MHB Confused by this probability question

[das1]

I came across this problem and I'm wondering if anyone can tell me what it means/how to do it:
"Given the formula below to model, what is the expected value of rolling two dice simultaneously?
$$E(X)= \sum_{i=1}^{\infty}x_i f(x_i)."$$

I've never seen this notation before; how does it work?

 

[Fermat1]

I came across this problem and I'm wondering if anyone can tell me what it means/how to do it:
"Given the formula below to model, what is the expected value of rolling two dice simultaneously?
E[X] = *the sum from 1 to infinity* of x_i f (x_i)"

I've never seen this notation before; how does it work?

You are dealing with a discrete random variable (the total sum on two dice is an integer) so the expectation won't be an integral, it will be a sum. Let $X$ be the sum of the two dice. $f_{X}(x)$ is a common notation for the density fuction. For discrete Random variables, as in this case we have $f_{X}(x)=P(X=x)$. $X$ can take values 1 to 12. The expectation is $1P(X=1)+2P(X=2)+..+12P(X=12)$

 

[das1]

Fermat, you're a legend, thank you.
One thing, shouldn't it be 2 to 12 because that's the lowest you can get when rolling 2 dice simultaneously?

 

[Jameson]

Fermat, you're a legend, thank you.
One thing, shouldn't it be 2 to 12 because that's the lowest you can get when rolling 2 dice simultaneously?

Good question! Normally you write all non-zero terms in the sum but you can actually use this formula for all discrete numbers. Why? Let's look at when the sum equals $-2$ for example. This part of the expected value sum would be $-2 \cdot P[X=-2]$. The probability the sum equals a negative number though is obviously 0, so this term becomes $-2 \cdot 0=0$. All other terms outside of the range of possible sums will also drop to zero, thus we usually only include the non-zero terms when writing out the work.

So yes, it would make more sense maybe to write $2P(X=2)+3P[X=3]..+12P(X=12)$ but $1P(X=1)+2P(X=2)+..+12P(X=12)$ isn't incorrect because $1P[X=1]=0$ :)