Given that I know:
1) $E(n)=2$ (or expected value of $n$ is $2$)
2) $n$ must be a positive integer ($n=1,2,3,\ldots$)
I am just wondering is there any method to evaluate $E(2^{n-1})$ (the expected value of $2^{n-1}$)
Frankly, I have no clue how this can be solved. But I think it involves some kind of distribution function (since exponential variable is involved here)
Thanks in advance.
$E(2^{n-1}) = \frac{1}{2} E(2^n) = \frac{1}{2} \sum_{k=1}^{\infty} 2^k P(k)$.
If $P(2) = 1$, then $E(n) =2, E(2^{n-1}) = 2$.
If $P(1) = 1/4, P(2) = 1/2, P(3) = 1/4$, then $E(n) = 2, E(2^{n-1}) = 5/4$, so it depends on the distribution function.