Expected value for 2 dice roll

Question

Imagine that we roll two fair six-sided dice (i.e., all six sides have equal probability). Let X1 and X2 be the random variables representing these outcomes. Now, imagine we take one of the dice rolls, say X1, and add a (possibly negative) constant c to the result. If this becomes less than zero, then we set it to zero; denote this by

(X + c)₊ = max(X + c, 0)

What is the expected value of (X₁ - 2)₊ * (X₂ + 1)₊?

My answer
E{(X₁ - 2)₊ * (X₂ + 1)₊} = $\frac{1+2+3+4}{6}$ * $\frac{2+3+4+5+6+7}{6}$ = 7.515

I don't know if my answer is right. For X₂, Am I suppose to divide by 6 or 36? Can anyone correct me if I'm wrong?

Answer 1

Assuming the dice are independent, we have $X_1$ and $X_2$ are independent.

$$E[\lfloor X_2+1\rfloor_+]=E[X_2+1]=E[X_2]+1=4.5$$

Your computation of $E[\lfloor X_1-2 \rfloor_+]$ is correct as well.

However, the answer is $\frac{10}{6}\cdot \frac92=7.5$

Answer 2

$X_1$ and $X_2$ are independent. Then so are $g(X_1)$ and $f(X_2)$ where $f,g$ are suitable functions.

Then $$\mathbb Ef(X_1)g(X_2)=\mathbb Ef(X_1)\mathbb Eg(X_2)$$

In your case $f$ is prescribed by $x\mapsto(x-2)_+$ and $g$ by $x\mapsto(x+1)_+$.

Your calculations of $\mathbb Ef(X_1)$ and $\mathbb Eg(X_2)$ are okay.

What I miss in your effort is an explanation for: $$\mathbb E[(X_1-2)_+\times(X_2+1)]=\text{first factor}\times\text{second factor}$$It should be mentioned that it rests on independence.