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I have a random set $\{a,b,c\}$ and a second set $\{e,d\}$ I draw one number first number and one from the second

Letting $X_1$ denote the first number and $X_2$ the second number find, $E(X_1)$ and $E(X_2)$ and $E(X_1+X_2)$

Please please help, it might be obvious but I just can't work it out!

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By uniform distribution do you mean that probability to pick a number from a set is equal to every number? i.e. $P(a)=P(b)=P(c)=1/3$? –  Tomas Dec 11 '12 at 20:47
    
Tomas - yes, I think he is referring to the Discrete uniform distribution (as opposed to the continuous one, that some of us are more used to). –  Conan Wong Dec 11 '12 at 20:48
    

2 Answers 2

$E[X_1]$ and $E[X_2]$ are simply the averages of all the elements in each set, respectively. $E[X_1+X_2]=E[X_1]+E[X_2]$.

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As Tomas points out, $P(a) = P(b) = P(c) = \frac{1}{3}$. Thus,

$$E(X_1) = aP(a) + bP(b) + cP(c)$$

$$=\frac{a+b+c}{3}$$

You can apply the same method to work out $E(X_2)$.

As Vincent points out, the linearity of Expected Value over these uniform distributions gives

$$E(X_1+X_2) = E(X_1) + E(X_2)$$

[Alternatively, you could go through all the possibilities of $X_1+X_2$, i.e. the set {$a+d, a+e, b+d,b+e,c+d,c+e$}. There are six elements in this set and the probability of getting each is $\frac{1}{6}$, because the distribution is uniform. And you could compute $E(X_1+X_2)$ in a similar fashion as above.]

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this is what I first thought but I figured it couldn't be that simple, thanks a lot :) –  Justin Dec 11 '12 at 20:59
    
You're welcome, Justin :-) –  Conan Wong Dec 11 '12 at 21:02

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