Hope to ask a dumb question.

$Y = aX$,with $a \in N_+$. Here, we know the correlation coefficient is 1.

Now, suppose $X \sim N(0,1)$.

  1. Here, we know $X, Y$ are not independent.
  2. Cov($X,Y$) = $E[XY] - E[X]E[Y] = aE[X^2] = a$ So, $X,Y$ are correlated.

Now, suppose $Y = aX^2$. Others remain the same.

  1. Here, we know $X,Y$ are not independent.
  2. Cov($X,Y$) = $E[X^3] = 0$

I am confused about both case, the upper one is correlated but the lower one is not.
How do I see it from an intuitive way?

  • $\begingroup$ Why is $Cov(X,Y)=1$? Shouldn't it be $a$? $\endgroup$ – user103828 Jan 28 '15 at 7:25
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    $\begingroup$ To summarize, IF some random variables are independent THEN their covariance is zero, but the inverse implication is false without some supplementary hypothesis. $\endgroup$ – Did Jan 28 '15 at 7:32
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    $\begingroup$ If you run through the algebra, I believe that $Cov(X, Y) = a Var(X)$. However, as $Var(Y) = a^2 Var(X)$ means that $\sigma_Y = a\sigma_X$ so the correlation coefficient $\large\rho = \frac{a Var(X)}{\sigma_x\cdot a\sigma_x} = 1$ which is what the OP meant, I believe. $\endgroup$ – Avraham Jan 28 '15 at 7:35

Correlation may have many meanings, but from the question, you are using the specific definition of the Pearson product-moment correlation coefficient. You are calling variables "correlated" when $\rho \neq 0$. That is solely when $\textrm{Cov}(X, Y) \neq 0$.

In the case of $Y = aX$, regardless of how $X$ is distributed, we can state the following: $$ \begin{align} Y &= aX\\ E(Y) &= aE(X)\\ E(XY) &= E(aX^2) = aE(X^2)\\ E(X)E(Y) &= E(X)E(aX) = aE(X)^2\\ \end{align} $$

So $$ \begin{align} \textrm{Cov}(X, Y) &= E(XY) - E(X)E(Y)\\ &= aE(X^2) - aE(X)^2\\ &= a\left(E(X^2) - E(X)^2\right)\\ &= a\textrm{Var}(X) \end{align} $$

Now $\textrm{Var}(Y) = \textrm{Var}(aX) = a^2\textrm{Var}(X)$ so $\sigma_Y = a\sigma_X$. This makes: $$ \begin{align} \rho_{X,Y} &= \frac{\textrm{Cov}(X, Y)}{\sigma_X\sigma_Y}\\ &=\frac{a\sigma^2_X}{\sigma_X a\sigma_X}\\ &= 1 \end{align} $$

So the variables are perfectly correlated. Now define $Y = aX^2$. Run through the same algebra: $$ \begin{align} Y &= aX^2\\ E(Y) &= aE(X^2)\\ E(XY) &= E(aX^3) = aE(X^3)\\ E(X)E(Y) &= E(X)E(aX^2) = aE(X)E(X^2)\\ \textrm{Cov}(X, Y) &= E(XY) - E(X)E(Y)\\ &= aE(X^3) - aE(X)E(X^2)\\ &= a\left[E(X^3) - E(X)E(X^2)\right] \end{align} $$ In general, I'm not sure we can say anything about the relationship between the two variables. However, we do happen to know the moments of the standard normal. In specific: $$ \begin{align} E(X) &= 0\\ E(X^2) &= 1 \textrm{ Since Var}(X) = 1 \textrm{ and } \mu = 0\\ E(X^3) &= 0 \textrm{ Since skewness of normal is } 0 \end{align} $$

So $$ \begin{align} \textrm{Cov}(X, Y) &= E(XY) - E(X)E(Y)\\ &= a\left[E(X^3) - E(X)E(X^2)\right]\\ &= a\left[0 - 0\times 1\right] = 0 \end{align} $$

As for intuition, perhaps the following. When $Y = aX$, graphing $Y$ against $X$ is a straight line; $X$ completely determines $Y$. However, when $Y = aX^2$, you have a parabola with the y-axis as the line of symmetry. $X$ no longer completely determines $Y$, as there is an $X$ of equal magnitude and opposite sign that can generate the same $Y$. The fact that it is equal and opposite may be the intuitive reason for the correlation to be 0, but the algebra is primary (if not only) reason.


$Var(X)=E[X^2]$ is the variance (measures spread around mean), which for a standard normal distribution is $1$.

$E[X^3]$ measures the skewness of the density and since a normal distribution is symmetric the skewness is $0$.

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    $\begingroup$ Why "$E[X^3]$ measures the symmetry of the density "? thx, $\endgroup$ – sleeve chen Jan 28 '15 at 7:35
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    $\begingroup$ To be more precise $E[(X^3-\mu)/\sigma]$ is a measure of the skewness (where $\mu$ is mean $\sigma$ is standard deviation) and when this is zero then the distribution is symmetric. Hence $E[X^2]$ and $E[X^3]$ measure different things. en.wikipedia.org/wiki/… $\endgroup$ – user103828 Jan 28 '15 at 7:48
  • $\begingroup$ $E(X^3)$ is the skewness of a standard normal, true, but here we are talking about the correlation between two random variables, $X$ and $Y = aX^2$. $\endgroup$ – Avraham Jan 28 '15 at 8:09
  • $\begingroup$ Agreed. I guess I was answering the question what is the difference between $E[X^2]$ and $E[X^3]$ rather than the relationship between $X$ and $Y$. $\endgroup$ – user103828 Jan 28 '15 at 8:14

I can't help you with the first case. What is your problem here?

But to the second one:

If two random variables are uncorrelated (i.e. covariance is zero), they are not necessarily independent. The example you have is the standard example to demonstrate this. Only if $X$ and $Y$ have a joint bivariate normal distribution, from $cov(X,Y)=0$ follows that they are independent.

For more see for example here in Wikipedia.


One approach to intuition is, since Y is always positive, the positiveness of X has no impact on the positiveness of Y. Therefore, X being more positive has as much impact on Y being larger than X being more negative. This is the antithesis of high correlation. I think that is all there is to it.


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