# Given the sample that not Independently and Identically Distributed

So I'm working on a question that wants me to consider the sample mean $\mu_1$ (estimator of $\mu$) when $n=3$ to be $μ_1=X_1+X_2+X_3$. Now first I assume that they are independently and identically distributed. I have to find whether it is unbiased and then the variance.

I did $E[\mu_1]=\dfrac{1}{3}E[X_1+X_2+X_3]=\dfrac{1}{3}(E[X_1]+E[X_2]+E[X_3])=\dfrac{1}{3}(3μ)=μ$ So there is no bias right?

For variance $V[μ_1]=\dfrac{1}{9}(V[X_1]+V[X_2]+V[X_3])=\dfrac{1}{9}(3σ^2)=\dfrac{1}{3}σ^2$

After that, I am told to assume that the sample is not i.i.d and I have to derive the bias and variance of the sample mean where the observations are independent but $V(X_3)=2V(X_1)=2V(X_2)$ So does that mean that $E[μ_1]$ is that same as above since the observations are independent. Im going off the thought that if $X$ and $Y$ are independent than $E[XY]=E[X]E[Y]$ and applying that to this, thinking if $X_1,X_2,X_3$ are independent observations than $E[μ_1]$ will just equal $μ$.

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You are right that $E(\mu_1)=\mu$ but this is not due to independence. This is due $E(X_1+X_2+X_3)=E(X_1)+E(X_2)+E(X_3)$. Again if $X$ and $Y$ are independent then \begin{align} E(XY)&=E(X)E(Y) \\ \implies Cov(X,Y)&=0 \end{align} So your variance will be $V(\mu_1)=\dfrac{1}{9}V[X_1+X_2+X_3]=\dfrac{1}{9} \Big(V[X_1]+V[X_2]+V[X_3] \Big)$, this is due independence.