I'm getting into the management of multiple independent random variables in determining expectation and variance, but cannot see where averaging of a linear combination of independent random variables applies, and what it specifically means.

For a basic linear combination of variables such as:

$$Y = X_{1}+X_{2}+...+X_{n}$$

...I can see getting $E(Y)$ and $Var(Y)$ for situations like the alloy composition in a given metal (e.g., zinc, tin, etc., mixed in iron ore, where each may have different probability density functions). It makes sense that $E(Y)$ will increase with additional independent variables, and that the same will happen to the variance:

$$E(Y) = E(X_{1})+E(X_{2})+...+E(X_{n})$$ $$Var(Y) = Var(X_{1})+Var(X_{2})+...+Var(X_{n})$$

However, I've run into the discussion of averaging these independent random variables, and am not seeing an application for them so its difficult to see what is going on.

In essence, they state the new random variable will be the following:

$$\bar Y = \frac{X_{1}+X_{2}+...+X_{n}}{n}$$

From here, $E(\bar Y)$ will be calculated to be the expectation of the average. I see how this may apply to independent events if $X_{1}$, $X_{2}$, etc., are all the same type (e.g., all are dice throws or all are coin tosses), but given that $Y = X_{1}+X_{2}+...+X_{n}$ works for situations like independent alloys in a metal, I cannot see what it would mean to take $\bar Y$ of different expectations for the various alloys. By looking at just two alloys (say Tin and Zinc) we would be averaging the expectation of each in the resulting metal, which would give us what?

It can't be the expectation of both Tin and Zinc, since this is established by calculations on $Y = X_{tin}+X_{zinc}$, which would be $E(Y) = E(X_{tin})+E(X_{zinc})$, etc.

The same question applies to a combination of a single dice throw and a single coin toss, where again the expectation and variance for the combination appears to be the sum of the separate expectations and variances (and by combining, these increase as expected), but what happens conceptually if you take the average of these two in a similar way as the Tin and Zinc above? What is the meaning of this calculation?


closed as unclear what you're asking by whuber Sep 21 '16 at 20:47

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    $\begingroup$ Can you formulate the simplest possible example that still captures the essence of your difficulty? Preferably with dice, because at the moment it's not clear whether the source of your difficulty is in the alloy domain or is it of statistical nature. $\endgroup$ – Aksakal May 5 '15 at 2:28

You have no issue when n=1 I take it? Otherwise, the averaging is just an application of the fact that expectation is a linear operator. In your domain of metals you might think of it simply as reducing the amount of inputs into your process by 1/n


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