Question on the properties of of the Statistical Variance So, I have a question on which I am not entirely sure on the answers and I would like to discuss it here. Given two random variables $X$ and $Y$, and two real numbers $a$ and $b$, then which of the following hold true?
1) Variance of $X$ is always non-negative
2) The Standard Deviation of $X$ is always non-negative
3) If $V(X) = V(Y)$, then $V(X+a)=V(Y+b)$
4) If $V(aX) = V(bX)$ for $a \neq 0$ and $b \neq 0$, then $a = b$
5) If $E[X] = E[Y]$ and $V(X) = V(Y)$, then $X = Y$
6) If $E[X] = E[Y]$ and $V(X) = V(Y)$, then $E[X^2] = E[Y^2]$
I know that 1) and 2) must be True for all X, because the formula of both Standard Deviation and Variance (also, in a logical sense: you cant measure the distance of a data point from the mean in negative values).
3) is also correct given that a translation of a Random Variable does not affect how far apart each data point lie from the mean.
Number 4) is where I'm a little bit less sure:
Given that
$$
V(zX) = z^2V(X)
$$
Then...
$$
V(aX) = V(bX) \\
a^2V(X) = b^2V(X) \\
a^2 = b^2 \\
\pm \sqrt{a^2} =\pm \sqrt{b^2}
$$
And this is where I'm a bit lost given the possible cases that can come up.
Number 5) seems logically enough for me to think it's true and number 6) unfortunately I am not sure.
Can anybody help?
 A: You have the right answer for $\#4.$
$\#5$ is false even when $X$ and $Y$ both have the same distribution. Suppose $X,Y$ are independent and both have the same continuous distribution, for example $X,Y \sim \operatorname{i.i.d. N}(0,1).$ In that case, $\Pr(X=Y) =0,$ and that's as far as you can get from $X=Y.$
But there's another question: If $\operatorname E(X) = \operatorname E(Y)$ and $\operatorname{var}(X) = \operatorname{var}(Y),$ then do $X$ and $Y$ both have the same distribution? The answer there is easily seen to be "no". For example, the $\operatorname{Poisson}(1)$ distribution has expectation $1$ and variance $1,$ and so does the $N(1,1)$ distribution, but one is continuous and the other is discrete, and they're nowhere near the same.
For $\#6$, recall that $\operatorname E(X^2) = \Big(\operatorname E(X)\Big)^2 + \operatorname{var}(X),$ so the answer here is affirmative.
A: For Q4, you've reduced it to knowing that $a^2 = b^2$, since this does not imply that $a=b$, the statement is clearly false
For Q5, consider the two distributions: 
\begin{align*}
X  &\sim N(\mu=50, \sigma^2=25)\\
Y &\sim Bin(n=100, p=0.5)
\end{align*}
Then $E(X) = 50 = E(Y) = np $ and $V(X) = 25 = V(Y) = np(1-p)$
For Q6:
\begin{align*}
V(X) = V(Y) &\implies E(X^2) - E(X)^2 = E(Y^2) - E(Y)^2
\end{align*}
and since we know that $E(X) = E(Y)$, we have that $E(X)^2 = E(Y)^2$ and so 
\begin{align*}
V(X) = V(Y) &\implies E(X^2) - E(X)^2 = E(Y^2) - E(Y)^2 \\
& \implies E(X^2) = E(Y^2)
\end{align*}
