# Estimation based on observing sum of two variables

Let $X_1,\dots,X_n$ are i.i.d normal $N(\mu,\sigma^2).$ Suppose that we only observe $$X_1+X_2,\dots,X_1+X_n,\dots,X_{n-1}+X_n,$$ i.e, $X_i+X_j$ for all $i<j.$

I wish to find the best estimator $\mu$ and $\sigma^2$ based on the above abservation. How will I do that?

As we know that if we observe $X$'s then it is easy to estimate $\mu$ and $\sigma^2$. However, since we observe ${n\choose 2}-$identically variables $X_i+X_j$ that are not independent one to the each other, then it would become more difficult. Any suggestion?

• Do you have in mind any parctical situation leading to data of that kind? – kjetil b halvorsen Oct 13 '14 at 8:00
• @kjetilbhalvorsen: no, I don't. It is actually a special case of my research – Jlamprong Oct 13 '14 at 8:12
• This seems to be a (very) special case of your previous question. If this is the only case in which you are interested, should we close the previous question? – whuber Oct 13 '14 at 17:02
• Note that one very straightforward way to obtain a good estimator of $\mu$ is to use an obvious one--such as $X_1+X_2$--and apply the Rao-Blackwell process. Whether that is "best" depends on unstated assumptions such as your loss function and any restrictions you might want to place on the estimator, such as being linear or unbiased. – whuber Oct 13 '14 at 17:07
• @whuber: Yes, indeed. It is a special case of my previous question. I asked it because I wanna get an illustration when $f(x,y)=x+y$ – Jlamprong Oct 13 '14 at 17:35

This is a usual linear model. You have $$\mathbf{Y}=(X_1+X_2,\ldots)'=\mathbf{A}(\mathbf{1}\mu) +\mathbf{\epsilon},$$ where $\mathbf{\epsilon} = (\epsilon_1,\ldots,\epsilon_n)=(X_i-\mu)_{i=1,\ldots ,n}$ is the error term. $\mathbf{A}$ is a $n \times {n \choose 2}$-matrix with twice the entry 1 in each row.
The rest is done by linear model estimation. You may try $\mu=(A'A)^{-}A'Y$, where $^-$ denotes a generalized inverse. Check if $(A'A)^-(A'A)=1$, then this is a BLUE. See Rao (1962).
Another approach for your special case: Clearly $E(Y_k)=2\mu$ and $Var(Y_k)=2\sigma^2$.
• And, I guess, $A$ has full rank, so you can recover the 's exactly. But even if that guess is wrong, you can do as above. – kjetil b halvorsen Oct 13 '14 at 9:14
• @Horst: Thanks for your help. Can we still recover $X$ if we observe $X_iX_j$ or $X_i^2+X_j$? – Jlamprong Oct 13 '14 at 9:22
• @Jlamprong: So you have a mapping of the vector $X$ to a long vector with the entries $X_i X_j$ and $X_i^2 + X_j$ and ask if this mapping is injective? – Horst Grünbusch Oct 13 '14 at 10:04
• @HorstGrünbusch: Isn't like $X_i+X_j?$ – Jlamprong Oct 13 '14 at 11:30