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I'm currently working on this exercise:

Suppose the random variables $X$ and $Y$ are independent and identically distributed. Let $Z = aX + Y$. If the correlation coefficient between $X$ and $Z$ is $\frac13$ , then what is the value of the constant a ?

My book says that the Convariance is defined as $Cov(X,Y) = (EXY) - (μxμy)$

So I know that $Cov(XZ) = 1/3$, how do I find the PDF from here? And what does it mean to be independent and identically distributed?

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  • $\begingroup$ Are you sure that the "correlation coefficient" is supposed to be $13$? A correlation coefficient $\rho$ must satisfy $|\rho|<1$. $\endgroup$ – Math1000 Oct 22 '19 at 18:55
  • $\begingroup$ Thank you for pointing out, it was supposed to be 1/3 $\endgroup$ – Pirategull Oct 22 '19 at 19:00
  • $\begingroup$ $X$ and $Y$ are independent so $Cov(X,Y)=0$. It is $Cov(X,Z) = 1/3$. $\endgroup$ – Theoretical Economist Oct 22 '19 at 19:05
  • $\begingroup$ thank you, I fixed that too $\endgroup$ – Pirategull Oct 22 '19 at 19:26
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Independent and identically distributed (or iid.) is an important thing to understand. It just means what it says however: $X$ and $Y$ are independent variables, which share the same distribution. In particular, $EX=EY$, $VX=VY$, and so on. An example would be two coin tosses; independent of each other, but having the same probability of heads/tails.

Also, it is the correlation coefficient, not the covariance, that is $1/3$. We have $$ Corr(X,Z):=\frac{Cov(X,Z)}{\sqrt{VXVZ}}=\frac13 $$ We can work out: $$ VZ = V(aX+Y) = a^2VX+VY = (a^2+1)VX $$ and $$ Cov(X,Z)= Cov(X,aX+Y) = aCov(X,X) + Cov(X,Y) = aVX+0 $$ I skipped some justifications that you can hopefully fill in. Can you finish from here?

If you want to work from the definitions $Cov(X,Y)=EXY-EXEY$ and $VX=EX^2-(EX)^2$, then remember that $E$ is linear, and that $E(XY)=EXEY$ for independent variables. (But I don't recommend that, since you should have the linearity theorems for variance and covariance available).

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Without loss of generality assume $X,\,Y$ have mean $0$ and variance $1$, so $Z$ has variance $1+a^2$ and $\Bbb EXZ=\Bbb E(aX^2+XY)=a$. Hence $\frac13=\frac{a}{\sqrt{1+a^2}}\implies a=\frac{1}{\sqrt{8}}$.

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