.e. Corr(x,y')?I mentioned y' in light of random perturbation, I would like to know what if I don't have that information, where I only know that y' is a estimation of y. Are there any literature that cover it?

Inferential StatisticsAnswered question

fabler107 2022-11-15

True or false?

The amount of hours you work babysitting and the amount of money you earn- has correlation, but no causation

The amount of hours you work babysitting and the amount of money you earn- has correlation, but no causation

Inferential StatisticsAnswered question

Cael Dickerson 2022-11-09

Correlation matrix with same pairwise correlation coefficientQuestion

Correlation matrix. Consider π random variables with the same pairwise correlation coefficient ${\mathrm{{\rm O}\x81}}_{n}$. Find the highest possible value of ${\mathrm{{\rm O}\x81}}_{n}$for

a) n=3

b) n=4

c) general, n $\beta \x89\u20af$2

HINT: Correlation matrix must be positive semi-definite.

My Workings

This is what I infer from "same pairwise coefficients":

$\left(\begin{array}{ccc}1& {\mathrm{{\rm O}\x81}}_{n}& {\mathrm{{\rm O}\x81}}_{n}\\ {\mathrm{{\rm O}\x81}}_{n}& 1& {\mathrm{{\rm O}\x81}}_{n}\\ {\mathrm{{\rm O}\x81}}_{n}& {\mathrm{{\rm O}\x81}}_{n}& 1\end{array}\right)$

Because a correlation matrix is positive semi-definite, all principal minors have to be positive

For n=3, the principal minors calculations yield

$\beta \x88\pounds {H}_{1}\beta \x88\pounds $=1

$\beta \x88\pounds {H}_{2}\beta \x88\pounds $=1-${\mathrm{{\rm O}\x81}}_{n}^{2}$ $\beta \x89\u20af0$ $\beta \x87\x92$${\mathrm{{\rm O}\x81}}_{n}$ $\beta \x89\u20ac1$

$\beta \x88\pounds {H}_{3}\beta \x88\pounds ={\mathrm{{\rm O}\x81}}_{n}^{2}$ - ${\mathrm{{\rm O}\x81}}_{n}$$\beta \x89\u20af0$ $\beta \x87\x92$ ${\mathrm{{\rm O}\x81}}_{n}$ $\beta \x89\u20af1$

$\beta \x88\pounds {H}_{4}\beta \x88\pounds =({1}^{3}$+2${\mathrm{{\rm O}\x81}}_{n}^{3}$)-(3${\mathrm{{\rm O}\x81}}_{n}^{2}$)$\beta \x89\u20af0$ . I solved for the roots and found 1

Conclusion: For n=3, max ${\mathrm{{\rm O}\x81}}_{n}$=1

I did the same method for n=4 and found max ${\mathrm{{\rm O}\x81}}_{n}$=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

${H}_{0}$

$\left(\begin{array}{cc}1& {\mathrm{{\rm O}\x81}}_{2}\\ {\mathrm{{\rm O}\x81}}_{2}& 1\end{array}\right)$

$\beta \x87\x92\beta \x88\x92\frac{1}{2\beta \x88\x921}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{2}\beta \x89\u20ac1$

${H}_{n}$: Pn: Suppose that for a nxn correlation matrix An with same pairwise coefficients ${\mathrm{{\rm O}\x81}}_{n}$, $\beta \x88\x92\frac{1}{n\beta \x88\x921}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{n}\beta \x89\u20ac1$ holds

${H}_{n+1}$

$\beta \x88\x92\frac{1}{n\beta \x88\x921}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{n}\beta \x89\u20ac1$

$\beta \x87\x90\beta \x87\x92$$\beta \x88\x92\frac{1}{(n+1)\beta \x88\x921}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{n+1}\beta \x89\u20ac1$

$\beta \x87\x90\beta \x87\x92$$\beta \x88\x92\frac{1}{n}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{n+1}\beta \x89\u20ac1$

And because for all $\beta \x88\x92\frac{1}{n}\beta \x89\u20af\beta \x88\x92\frac{1}{n\beta \x88\x921}$

$\beta \x88\x92\frac{1}{n\beta \x88\x921}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{n+1}\beta \x89\u20ac1$

Correlation matrix. Consider π random variables with the same pairwise correlation coefficient ${\mathrm{{\rm O}\x81}}_{n}$. Find the highest possible value of ${\mathrm{{\rm O}\x81}}_{n}$for

a) n=3

b) n=4

c) general, n $\beta \x89\u20af$2

HINT: Correlation matrix must be positive semi-definite.

