Question on the properties of of the Statistical Variance

Question

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?

Answer 1

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*}

Answer 2

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.