Conditional variance of discrete random variables Given  and  are independent discrete random variables with
$$\mathbb{E}[]=0, \mathbb{E}[]=1, \mathbb{E}[^2]=8, \mathbb{E}[^2]=10$$
and
$$Var()=Var()=8$$
Let $=$ and $=+$.
To find $\mathbb{E}[AB]$, then
$$\mathbb{E}[AB]=\mathbb{E}[XY(X+Y)]=\mathbb{E}[X^2Y + XY^2]=\mathbb{E}[X^2Y] + \mathbb{E}[XY^2]=\mathbb{E}[X^2]\mathbb{E}[Y]+\mathbb{E}[X]\mathbb{E}[Y^2]=8$$
But I get a different value using the following approach
$$\mathbb{E}[AB]=\mathbb{E}[A]\mathbb{E}[B]=\mathbb{E}[XY]\mathbb{E}[X+Y]=\mathbb{E}[X]\mathbb{E}[Y]*(\mathbb{E}[X]+\mathbb{E}[Y])=0$$
Out of curiosity, why is this the case?
I want to believe that my first approach is correct, thus $\mathbb{E}[AB]=8$.
Anyways, I will proceed with my actual question.
To find $(A)$, given $()=[()^2]−([])^2$, then
$$\mathsf{Var}(A)=\mathbb{E}[(XY)^2]-(\mathbb{E}[XY])^2=\mathbb{E}[X^2Y^2]-(\mathbb{E}[X]\mathbb{E}[Y])^2=\mathbb{E}[X^2]\mathbb{E}[Y^2]=8*10=80$$
So far so good, however, I am having troubles finding the conditional probability for $(|Y=1)$.
I know that the conditional variance of a random variable is determined with
$$\mathsf{Var}(X|Y)=\mathbb{E}[(X-\mathbb{E}[X|Y])^2|Y]$$
By substituting in the respective parameters, then
$$\mathsf{Var}(XY|Y=1)=\mathbb{E}[(XY-\mathbb{E}[XY|Y=1])^2|Y=1]$$
And now what? There is a bunch of nested conditional expectations.
Good thing is, there is a formula for conditional expectations:
$$µ_{X | Y =y} = \mathbb{E}(X | Y = y) = \sum xf_{X | Y} (x | y).$$
Sad thing is, I don't know what to do with it. Am I overcomplicating things?
What I do know is that $\mathbb{E}(X | Y = y)$ is the mean value of $X$, when $Y$ is fixed at $y$. I already found out the value for $(A)$ which I don't know if it's useful to find the conditional one or not. Also, $\mathbb{E}[XY]=0$.

*

*From here onwards, how do I calculate the conditional variance?

*And is there an easier perhaps more straightforward way to evaluate it?

Hopefully someone can help me figure this out. Thanks!
 A: When $A$ and $B$ are not independent, $E[AB]\overset{?}{=}E[A]E[B]$ does not necessarily hold.

You wrote down the definition of conditional variance incorrectly.
The expression $E[(Y - E[Y \mid X])^2 \mid X]$ is the definition of $\text{Var}(Y \mid X)$, not $\text{Var}(X \mid Y)$.
So, $\text{Var}(XY \mid Y=1) = E[(XY - E[XY \mid Y=1])^2 \mid Y=1]$. If you do the computation you'll end up with $\text{Var}(X)$ which makes sense: since $X$ and $Y$ are independent, $XY$ simply becomes $X$ when conditioning on $Y=1$.
A: This equation $$\mathsf{Var}(A)=\mathbb{E}[XY]^2-(\mathbb{E}[XY])^2=\mathbb{E}[X^2Y^2]-(\mathbb{E}[X]\mathbb{E}[Y])^2=\mathbb{E}[X^2]\mathbb{E}[Y^2]=8*10=80$$
contains typographical errors.  The corrected equation, with corrections in red, should look like this:
$$\mathsf{Var}(A)=\mathbb{E}[\color{red}{(}XY\color{red}{)^2}]-(\mathbb{E}[XY])^2=\mathbb{E}[X^2Y^2]-(\mathbb{E}[X]\mathbb{E}[Y])^2=\mathbb{E}[X^2]\mathbb{E}[Y^2]=8*10=80.$$
The reason is akin to the reason why $f(x)^2$ is taken to mean $(f(x))^2$, rather than $f(x^2)$.
The conditional variance $$\mathsf{Var}(A \mid Y = 1)$$ is straightforward:  Given that $Y = 1$, then $A = XY = X$, so $$\mathsf{Var}(A \mid Y = 1) = \mathsf{Var}(X \mid Y = 1) = \mathsf{Var}(X),$$ where the last equality that states that the conditional variance of $X$ given $Y = 1$ is equal to the unconditional variance of $X$, holds because $X$ and $Y$ are independent.