My Workings

This is what I infer from "same pairwise coefficients":

$\left(\begin{array}{ccc}1& {\mathrm{{\rm O}\x81}}_{n}& {\mathrm{{\rm O}\x81}}_{n}\\ {\mathrm{{\rm O}\x81}}_{n}& 1& {\mathrm{{\rm O}\x81}}_{n}\\ {\mathrm{{\rm O}\x81}}_{n}& {\mathrm{{\rm O}\x81}}_{n}& 1\end{array}\right)$

Because a correlation matrix is positive semi-definite, all principal minors have to be positive

For n=3, the principal minors calculations yield

$\beta \x88\pounds {H}_{1}\beta \x88\pounds $=1

$\beta \x88\pounds {H}_{2}\beta \x88\pounds $=1-${\mathrm{{\rm O}\x81}}_{n}^{2}$ $\beta \x89\u20af0$ $\beta \x87\x92$${\mathrm{{\rm O}\x81}}_{n}$ $\beta \x89\u20ac1$

$\beta \x88\pounds {H}_{3}\beta \x88\pounds ={\mathrm{{\rm O}\x81}}_{n}^{2}$ - ${\mathrm{{\rm O}\x81}}_{n}$$\beta \x89\u20af0$ $\beta \x87\x92$ ${\mathrm{{\rm O}\x81}}_{n}$ $\beta \x89\u20af1$

$\beta \x88\pounds {H}_{4}\beta \x88\pounds =({1}^{3}$+2${\mathrm{{\rm O}\x81}}_{n}^{3}$)-(3${\mathrm{{\rm O}\x81}}_{n}^{2}$)$\beta \x89\u20af0$ . I solved for the roots and found 1

Conclusion: For n=3, max ${\mathrm{{\rm O}\x81}}_{n}$=1

I did the same method for n=4 and found max ${\mathrm{{\rm O}\x81}}_{n}$=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

${H}_{0}$

$\left(\begin{array}{cc}1& {\mathrm{{\rm O}\x81}}_{2}\\ {\mathrm{{\rm O}\x81}}_{2}& 1\end{array}\right)$

$\beta \x87\x92\beta \x88\x92\frac{1}{2\beta \x88\x921}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{2}\beta \x89\u20ac1$

${H}_{n}$: Pn: Suppose that for a nxn correlation matrix An with same pairwise coefficients ${\mathrm{{\rm O}\x81}}_{n}$, $\beta \x88\x92\frac{1}{n\beta \x88\x921}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{n}\beta \x89\u20ac1$ holds

${H}_{n+1}$

$\beta \x88\x92\frac{1}{n\beta \x88\x921}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{n}\beta \x89\u20ac1$

$\beta \x87\x90\beta \x87\x92$$\beta \x88\x92\frac{1}{(n+1)\beta \x88\x921}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{n+1}\beta \x89\u20ac1$

$\beta \x87\x90\beta \x87\x92$$\beta \x88\x92\frac{1}{n}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{n+1}\beta \x89\u20ac1$

And because for all $\beta \x88\x92\frac{1}{n}\beta \x89\u20af\beta \x88\x92\frac{1}{n\beta \x88\x921}$

$\beta \x88\x92\frac{1}{n\beta \x88\x921}\beta \x89\u20ac{\mathrm{{\rm O}\x81}}_{n+1}\beta \x89\u20ac1$

Inferential StatisticsAnswered question

Alice Chen 2022-11-07

Correlation Function

Let X,Y be random variables. If $\mathrm{{\rm O}\x81}(X,Y)=a$ (Correlation), where $a\beta \x88\x88(0,1)$, what can be said about the relationship between X and Y? Is it true that Y=bX+c+Z, where Z is a random variable? If it is true then how is $|Z|$ related to the correlation a?

Let X,Y be random variables. If $\mathrm{{\rm O}\x81}(X,Y)=a$ (Correlation), where $a\beta \x88\x88(0,1)$, what can be said about the relationship between X and Y? Is it true that Y=bX+c+Z, where Z is a random variable? If it is true then how is $|Z|$ related to the correlation a?

Inferential StatisticsAnswered question

reevelingw97 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.

Inferential StatisticsAnswered question

varsa1m 2022-10-28

Correlation and heteroscedasticityI'm studying a dataset and observed a positive correlation between two variables, but when I plot them, it seems that they are heteroscedastics, what conclusions can I get from it ? (Can I really assume that the positive correlation is real ?)

Inferential StatisticsAnswered question

snaketao0g 2022-10-24

It is important to know that "Causation is not the same as Correlation." Which of the following gives a reason that causation is not the same as correlation?

A) Correlation is only a number between -1 and 1.

