Let's say we have two random variables $Y$ and $X$ used to form regression model $$Y=\alpha+\beta X+\mu$$ It also holds that $E(\mu)=0$, $\text{Var}(\mu)=\sigma_{\mu}^2$, $\text{Var}(X)=\sigma_{X}^2$, $\text{Var}(Y)=\sigma_{Y}^2$, $\text{Corr}(X,Y)=r$ and $\text{Corr}(X,\mu)=r_{X\mu}$. Find $\beta$. I tried to solve this as follows:
For simple linear regression $\beta=\dfrac{\text{Cov}(X,Y)}{\text{Var}(X)}$ and $\text{Corr}(X,Y)=\dfrac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y}=r$ so that: $$\beta=\frac{\text{Corr}(X,Y) \cdot \sigma_X \sigma_Y}{\sigma_X^2}=r\frac{\sigma_Y}{\sigma_X}$$ Is this as simple as this?
Since you probably want the answer for the true $\beta$ given all moments let's be detailed. $$ E(Y)=\alpha+\beta E(X) \\ XY=X\alpha+ \beta X^2 +X\mu \\ E(XY)=E(X)\alpha+\beta E(X^2)+E(X\mu)\\ Cov(XY)=E(XY)-E(X)E(Y)=E(X)\alpha+\beta E(X^2)+E(X\mu)-E(X)\alpha\\ -\beta E(X)^2=\beta Var(X)+Cov(X\mu)\\ So\\ r\sigma_X\sigma_Y=\beta \sigma_X^2+r_{X_{\mu}}\sigma_X \sigma_\mu $$ Thus (FIXED) $$ \beta=\frac{r\sigma_Y-r_{X_{\mu}}\sigma_\mu}{\sigma_X} $$