Given π and π are independent discrete random variables with
$$\mathbb{E}[π]=0, \mathbb{E}[π]=1, \mathbb{E}[π^2]=8, \mathbb{E}[π^2]=10$$
and
$$Var(π)=Var(π)=8$$
Let $π΄=ππ$ and $π΅=π+π$.
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 $π΅πΊπ(A)$, given $π΅πΊπ(π)=πΌ[(π)^2]β(πΌ[π])^2$, 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 $π΅πΊπ(π΄|Y=1)$.
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:
$$Β΅_{X | Y =y} = \mathbb{E}(X | Y = y) = \sum xf_{X | Y} (x | y).$$
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 $π΅πΊπ(A)$ 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!
Best Answer
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.