B) Sometimes the two variables with a correlation are both the results of a third factor or variable.

C) Even with a strong statistical correlation, there are usually some people who don't fit the pattern.

D) All of the above.

A) Correlation is only a number between -1 and 1.

B) Sometimes the two variables with a correlation are both the results of a third factor or variable.

C) Even with a strong statistical correlation, there are usually some people who don't fit the pattern.

D) All of the above.

Inferential StatisticsAnswered question

Jaelyn Payne 2022-10-15

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

Inferential StatisticsAnswered question

erwachsenc6 2022-10-15

Correlation Coefficient - statementsThis is in relation to a scatterplot of 400 data points, with a very strong negative correlation.

For the statements, state whether they are correct or incorrect and why.

If each value of the y variable is divided by four, then the value of the correlation coefficient will also be divided by four. - Assume this one is true? As the correlation coefficient then wouldn't change?

The scatterplot shows that the variable x causes the variable y to take the values it does. - not too sure about this one?

For the statements, state whether they are correct or incorrect and why.

If each value of the y variable is divided by four, then the value of the correlation coefficient will also be divided by four. - Assume this one is true? As the correlation coefficient then wouldn't change?

The scatterplot shows that the variable x causes the variable y to take the values it does. - not too sure about this one?

Inferential StatisticsAnswered question

Janessa Benson 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?

Inferential StatisticsAnswered question

Aidyn Crosby 2022-09-29

What is the main difference between correlation and causation? (Answer the question in a short paragraph (roughly 3-5 sentences). If necessary, explain the concept and/or give examples)

Inferential StatisticsAnswered question

Melina Barber 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

$\mathrm{{\rm O}\x81}(X,Y)=\frac{\underset{i=1}{\overset{20}{\beta \x88\x91}}({x}_{i}\beta \x88\x92\stackrel{{\rm B}\u2015}{x})({y}_{i}\beta \x88\x92\stackrel{{\rm B}\u2015}{y})}{\sqrt{(\underset{i=1}{\overset{20}{\beta \x88\x91}}({x}_{i}\beta \x88\x92\stackrel{{\rm B}\u2015}{x}{)}^{2})(\underset{i=1}{\overset{20}{\beta \x88\x91}}({y}_{i}\beta \x88\x92\stackrel{{\rm B}\u2015}{y}{)}^{2})}}$

If $\mathrm{{\rm O}\x81}(X,Y)=1$, we can say that X and Y have a linear correlation. If $0.7\beta \x89\u20ac\mathrm{{\rm O}\x81}(X,Y)<1$ then X and Y has a strong linear correlation, if $0.5\beta \x89\u20ac\mathrm{{\rm O}\x81}(X,Y)<0.7$ then X and Y has a modest linear correlation, and if $0\beta \x89\u20ac\mathrm{{\rm O}\x81}(X,Y)<0.5$ then X and Y has a weak linear correlation. Using this formula, we get $\mathrm{{\rm O}\x81}$(X,Y)=0.9 and $\mathrm{{\rm O}\x81}$(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 $\mathrm{{\rm O}\x81}$(X,Z)=0.5 indicate the modest correlation coefficient, what is another intepretation of this value? What is the difference between $\mathrm{{\rm O}\x81}$(X,Y) and $\mathrm{{\rm O}\x81}$(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

$\mathrm{{\rm O}\x81}(X,Y)=\frac{\underset{i=1}{\overset{20}{\beta \x88\x91}}({x}_{i}\beta \x88\x92\stackrel{{\rm B}\u2015}{x})({y}_{i}\beta \x88\x92\stackrel{{\rm B}\u2015}{y})}{\sqrt{(\underset{i=1}{\overset{20}{\beta \x88\x91}}({x}_{i}\beta \x88\x92\stackrel{{\rm B}\u2015}{x}{)}^{2})(\underset{i=1}{\overset{20}{\beta \x88\x91}}({y}_{i}\beta \x88\x92\stackrel{{\rm B}\u2015}{y}{)}^{2})}}$

If $\mathrm{{\rm O}\x81}(X,Y)=1$, we can say that X and Y have a linear correlation. If $0.7\beta \x89\u20ac\mathrm{{\rm O}\x81}(X,Y)<1$ then X and Y has a strong linear correlation, if $0.5\beta \x89\u20ac\mathrm{{\rm O}\x81}(X,Y)<0.7$ then X and Y has a modest linear correlation, and if $0\beta \x89\u20ac\mathrm{{\rm O}\x81}(X,Y)<0.5$ then X and Y has a weak linear correlation. Using this formula, we get $\mathrm{{\rm O}\x81}$(X,Y)=0.9 and $\mathrm{{\rm O}\x81}$(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 $\mathrm{{\rm O}\x81}$(X,Z)=0.5 indicate the modest correlation coefficient, what is another intepretation of this value? What is the difference between $\mathrm{{\rm O}\x81}$(X,Y) and $\mathrm{{\rm O}\x81}$(X,Z), noting that Y and Z is not a linear function of X.

