# What does E(X + Y) mean?

I am working on some problems in probability theory and keep getting stuck on some of the concepts regrading expected values. I understand that if you have one dice roll you would have a distribution of $$X$$ with values taking ie. $$P(X=1) = {1 \over 6}$$, but how do you interpret this when you are considering two die rolls? Understanding that these are independent you would essentially get the same distribution.

Question: What is the distribution of $$(X + Y)$$ and how would you calculate $$E(X + Y)$$ if $$X$$ represents the first roll and $$Y$$ represents the second roll?

Is this just the same as $$E(X) + E(Y)$$?

Part 2: If you are trying to find $$E(2X - 2)$$ do you subtract two from each different value? For example $$[2 * 1 * {1 \over 6} - 2] + [2 * 2 * {1 \over 6} -2]$$...

• Expectation is linear, so $E[X+Y]=E[X]+E[Y]$ and $E[\alpha X]= \alpha E[X]$ Sep 25, 2018 at 16:01

No matter whether X and Y are dependent or independent, $$E(X+Y)=E(X)+E(Y)$$

Coming to $$E(2X-2)$$, it can be easily written as $$E(2X) - E(2)$$ which in turn equals to $$2E(X)-2$$.

• That answers my question perfectly, thank you so much! Sep 25, 2018 at 16:02
• No. You cannot add the distributions. Considering $Z=X+Y$, $P(Z=z)$ can be written as $\sum_{x+y=z} P_{X,Y}(x,y) =\sum_{x+y=z} P_{X,Y}(x,z-x) = \sum_{x+y=z} P_{X}(x)P_{Y}(z-x)$. Sep 25, 2018 at 16:26
• For example, $P(X+Y=3)=P(Z=3)=P_{X}(1)P_{Y}(2)+P_{X}(2)P_{Y}(1) = 2/36$ Sep 25, 2018 at 16:30
$$E[X+Y] = E[X] + E[Y]\\ E[aX] = aE[X]\\ E[2X + Y + 2] = 2E[X] + E[Y] + 2$$