Correlation and multiple random variables

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?

• Are you sure that the "correlation coefficient" is supposed to be $13$? A correlation coefficient $\rho$ must satisfy $|\rho|<1$. – Math1000 Oct 22 '19 at 18:55
• Thank you for pointing out, it was supposed to be 1/3 – Pirategull Oct 22 '19 at 19:00
• $X$ and $Y$ are independent so $Cov(X,Y)=0$. It is $Cov(X,Z) = 1/3$. – Theoretical Economist Oct 22 '19 at 19:05
• thank you, I fixed that too – Pirategull Oct 22 '19 at 19:26

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).
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 $$\frac13=\frac{a}{\sqrt{1+a^2}}\implies a=\frac{1}{\sqrt{8}}$$.