Inferential StatisticsAnswered question

malaana5k 2022-09-26

Meaning of 'obligatory disclaimer'

Can you explain to me what 'obligatory disclaimer' means in probability. Is it like a lack of information or what?

The context is the following:

Second, we found that the marginal distribution of Y is Bern(0.08), whereas the conditional distribution of Y given X = 1 is Bern(0.2) and the conditional distribution of Y given X = 0 is Bern(0.04). Since conditioning on the value of X alters the distribution of Y , X and Y are not independent: learning whether or not the sampled individual is a current smoker gives us information about the probability that he will develop lung cancer. This example comes with an obligatory disclaimer. Although we have found that X and Y are dependent, we cannot make conclusions about whether smoking causes lung cancer based on this association alone. (Joseph K. Blitzstein, Jessica Hwang--Introduction to Probability)

Can you explain to me what 'obligatory disclaimer' means in probability. Is it like a lack of information or what?

The context is the following:

Second, we found that the marginal distribution of Y is Bern(0.08), whereas the conditional distribution of Y given X = 1 is Bern(0.2) and the conditional distribution of Y given X = 0 is Bern(0.04). Since conditioning on the value of X alters the distribution of Y , X and Y are not independent: learning whether or not the sampled individual is a current smoker gives us information about the probability that he will develop lung cancer. This example comes with an obligatory disclaimer. Although we have found that X and Y are dependent, we cannot make conclusions about whether smoking causes lung cancer based on this association alone. (Joseph K. Blitzstein, Jessica Hwang--Introduction to Probability)

Inferential StatisticsAnswered question

Jazmyn Pugh 2022-09-25

Unable to understand correlation coefficient / auto correlation

Suppose I have a vector say:

[5 5 5 5 4 5]

then common sense says that there is a very high auto-correlation for the vector because it is more or less the same values. But when I try to calculate the auto-correlation coefficient, I'm getting a very low value(<0.3) for all lags. What does this mean? shouldn't it be higher because the series is very similar?Am I missing something?does correlatiom mean not similarity but similarity in rate(Rate of change)?

Suppose I have a vector say:

[5 5 5 5 4 5]

then common sense says that there is a very high auto-correlation for the vector because it is more or less the same values. But when I try to calculate the auto-correlation coefficient, I'm getting a very low value(<0.3) for all lags. What does this mean? shouldn't it be higher because the series is very similar?Am I missing something?does correlatiom mean not similarity but similarity in rate(Rate of change)?

Inferential StatisticsAnswered question

Haiphongum 2022-09-20

What can we conclude from correlation?I just got my statistics test back and I am totally confused about one of the questions!

A study was done that took a simple random sample of 40 people and measured whether the subjects were right-handed or left-handed, as well as their ages. The study showed that the proportion of left-handed people and the ages had a strong negative correlation. What can we conclude? Explain your answer.

I know that we can't conclude that getting older causes people to become right-handed. Something else might be causing it, not the age. If two things are correlated, we can only conclude association, not causation. So I wrote:

We can conclude that many people become right-handed as they grow older, but we cannot tell why.

A study was done that took a simple random sample of 40 people and measured whether the subjects were right-handed or left-handed, as well as their ages. The study showed that the proportion of left-handed people and the ages had a strong negative correlation. What can we conclude? Explain your answer.

I know that we can't conclude that getting older causes people to become right-handed. Something else might be causing it, not the age. If two things are correlated, we can only conclude association, not causation. So I wrote:

We can conclude that many people become right-handed as they grow older, but we cannot tell why.

Inferential StatisticsAnswered question

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 $P({T}_{1},{T}_{2})={0.95}^{2}\beta \x88\x970.01+{0.05}^{2}\beta \x88\x970.99=0.0115$, because the tests are independent.

