Find $Y=X^2$ from given $f(X)$ and find the $\rho(x,y)$ correlation coefficient

Question

I have a question.

\begin{equation} f_X(x) = \begin{cases} 1-x, &{0 \leq x \leq 1}\\ 1+x, &{-1 \leq x < 0}\\ 0 , &\text{otherwise}\\ \end{cases} \end{equation}

This is the given pdf and $Y=X^2$ given also. I need to find $\rho(x,y)$ (correlation coefficient) and also check if $X$ and $Y$ are independent or not. I can do these parts but I don't know how to find $f_Y(y)$ from given pdf.

Thanks.

Answer 1

I can do these parts but I don't know how to find $f_Y(y)$ from given pdf.

You can derive $f_Y(y)$ by a simple transformation:

$$f_Y(y)=\left[\frac{1}{\sqrt{y}}-1 \right]\cdot\mathbb{1}_{(0;1]}(y)$$

...but this is not needed to solve your problem.

Obviously $(X;X^2)$ are not independent because one variable is expressed in function of the other one.

To calculate their correlation coefficient just use the definition

• First calculate covariance

$$\mathbb{E}[XY]-\mathbb{E}[X]\cdot\mathbb{E}[Y]$$

• then, in necessary, by definition

$$\rho_{XY}=\frac{\mathbb{Cov}[XY]}{\sigma_X\cdot \sigma_Y}$$

any needed quantity can be calculated starting from $f_X(x)$ without deriving expressly $f_Y(y)$

This because any quantity can be expressed in

$$\mathbb{E}[X^n]=\int x^n f_X(x)dx$$

---

After easy calculations you will see that $\mathbb{Cov}[X;X^2]=0$ thus you have to show that, anyway, they are not independent.

To show that these rv's are not independent consider that an alternative definition of independence is, $\forall x,y$

$$f(y|x)=f(y)$$

now consider that, $X \in[-1;1]$ and $Y \in(0;1]$ but when, for example, $|X|<0.5$ also $(Y|X)$ support changes thus, i.e.

$$\mathbb{P}[Y>0.5||X|<0.5]=0$$

which is different from

$$\mathbb{P}[Y>0.5]$$

that is positive. This is a clear contradiction w.r.t. definition of independence

In other words, the occurrence of $Y$ strictly depends on what happens in $X$