# Finding correlation coefficient of $X$ and $XY$

Let $$X$$ and $$Y$$ be independent random variables with nonzero variances. I'm looking to find the correlation coefficient $$\rho$$ of $$Z=XY$$ and $$X$$ in terms of the means and variances of $$X$$ and $$Y$$, i.e. $$\mu_X, \mu_Y, \sigma^2_X, \sigma^2_Y$$.

(I have searched different methods online, including Correlation between X and XY. However, I'm wondering if I could use a simple calculation-approach rather than using moments as well.)

The result I obtained, along with the steps I've used, is the following:

\begin{align} \rho & = \frac{\text{Cov}(Z,X)}{\sigma_Z\sigma_X}\\[1em] & = \frac{E\left[\left(Z-\mu_Z\right)\left(X-\mu_X\right)\right]}{\sigma_Z\sigma_X} \\[1em] & = \frac{E\left[\left(XY-\mu_X\mu_Y\right)\left(X-\mu_X\right)\right]}{\sqrt{E\left[\left(XY\right)^2\right]-\left[E\left(XY\right)\right]^2}\cdot\sigma_X} \\[1em] & = \frac{E\left(X^2Y\right)-\mu_X^2\mu_Y}{\sqrt{E\left(X^2\right)E\left(Y^2\right)-\left[E\left(X\right)\right]^2\left[E\left(Y\right)\right]^2}\cdot\sigma_X} \\[1em] & = \frac{E\left(X^2\right)E\left(Y\right)-\mu_X^2\mu_Y}{\sqrt{\left(\sigma_X^2+\mu_X^2\right)\left(\sigma_Y^2+\mu_Y^2\right)-\mu^2_X\mu^2_Y}\cdot\sigma_X} \\[1em] & = \frac{\mu_Y\left[E\left(X^2\right)-\mu^2_X\right]}{\sqrt{\sigma^2_X\sigma^2_Y+\sigma_X^2\mu_Y^2+\sigma_Y^2\mu_X^2}\cdot\sigma_X} \\[1em] & = \frac{\mu_Y\sigma_X^2}{\sqrt{\sigma^2_X\sigma^2_Y+\sigma_X^2\mu_Y^2+\sigma_Y^2\mu_X^2}\cdot\sigma_X} \\[1em] & = \frac{\mu_Y\sigma_X}{\sqrt{\sigma^2_X\sigma^2_Y+\sigma_X^2\mu_Y^2+\sigma_Y^2\mu_X^2}} \end{align}

which is seemingly different from the result from the moment approach used in Correlation between X and XY. In which step has an error in my calculation occurred (if any), and how can I obtain $$\rho$$ from the approach I am trying to use?

• Why did you use $\mu_z = \mu_x\mu_y$? Nov 27, 2020 at 4:20
• @helperFunction since $X$ and $Y$ are independent variables and $Z=XY$, I put $\mu_Z=E(Z)=E(XY)=E(X)E(Y)=\mu_X\mu_Y$. Nov 27, 2020 at 4:23
• In the fourth equality, I think you're missing two cross terms that come to $-2\mu_X^2\mu_Y$, so you end up with $-\mu^2_X\mu_Y$ rather than $+\mu^2_X\mu_Y$ Nov 27, 2020 at 5:00
• @ThomasLumley you are absolutely right; I've edited my result, which simplifies a bit but I still am unsure if that should be the final form. Do you think there are further steps to this? Nov 27, 2020 at 5:12
• for what applications would someone want to multiply $X$ and $Y$ together? Nov 27, 2020 at 7:25

The simplest example I can think of for this is $$Y$$ being a constant that isn't 0, 1 or -1. So, let $$Y=\mu_Y$$ be a positive constant that isn't 1, and $$\sigma^2_Y=0$$.
The first three equalities are just expanding definitions, so the fourth is the first time something could go wrong. And it does. The numerator in the third line simplifies to $$\mu_Y\mathrm{var}[X]$$. The numerator in the fourth line doesn't. Or didn't when I wrote this; it has now been edited.
What you've written is the same as the expression in the link. In the link, there is a typo in the denominator, as $$\mu_2(Y)^2$$ should be $$\mu_1(Y)^2$$.
$$\begin{eqnarray}\text{Cor}(X,XY) &=& \frac{\mu_2(X)\mu_1(Y) - \mu_1(X)^2\mu_1(Y)}{\sqrt{(\mu_2(X)-\mu_1(X)^2)(\mu_2(X)\mu_2(Y) - \mu_1(X)^2\mu_1(Y)^2)}} \\ &=& \frac{E[X^2]\mu_Y - \mu_X^2\mu_Y}{\sqrt{\sigma_X^2(E[X^2]E[Y^2]-\mu_X^2\mu_Y^2)}}\\ &=& \frac{\sigma_X\mu_Y}{\sqrt{(\sigma_X^2+\mu_X^2)(\sigma_Y^2+\mu_Y^2)-\mu_X^2\mu_Y^2}}\\ &=& \frac{\sigma_X\mu_Y}{\sqrt{\sigma_X^2\sigma_Y^2 + \sigma_X^2\mu_Y^2 + \mu_X^2\sigma_Y^2}}\\ \end{eqnarray}$$