Question

Prove the following claims. Here, X, Y are discrete random variables on the same sample space, and a, b ? ?. (a) ? (aX + b) = a? X + b (b) ??²(aX + b) = a²?²(X) (c) ?²(X) = ?(X²) - (? X)²

          Prove the following claims. Here, X, Y are discrete random variables on the same sample space, and a, b ? ?.
(a) ? (aX + b) = a? X + b
(b) ??²(aX + b) = a²?²(X)
(c) ?²(X) = ?(X²) - (? X)²

Prove the following claims. Here, X, Y are discrete random variables on the same sample space, and a, b ? ?.
(a) ? (aX + b) = a? X + b
(b) ??²(aX + b) = a²?²(X)
(c) ?²(X) = ?(X²) - (? X)²

Added by Rachel M.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Prove the following claims. Here, X and Y are discrete random variables on the same sample space, and a and b ∈ R: (a) E(aX + b) = aE(X) + b (b) Var(aX + b) = a^2Var(X) (c) Cov(X, Y) = E(XY) - E(X)E(Y)

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Transcript

00:01 All right, so for part a here, we have that the expected value of ax plus b is going to be equal to the sum over all possible values of x.

00:15 So we'll just say sum over i of ax plus b times the probability that ax plus b takes a particular value, or really, that's just going to be dependent on what value x takes on.

00:34 So we have that this is then going to be possible to express as the sum over i of p of xi times a plus, or pardon me, that would be the sum of p of xi times a times xi, excuse me, plus the sum of p of xi as we go over the different possible values for x.

01:13 And so, we can bring a in front of that first sum, so this becomes a times the sum over i of p of xi times xi, and then we'd have, for the second summation there, plus b times the sum of p of xi.

01:29 Now, we should be able to recognize that sum of p of xi times xi is just going to be the expected value of x, and then we know that the sum of the probabilities of all possible outcomes for a discrete variable, it's going to be simply equal to 1.

01:44 So we'd have plus b times 1, and therefore we have e of ax plus b equals a, x plus b.

01:51 Then for part b, sigma squared of ax plus b would then be equal to the expected value of ax plus b squared minus the square of the expected value of ax plus b.

02:14 Now we know that that first expression is going to become the expected value of a squared x squared plus 2axb plus b squared minus we already know what the expected value of ax plus b would be.

02:37 We know that that's going to be a, ex plus b squared.

02:44 So we have that that first term is going to expand out applying linear linear.

02:49 Now because we have proven it a squared times e of x squared plus 2ab times e of x plus b squared because we're just taking the expected value of a constant then we have minus a squared x squared minus 2ab e x minus b squared.

03:20 So we can see that we have, let's see here, one moment, plus b squared minus b squared, plus 2ab x minus 2ab e x, minus 2ab e x...

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