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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 $[2 * 1 * {1 \over 6} - 2] + [2 * 2 * {1 \over 6} -2]$...

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

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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 $2E(X)-2$.

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  • $\begingroup$ That answers my question perfectly, thank you so much! $\endgroup$
    – Ethan
    Sep 25, 2018 at 16:02
  • $\begingroup$ glad to help you $\endgroup$ Sep 25, 2018 at 16:04
  • $\begingroup$ One further point of clarification, do the distributions follow this same rule? So distribution of (X + Y) is 1/6 + 1/6 for each outcome? $\endgroup$
    – Ethan
    Sep 25, 2018 at 16:14
  • $\begingroup$ 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) $. $\endgroup$ Sep 25, 2018 at 16:26
  • $\begingroup$ For example, $P(X+Y=3)=P(Z=3)=P_{X}(1)P_{Y}(2)+P_{X}(2)P_{Y}(1) = 2/36$ $\endgroup$ Sep 25, 2018 at 16:30
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The expected value operator is "linear"

$E[X+Y] = E[X] + E[Y]\\ E[aX] = aE[X]\\ E[2X + Y + 2] = 2E[X] + E[Y] + 2$

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