The Correlation of two Random Variables

Question

I would expect the answer to the problem below to be at least $\frac{1}{2}$. However, it is not. Therefore, I am thinking I did something wrong. Did I?

Problem:
Suppose that $X$ is a uniformly distributed random variable over the interval $[-1,1]$. That is, the density function for $X$ is: $$ f(x) = \begin{cases} \frac{1}{2}, & \text{for } {-1} \leq x \leq 1 \\ 0, & \text{otherwise } \end{cases} $$ Now $Y$ is a random variable that follows the triangle distribution. That is, the density function for $Y$ is: $$ g(y) = \begin{cases} y + 1, & \text{for } {-1} \leq y \leq 0 \\ 1 - y, & \text{for } 0 \leq y \leq 1 \\ 0, & \text{otherwise } \end{cases} $$ We generate a value $x_0$ using the density function $f(x)$. We then generate a value $y_0$ using the density function $g(y)$ such that $$ \int_{-1}^{x_0} \frac{1}{2} \, dx = \int_{-1}^{y_0} g(y) \, dy $$ What is the correlation between $X$ and $Y$?

Answer:

The formula for correlation is: $$\rho = \frac{\sigma_{xy}}{ \sigma_x \sigma_y } $$ Observe that the $E(X) = E(Y) = 0$. \begin{align*} E(X^2) &= \int_{-1}^{1} \frac{x^2}{2} \, dx = \frac{x^3}{6} \Big|_{-1}^{1} = \frac{1}{6} - \frac{-1}{6} \\ E(X^2) &= \frac{1}{3} \\ \sigma_x^2 &= E(X^2) - (E(x))^2 = \frac{1}{3} \\ \sigma_x &=\frac{ \sqrt{3} }{3} \\ E(Y^2) &= \int_{-1}^{1} y^2 f(y) \, dy = \int_{-1}^{0} y^2 f(y) \, dy + \int_{0}^{1} y^2 f(y) \, dy \\ \int_{-1}^{0} y^2 (y+1) \, dy &= \int_{-1}^{0} y^3 + y^2 \,\, dy = \frac{y^4}{4} + \frac{y^3}{3} \Big|_{-1}^{0} \\ \int_{-1}^{0} y^2 (y+1) \, dy &= 0 - \left( \frac{1}{4} - \frac{1}{3} \right) \\ \int_{-1}^{0} y^2 (y+1) \, dy &= \frac{1}{12} \\ \sigma_y^2 &= E(Y^2) - (E(y))^2 = \frac{1}{12} \\ \sigma_y &= \frac{1}{\sqrt{12}} = \frac{ \sqrt{3} } {6} \end{align*} Now we need to find $\sigma_{xy}$.To do this, we need to find $E(XY)$. Observe that if $-1\leq x \leq 0$ then the area under the $f(x)$ between $x = -1$ and $x = 0$ is $\left( \frac{1}{2}\right) (x + 1)$. Now we need to find $y$ in terms of $x$. \begin{align*} \left( \frac{1}{2}\right) (x + 1) &= \int_{-1}^{y} y + 1 \, dy = \frac{y^2}{2} + y \Big|_{-1}^{y} \\ \left( \frac{1}{2}\right) (x + 1) &= \frac{y^2}{2} + y - \left( \frac{1}{2} - 1 \right) \\ \left( \frac{1}{2}\right) (x + 1) &= \frac{y^2}{2} + y + \frac{1}{2} \\ x + 1 &= y^2 + 2y + 1 \\ x &= y^2 + 2y \\ E(XY) &= 2 \int_{-1}^{0} (y^2+y)(y)(y+1) \, dy \\ \int_{-1}^{0} (y^2+y)(y)(y+1) \, dy &= \int_{-1}^{0} y^4 + y^3 + y^3 + y^2 \, dy \\ \int_{-1}^{0} (y^2+y)(y)(y+1) \, dy &= \int_{-1}^{0} y^4 + 2y^3 + y^2 \, dy \\ \int_{-1}^{0} (y^2+y)(y)(y+1) \, dy &= \frac{y^5}{5}+ \frac{2y^4}{4} + \frac{y^3}{3} \Big|_{-1}^{0} \\ \int_{-1}^{0} (y^2+y)(y)(y+1) \, dy &=0 - \left( -\frac{1}{5} + \frac{2}{4} - \frac{1}{3} \right) = \frac{1}{5} - \frac{2}{4} + \frac{1}{3} \\ \int_{-1}^{0} (y^2+y)(y)(y+1) \, dy &= \frac{1}{30} \\ E(XY) &= \frac{1}{15} \\ \rho &= \frac{ \left( \frac{1}{15} \right) }{ \left( \frac{ \sqrt{3} }{3} \right) \left( \frac{ \sqrt{3} } {6} \right) } = \frac{ \frac{1}{15} } { \frac{3}{18} } \\ \rho &= \frac{2}{5} \end{align*}

Answer 1

For $E[XY]$, if I understood correctly, there is a mistake. You are first choosing $X$ and than $Y=h(X)$, with $h$ deterministic. So you have to evaluate:

$E[XY]=\int f_X(x) xh(x)dx$

Instead it looks that you integrated over $y$. These would correspnd to the situation where you first choose $Y$ and than $X$ accordingly. But consider than that when $h$ is invertible:

$\int f_X(x) xh(x)dx \ne \int f_Y(y) h^{-1}(y)ydy$ [0]

An identity clearly cannot be generally true since the left member is independent of the density of $Y$ , $f_Y$.

UPDATE: Actually it is a nice result that that, for the particular $h$ chosen in this problem, which depends on the c.d.f. of the distributions, the situation is symmetric and the method of the OP should be correct (even if maybe the OP did not do it on purpose?). Let's try to show this:

Denoting with $F_X$ the c.f.d. we have that:

$E[XY]=E[XF^{-1}_Y[F_X(X)]]=\int dx f_X(x) x F^{-1}_Y[F_X(x)]$ [1]

Let me know if [1] is not clear. Changing variables $F_X(x)=z$ we have that (remember that the derivative of the c.d.f. is equal to the p.d.f.):

$E[XY]=\int F_X^{-1}(z)F_Y^{-1}(z)dz$ [2]

This shows the symmetric role of $X$ and $Y$ in this particular setting where [0] holds as an equality. This means that [0] holds for the particular $h$ of this problem as an equality and that the method of the OP must give the right result. (NB: expression [2] is purely theoretical and I will not use it in this post)

I think at the moment therefore that the OP merely forgot a factor of 2 in his calculations. In the specific problem the left member of [0] becomes:

$\int f_X(x) xh(x)dx = 1/2\times2 \times\int_{-1}^0 x(\sqrt{1-x}-1)dx$

and the right member:

$\int f_Y(y) h^{-1}(y)ydy=2\int_{-1}^0 y(y^2+2y)(y+1) dy$

and both expressions equal $E[XY]=7/30$, as Wolfram alpha can easily verify:

https://www.wolframalpha.com/input/?i=int_%7B-1%7D%5E0+x+%28%5Csqrt%281%2Bx%29-1%29

and

https://www.wolframalpha.com/input/?i=2*int_%7B-1%7D%5E0+y%28y%5E2%2B2y%29%28y%2B1%29

Note that the second integral is almost equal to the one used by the OP but differs from that by a factor of 2 in $y^2+2y$.

I have to say that similar problematics affect the calculation of the variance of $Y$ from the OP calculations... In that case I think the situation is not symmetric and an identity like [0] does not hold for the variance of $Y$...