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I have this question:

What's the expected value of a random variable $X$ if $P(X=1)=1/3$, $P(X=2)=1/3$, and $P(X=6)=1/3$?

I am very confused as to how I can work this problem out. I was thinking it would be something like:

$$E[X] = P(X=1) \cdot (1/3) + P(X=2) \cdot 1/3 + P(X=6) \cdot 1/3.$$

I am not sure this is correct because then I do not have values for $P(X=1)$, $P(X=2)$, and $P(X=6)$. Should I just do the calculation like this:

$$E[x] = (1/3)+(1/3)+(1/3)$$

I am not sure exactly how Expected value for random variables should be calculated. Should $E[x]$ always add up to $1$?

Thank you.

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  • $\begingroup$ Your first expression is not right. You want $1\cdot\Pr(X=1)+2\cdot\Pr(X=2)+6\cdot\Pr(X=6)$. $\endgroup$ – André Nicolas Mar 3 '14 at 21:51
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The expected value is simply

$$1\cdot 1/3 + 2\cdot 1/3 + 6\cdot 1/3 = 3 $$

because it is defined by

$$E[X] = \sum_i P(X_i)X_i$$

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  • $\begingroup$ Can you expand more on why you did this? $\endgroup$ – userunknown Mar 3 '14 at 21:49
  • $\begingroup$ I added some more information to my answer. $\endgroup$ – Matt L. Mar 3 '14 at 21:52
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Matt's answer is correct. The expected value is by definition what you expect to get. In investments in gambling you're expected value would be want you expect to run home with after many trials. Most likely the expected value is then negative. That's the concept behind it.

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