Let $X_1, X_2,\dots, X_n$ be a sample of size $n$ from a distribution with unknown mean $−\infty<\mu<\infty$, and unknown variance $\sigma^2 > 0$.

Show that the $Y = (X_1 + 2X_2 + 3X_3 +\cdots+ nX_n) / (1 + 2 + 3+\cdots+n )$ is an unbiased estimator of $\mu$.


Expectation is linear: $\mathbb E(U+V)=\mathbb EU+\mathbb EV$ and $\mathbb EaU=a\mathbb EU$.

Application of that here leads to: $\mathbb EY=\mu$.

  • $\begingroup$ Hmm... but how exactly do you do it here? $\endgroup$
    – M31
    Dec 1 '15 at 10:11
  • $\begingroup$ $Y=a^{-1}Z$ with $a=1+\cdots+n$ and $Z=X_1+\cdots+nX_n$, so $\mathbb EY=a^{-1}\mathbb EZ$. Next: $\mathbb EZ=\mathbb EX_1+\cdots+n\mathbb EX_n=\mu.a$ $\endgroup$
    – drhab
    Dec 1 '15 at 10:17

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