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(Linearity of expectation I) Let X be a random variable, and a, b be constants. Use properties ofintegration/summation to show that:E(aX + b) = aE[X + b]

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Solution

The statement is incorrect. The correct statement should be E(aX + b) = aE[X] + b. Here's the proof:

The expectation of a random variable is defined as the sum of the product of each outcome and its probability. For a discrete random variable X, this is:

E[X] = ΣxP(X=x)

where the sum is over all possible outcomes x.

If we multiply X by a constant a and add a constant b, we get a new random variable aX + b. The expectation of this new random variable is:

E[aX + b] = Σ(a*x + b)P(X=x)

We can split this sum into two parts:

E[aX + b] = ΣaxP(X=x) + ΣbP(X=x)

The first part is a times the expectation of X:

ΣaxP(X=x) = aΣxP(X=x) = aE[X]

The second part is b times the sum of the probabilities of all outcomes, which is just 1:

ΣbP(X=x) = bΣP(X=x) = b*1 = b

So we have:

E[aX + b] = aE[X] + b

which is the correct statement.

This problem has been solved

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