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

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, E(aX + bY + c) is the sum or integral of (aX + bY + c) times its probability.

2. By the properties of integration or summation, this can be broken down into three separate sums/integrals: a times the sum/integral of X times its probability, plus b times the sum/integral of Y times its probability, plus the sum/integral of c times its probability.

3. The sum/integral of X times its probability is just the expectation of X, E[X], and similarly for Y. The sum/integral of c times its probability is just c, because the probability sums/integrates to 1.

4. Therefore, E(aX + bY + c) = aE[X] + bE[Y] + c, which completes the proof.

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

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, E(aX + bY + c) is the sum or integral of (aX + bY + c) times its probability.

2. By the properties of integration or summation, this can be broken down into three separate sums or integrals: a times the sum or integral of X times its probability, plus b times the sum or integral of Y times its probability, plus the sum or integral of c times its probability.

3. The sum or integral of X times its probability is just the expectation of X, E[X], and similarly for Y. The sum or integral of c times its probability is just c, because the sum or integral of a constant times a probability is just the constant (since the sum or integral of the probabilities is 1).

4. Therefore, E(aX + bY + c) = aE[X] + bE[Y] + c, as required.