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Weighted average

The definition of the variance of a random variable can be used to show a number of additional results.

a. Show that Var(x) =E(x2) – [E(x)]2.
b. Use Markov’s inequality to show that if x can take on only non-negative values,

This result shows that there are limits on how often a random variable can be far from its expected value. If k=hσ this result also says that

Therefore, for example, the probability that a random variable can be more than two standard deviations from its expected value is always less than 0.25. The theoretical result is called Chebyshev’s inequality.

c. Equation 2.197 showed that if two (or more) random variables are independent, the variance of their sum is equal to the sum of their variances. Use this result to show that the sum of n independent random variables, each of which has expected value μ and variance σ2, has expected value nμ and variance nσ2. Show also that the average of these n random variables (which is also a random variable) will have expected value μ and variance σ2/n. This is sometimes called the law of large numbers — that is, the variance of an average shrinks down as more independent variables are included.

d. Use the result from part (c) to show that if xand x2 are independent random variables each with the same expected value and variance, the variance of a weighted average of the two X=kx1+(1−k)x2, 0≤k≤1 is minimised when k= 0.5. How much is the variance of this sum reduced by setting k properly relative to other possible values of k?

e. How would the result from part (d) change if the two variables had unequal variances?

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