Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that X and Y are random variables such that E(Y | X) = 7 - (1/4)x and E(X | Y) = 10 - Y . Determine the correlation of X and Y .

Edit:

So far I've got

E(x)=4 E(y)=6

Now I'm trying to find

E(xy) to use in cov(x,y)=E(xy)-E(x)E(y)

V(x)

V(y)

all to use in cor(x,y)=cov(x,y)/(v(x)v(y))^.5

share|improve this question
    
Correlation as measured by...? –  Xodarap Dec 5 '12 at 3:15
    
cor(x,y) = cov(x,y) / (v(x)v(y))^.5 ...Im not sure I understand your question though –  Dan Dec 5 '12 at 3:20
    
Pearson's is a fine way to define correlation; it's just not the only way. –  Xodarap Dec 5 '12 at 3:22

1 Answer 1

Hint: $E[Y\mid X]$ is the minimum-mean-square-error estimator of $Y$ given the value of $X$. The linear minimum-mean-square-error estimator of $Y$ given the value of $X$ is

$$\hat{Y} = \mu_Y + \frac{\rho\sigma_Y}{\sigma_X}(X-\mu_X).$$

Similar statements apply to $E[X\mid Y]$ etc. Just interchange $X$ and $Y$ in the above formulas.

Now, if $E[Y\mid X]$ is a linear function of $X$ and $E[X\mid Y]$ is a linear function of $Y$, can you use the known forms of the linear minimum-mean-square-error estimators to deduce the value of $\rho$?

share|improve this answer
    
yeah Ive found E(y)=6. Any other hints? –  Dan Dec 5 '12 at 3:28
1  
You don't need to find $E[Y]$ or $E[X]$. Hint: show that the coefficients of $X$ and $Y$ in the equations given to you are $\rho\sigma_Y/\sigma_X$ and $\rho\sigma_X/\sigma_Y$ and use the information given to you to deduce that $\rho=0.5$ without needing to find $E[X]$ or $E[Y]$ or $\text{var}(X)$ or $\text{var}(Y)$ or $\text{cov}(X,Y)$ or $E[XY]$. –  Dilip Sarwate Dec 5 '12 at 3:35
    
@DilipSarwate (+1), but I think you meant to write $\rho = -1/2$ in your comment. –  r.e.s. Dec 5 '12 at 4:16
    
@r.e.s. Yes, you are correct. I realized that I had neglected to take the sign of $\rho$ into account after it was too late to edit the comment and was going to write a further comment. Thanks for the upvote. –  Dilip Sarwate Dec 5 '12 at 11:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.