Is this true that $E(XY) =E(X) E(Y)$ implies the statistical independence of two normal random variables? Suppose $X$ and $Y$ are two normal variables and $$E(XY) = E(X) E(Y),$$ then is it true that $X$ and $Y$ are independent?
I know that when $(X,Y)$ obeys bivariate normal distribution, the statement is true. But in general, I am not sure. If this statement is wrong, then is there some easy ways to prove the independence of two normal variables?
 A: No. 
Suppose $X$ is normally distributed $N(0,1)$, and $Z$ is uniformly distributed on $\{-1,+1\}$ and independent of $X$.
Set $Y=XZ$. Then $X$ and $Y$ are both normal distributed, so $E(X)E(Y)=0\cdot 0 = 0$, but $E(XY)=E(X^2Z)$ is clearly also $0$. And $X$ and $Y$ are far from independent.
A: ok, You want normal distributions.  Suppose $Z\sim N(0,1)$ and
$$
X = \begin{cases} \phantom{-}Z & \text{if } -c<Z<c, \\  -Z & \text{if } Z>c \text{ or } Z<-c. \end{cases}
$$
If $c$ is very large then $\operatorname{cor}(Z,X)<0$ and if $c$ is very small (but positive) then $\operatorname{cor}(Z,X)<0$.  Somewhere in between you get uncorrelated variables, but they are not independent.  Showing that $X$ is normally distributed I leave as an excercise.
Here's one of the simplest examples:
$$
X = \begin{cases} \phantom{-}1 \\ \phantom{-}0 & \text{each with probability } \dfrac 1 3, \\ -1 \end{cases} \quad \text{and} \quad Y =X^2.
$$
Then $\operatorname{E}(X) = 0$ and $\operatorname{E}(Y) = \dfrac 2 3$ and $\operatorname{E}(XY) = 0.$
