I have this random variable $X$, that has its values in $\{0,1,2,...,2n\}$.
It is known that $P(X = n-i) = P(X = n+i)$ for any $i$, $i \in [1,n]$.
How should I calculate the expected value of the random variable?
The correct result is $n$.
Expected Value of Random Variable $\{0,1,...,2n\}$
0
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probability
statistics
random-variables
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0I see no reason why it should be $n$ just based on what you have there. if, for example, the distribution is uniform, then the expectation would be $2n-1$ – 2017-01-12
1 Answers
4
It becomes more clear if you instead consider the expected value of $Y = X-n$. You then have $P(Y = -i) = P(Y = i)$. The contributions to the expected value from $\pm i$ will cancel out exactly, leaving $E(Y) = 0$. And thus $E(X) = n$.
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0I am not really sure about the part E(X) = n. Why is that? – 2017-01-12