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Expectation and Variance – Solutions
STAT-UB.0103 – Statistics for Business Control and Regression Models
Random variables (review)
- Let X be a random variable describing the number of cups of coffee a randomly-chosen NYU undergraduate drinks in a week. Suppose that there is a 10% chance that the student has one cup of coffee, 30% chance that the student has two cups of coffee, 40% chance that the student has 3 cups of coffee, and a 20% chance stat the student has four cups of coffee. (a) Let p(x) be the probability distribution function of X. Fill in the following table: x 1 2 3 4 p(x)
Solution: x 1 2 3 4 p(x) .10 .30 .40.
(b) Find E(X), the expectation of X.
Solution:
E(X) = (.10)(1) + (.30)(2) + (.40)(3) + (.20)(4) = 2. 7.
(c) What is the interpretation of the expectation of X?
Solution: The long-run sample mean. If we performed the random experiment upon which X is measured many times, getting a different value of X each time, then the sample mean would be very close to the expectation of X.
Variance and Standard Deviation
- This is a continuation of problem 1.
(a) Find var(X) and sd(X), the variance and standard deviation of X.
Solution: var(X) = (.10)(1 − 2 .7)^2 + (.30)(2 − 2 .7)^2 + (.40)(3 − 2 .7)^2 + (.20)(4 − 2 .7)^2 =. 81. The standard deviation of X is given by sd(X) =
var(X) = 0. 9.
(b) What is the interpretation of the standard deviation of X?
Solution: The long-run sample standard devation. If we performed the random ex- periment upon which X is measured many times, getting a different value of X each time, then the sample standard deviation would be very close to the standard devia- tion of X. If the PDF of X is symmetric and mound-shaped, we can use the empirical rule to make predictions about the value of X.
(c) What is the standard deviation of W?
Solution: Using the PDF computed in part (a), and the expected value computed in part (b), we compute the variance of W as
σ^2 = (.50)(− 6 − 1)^2 + (.25)(0 − 1)^2 + (.25)(16 − 1)^2 = 81.
Thus, the standard deviation of W is
σ =
(d) What are the interpretations of the expectation and standard deviation of W?
Solution: If we played the game many many times, then the average of our winnings over all times we played would be close to the $1, and the standard deviations of our winnings over all times we played would be close to $9.
Properties of Expectation and Variance
- Affine Transformations. Let X be a random variable with expectation μX = 2 and standard deviation σX = 3. (a) What is the expectation of 5X + 2?
Solution: 5 μX + 2 = 12.
(b) What is the standard deviation of 5X + 2?
Solution: | 5 |σX = 15.
- Sums of Independent Random Variables. Let X and Y be independent random variables with μX = 1, σX = 3, μY = −5, σY = 4. (a) What is E(X + Y )?
Solution: E(X + Y ) = μX + μY = 1 + (−5) = − 4.
(b) Find var(X + Y ) and sd(X + Y ).
Solution:
var(X + Y ) = σ^2 X + σ^2 Y = (3)^2 + (4)^2 = 25, sd(X + Y ) =
var(X + Y ) = 5.
- Let X and Y be independent random variables with μX = −2, σX = 1, μY = 3, σY = 4. (a) Find the expectation and standard deviation of − 3 X + 2.
Solution:
E(− 3 X + 2) = − 3 μX + 2 = −3(−2) + 2 = 8, sd(− 3 X + 2) = | − 3 |σX = 3(1) = 3.
Advanced Problems
- Bernoulli random variable. Suppose you flip a biased coin that lands Heads with probability p and lands tails with probability 1 − p. Define the random variable
X =
1 if the coin lands Heads; 0 if the coin lands Tails.
This random variable is called a “Bernoulli random variable with success probability p.” (a) What is the PDF of X?
Solution: x 0 1 p(x) 1 − p p
(b) Find μ, the expectation of X
Solution: μ = (1 − p)(0) + (p)(1) = p.
(c) Find σ^2 , the variance of X.
Solution: σ^2 = (1 − p)(0 − p)^2 + (p)(1 − p)^2 = p(1 − p).
- Suppose you have a biased coin that lands Heads with probability p and lands Tails with probability 1 − p. You flip the coin 2 times. Let Y be the number of times the coin lands Heads. (a) What is E(Y )?
Solution: E(Y ) = p + p = 2p.
(b) What is var(Y )? Hint: Y = X 1 + X 2 , where X 1 and X 2 are independent Bernoulli random variables corre- sponding to the 2 coin flips. Use the answer to problem 7(c).
Solution: var(Y ) = p(1 − p) + p(1 − p).
(c) Suppose instead that you flip the coin n times, and let Y count the number of Heads. What are the expectation and variance of Y? Hint: Y = X 1 + X 2 + · · · + Xn.
Solution: E(Y ) = np; var(Y ) = np(1 − p).