By law of total probability we know that:

$P({T}_{1},{T}_{2})={0.95}^{2}\beta \x88\x970.01+{0.05}^{2}\beta \x88\x970.99=0.0115$

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

$P({T}_{1},{T}_{2})={0.95}^{2}\beta \x88\x970.01+{0.05}^{2}\beta \x88\x970.99=0.0115$

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:

$P({T}_{1})=0.95\beta \x88\x970.01+0.05\beta \x88\x970.99\beta \x89\x880.059$

$P({T}_{1})=0.95\beta \x88\x970.01+0.05\beta \x88\x970.99\beta \x89\x880.059$

So we have:

$P({T}_{1},{T}_{2})=P({T}_{1}{)}^{2}\beta \x89\x88{0.059}^{2}\beta \x89\x880.003481$

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 $P({T}_{1},{T}_{2})={0.95}^{2}\beta \x88\x970.01+{0.05}^{2}\beta \x88\x970.99=0.0115$, because the tests are independent.

By law of total probability we know that:

$P({T}_{1},{T}_{2})={0.95}^{2}\beta \x88\x970.01+{0.05}^{2}\beta \x88\x970.99=0.0115$

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

$P({T}_{1},{T}_{2})={0.95}^{2}\beta \x88\x970.01+{0.05}^{2}\beta \x88\x970.99=0.0115$

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:

$P({T}_{1})=0.95\beta \x88\x970.01+0.05\beta \x88\x970.99\beta \x89\x880.059$

$P({T}_{1})=0.95\beta \x88\x970.01+0.05\beta \x88\x970.99\beta \x89\x880.059$

So we have:

$P({T}_{1},{T}_{2})=P({T}_{1}{)}^{2}\beta \x89\x88{0.059}^{2}\beta \x89\x880.003481$

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.

Inferential StatisticsOpen question

Jaydan Ball 2022-08-31

Do future events influence the probability of past events?

This question came to mind when I was dealing with this question: A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be clubs. Find the probability that the lost card is a club.

My query: The probability that a club is lost is 13/52. The next event of taking two cards from this incomplete set of cards happens after one of the cards is lost. How can this affect a past event of losing a card from the pack?

This question came to mind when I was dealing with this question: A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be clubs. Find the probability that the lost card is a club.

My query: The probability that a club is lost is 13/52. The next event of taking two cards from this incomplete set of cards happens after one of the cards is lost. How can this affect a past event of losing a card from the pack?

Inferential StatisticsOpen question

Licinilg 2022-08-23

Determining correlation given 2 other correlation

Hi I'm given a rather ambigious question on correlations. My question is how do we determine the correlation between 2 varibles given their correlation with other variables?

Correlation between chocolate consumption per capita and number of Nobel laureates per 10 million persons for a broader list of 90 countries = 0.45

Perfect positive association bewteen chocolate consumption and chili consumption per capita (correlation = 1)

Correlation between chili consumption per capita and nobel laureates per 10 million persons: ?

1) > 0.45

2) = 0.45

3) < 0.45

4) cannot determine

Hi I'm given a rather ambigious question on correlations. My question is how do we determine the correlation between 2 varibles given their correlation with other variables?

Correlation between chocolate consumption per capita and number of Nobel laureates per 10 million persons for a broader list of 90 countries = 0.45

Perfect positive association bewteen chocolate consumption and chili consumption per capita (correlation = 1)

Correlation between chili consumption per capita and nobel laureates per 10 million persons: ?

1) > 0.45

2) = 0.45

3) < 0.45

4) cannot determine

Inferential StatisticsOpen question

brasocas6 2022-08-19

Statistics for Behavioral Sciences - Correlation

What does a negative sign tell you about the relationship between two variables, X and Y?

Select all that apply

1) a negative correlation means X and Y change in the same direction

2) a negative correlation means as X increases, Y decreases

3) a negative correlation means X and Y change in opposite directions

4) a negative correlation means that X and Y are weakly related to each other

What does a negative sign tell you about the relationship between two variables, X and Y?

Select all that apply

1) a negative correlation means X and Y change in the same direction

2) a negative correlation means as X increases, Y decreases

3) a negative correlation means X and Y change in opposite directions

4) a negative correlation means that X and Y are weakly related to each other

Inferential StatisticsOpen question

sublimnes9 2022-08-18

StatisticsοΌHow to prove that one variable's change is influenced by another variable,that is to say ,they are related?

I major in Bioinformatics. Now,I am in a problem: we all know that temperature changes during a year , I find that a disease incidence is really high when temperature is relatively high , while it becomes really low when the temperature is relatively low , that is to say ,they are related. So ,I want to find out a way to prove that they are related ,not just intuitively feel that they are related. So, any suggestions?

I major in Bioinformatics. Now,I am in a problem: we all know that temperature changes during a year , I find that a disease incidence is really high when temperature is relatively high , while it becomes really low when the temperature is relatively low , that is to say ,they are related. So ,I want to find out a way to prove that they are related ,not just intuitively feel that they are related. So, any suggestions?

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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.