$\mathbb{E}[X^2-Y^2]$ and $\sigma(2X+3Y)$ From Given Expected Values

Question

For a homework problem, we are given $\mathbb{E}[X]=2$, $\mathbb{E}[Y]=6$, $\sigma^2(X)=9$, and $\sigma^2(Y)=16$. I don't quite understand expected values, co/variance, and deviations.

As per my textbook, $\sigma^2(Y)= \mathbb{E}[Y^2]-\mathbb{E}[Y]^2$, but I don't know how to calculate $\mathbb{E}[Y^2]$. I know $\mathbb{E}[Y]^2$ is just $6^2=36$, but according to my textbook $\mathbb{E}[Y^2]$ is NOT equivalent to $\mathbb{E}[Y] \times \mathbb{E}[Y]$. It is only the product of expectations when there are 2 independent random variables, and here there is only 1.

Also, even if I was able to calculate $\mathbb{E}[Y^2]$ (and X^2), it doesn't explain what to do when calculating $\mathbb{E}[X^2-Y^2]$. We aren't given the breakdown of the different independent variables, only what their expected values are. Additionally, for $\sigma$, it says it's the square root of the variance, $\sigma^2$. So I assume to compute $\sigma(2X+3Y)$, we need to first compute $\sigma^2(2X+3Y)$. But, I only understand how to compute the variance when there's 1 independent variable, not 2.

I'm sure this can't be that difficult, and I'm probably severely overcomplicating this. If someone can please help me out I'd greatly appreciate it. Thank you!

Answer 1

Hint: It is straightforward! $$\mathbb{E}(Y^2) = \sigma^2[Y] + \mathbb{E}(Y)^2 = 16 + 6^2 = 52$$

Also, expected value is linear. So, $\mathbb{E}(X+Y) = \mathbb{E}(X) + \mathbb{E}(Y)$.

In Addition: $$\sigma^2(\sum_{i=1}^nX_i) = \sum_{i=1}^n\sigma^2(X_i) + \sum_{i\neq j}Cov(X_i,X_j)$$ As we know $X_1$ and $X_2$ are independent, so $Cov(X_1,X_2) = 0$. Hence, $$\sigma^2(X_1 + X_2) = \sigma^2(X_1) + \sigma^2(X_2)$$ Moreover, we know $\sigma^2(aX) = a^2\sigma^2(X)$. So,

$$\sigma^2(2X + 3Y) = 4\sigma^2(X) + 9 \sigma^2(Y) = 4\times 9 + 9 \times 16 = 180$$

Answer 2

First, you have $$ E(X^2-Y^2)=E(X^2)-E(Y^2)=\Big(\sigma^2(X)+[E(X)]^2\Big)-\Big(\sigma^2(Y)+[E(Y)]^2\Big). $$ Next, because you say "there's 1 independent variable, not 2," I'll assume $X$ and $Y$ are independent, else there wouldn't be enough info to compute $\sigma(3X+3Y)$. Under the assumption, however, you have: $$ \sigma^2(2X+3Y)=\sigma^2(2X)+\sigma^2(3Y)=4\sigma^2(X)+9\sigma^2(Y). $$ The first equality above is where the independence between $X$ and $Y$ is used: independence implies zero covariance (the reverse isn't true however but this problem doesn't concern that).