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(Linearity of expectation II) Let X, Y be random variables and a, b, and c be constants. Use propertiesof integration/summation to show that:E(aX + bY + c) = aE[X] + bE[Y ] + c

Question

(Linearity of expectation II) Let X, Y be random variables and a, b, and c be constants. Use propertiesof integration/summation to show that:E(aX + bY + c) = aE[X] + bE[Y ] + c

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Solution

Sure, here is the step-by-step proof:

  1. The expectation of a random variable is defined as the sum (for discrete variables) or the integral (for continuous variables) of the variable multiplied by its probability. So, we have:

    E(aX + bY + c) = ∫(aX + bY + c)f(x,y) dx dy (for continuous variables)

    or

    E(aX + bY + c) = Σ(aX + bY + c)P(X=x, Y=y) (for discrete variables)

  2. We can split the integral or sum into separate parts:

    E(aX + bY + c) = a∫Xf(x,y) dx dy + b∫Yf(x,y) dx dy + c∫f(x,y) dx dy (for continuous variables)

    or

    E(aX + bY + c) = aΣXP(X=x, Y=y) + bΣYP(X=x, Y=y) + cΣP(X=x, Y=y) (for discrete variables)

  3. The integral or sum of the joint probability density or mass function over all possible values of X and Y is 1. So, we have:

    ∫f(x,y) dx dy = 1 (for continuous variables)

    or

    ΣP(X=x, Y=y) = 1 (for discrete variables)

  4. Therefore, we can simplify the equation to:

    E(aX + bY + c) = aE[X] + bE[Y] + c

This shows that the expectation is a linear operator, which is the property we wanted to prove.

This problem has been solved

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