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We have $N$ realisations of five individual, IID random variables $X_1$, $X_2$, $X_3$, $X_4$ and $X_5$. We define another random variable $S = X_1+X_2+X_3+X_4+X_5$. Now, for a given $S$ generated from the same process, the individual components of which are unknown, how will we find the best estimates of $X_1$, $X_2$, $X_3$, $X_4$ and $X_5$?

EDIT 1: Each of the RV's $X_1$, $X_2$ up to $X_5$ are independent, but not identically distributed.

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  • $\begingroup$ Are they normal? $\endgroup$ – yoki Nov 20 '14 at 13:26
  • $\begingroup$ @yoki they are not necessarily random. In fact, four of them are known to be skewed distributions, possibly Beta. $\endgroup$ – Ayyappadas Nov 20 '14 at 13:33
  • $\begingroup$ Your question states them to be iid. Do you mean instead that $X_1,...,X_5$ are not iid, but instead that if $V=(X_1,X_2,...,X_5)$ and we observe $V_1, ..., V_N$ that the $V_i$ are iid? $\endgroup$ – Glen_b Nov 20 '14 at 13:40
  • $\begingroup$ Can you define what you mean by 'best' in 'best estimates'? What's your criterion? $\endgroup$ – Glen_b Nov 20 '14 at 14:16
  • $\begingroup$ Well, I would put it this way- given the process data, i.e $N$ realisations of ${X_1,X_2,X_3,X_4,X_5}$ and $S=X_1+X_2+X_3+X_4+X_5$, if I happen to know that there is a realisation $S_i$, what will be the set of mean ${ X_{1i}, X_{2i}, X_{3i},X_{4i},X_{5i}}$ and the corresponding set of SD's. I hope that makes it clear. $\endgroup$ – Ayyappadas Nov 20 '14 at 15:09
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In the general case, you can use Bayes' theorem. The optimal (MMSE) estimator is given by $\hat{X} = E[X|S]$. Let $X=X_1$ (it doesn't matter which index as they are iid), so: $$ f_{X|S}(x,s) = \frac{f_{S|X}(s,x)f_X(x)}{f_S(s)}$$ $f_X(x)$ is given, $f_S(s)$ is also given by an immediate calculation, and $f_{S|X}$ can be calculated via $P(S<s|X=x)=P(x+X_2+...+X_5<s)=P(X_2+...+X_5<s-x)=F_{S'}(s-x),$ where $S'$ is the partial sum of only four variables.

Finally, calculate via the conditional expectation definition: $\hat{X}=\int x f_{X|S}(x,s)dx$.

If your distribution is stable then you can probably get easier equations due to the iid sums present.

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  • $\begingroup$ Thanks @yoki. But I made a mistake with the IID part. They are independent, but not identically distributed. $\endgroup$ – Ayyappadas Nov 20 '14 at 13:45
  • $\begingroup$ It does not matter for applying Bayes' theorem. You just have to keep track of the components. $\endgroup$ – Xi'an Nov 20 '14 at 16:38

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