Difference between $E[X^2]$ and $E[X^3]$ Hope to ask a dumb question. 
$Y = aX$,with $a \in N_+$. Here, we know the correlation coefficient is 1. 
Now, suppose $X \sim N(0,1)$.   


*

*Here, we know $X, Y$ are not independent.  

*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.  


*

*Here, we know $X,Y$ are not independent. 

*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? 
 A: 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.
A: $Var(X)=E[X^2]$ 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 $0$. 
A: 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.
A: 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.
