Linear transformations of a random variable
Every value of a random variable is changed by the same rule, as when a day’s count of rentals becomes a day’s profit. Do the mean and standard deviation have to be worked out again from scratch, and does anything else about the distribution change?
Where you are 11 in Random variables
Why this exists
A random variable X comes with a Random variables and probability distributions, a mean \mu_X = \sum x\,P(x), and a variance \operatorname{Var}(X) = \sum (x - \mu_X)^2 P(x) with standard deviation \displaystyle \sigma_X = \sqrt{\operatorname{Var}(X)}. Often the number that matters is not X itself but something computed from it. Leire rents kayaks at a lake dock for 12 dollars each and pays 30 dollars a day for her spot, so her daily profit is Y = 12X - 30, where X is the number of rentals. A time in minutes becomes a time in hours; a score becomes a z-score.
A new table for Y and every weighted sum recomputed would work, but for a rule of the form a + bX the mean and standard deviation of Y follow from those of X in one line each. The same fact for a list of data was the subject of Shifting and rescaling a data set; here it is proved for a random variable from the weighted sums, extended to negative multipliers, and used to standardize a random variable.
The lesson runs over two days. Day 1 is the rules for a + bX; Day 2 is standardizing and money models. All settings are made up: Leire’s kayak dock, where Haizea leads guided trips and Ekaitz runs a snack cart.
Definition
A linear transformation of a random variable X is a new random variable
where a and b are constants, the same for every outcome. Whenever X takes the value x, Y takes the value a + bx. The constant a is added to every value; b multiplies every value.
Standardizing X (when \sigma_X > 0) means forming
This is a linear transformation with b = \dfrac{1}{\sigma_X} and a = -\dfrac{\mu_X}{\sigma_X}. Each value of Z is the z-score of the matching value of X: how many standard deviations that value lies above (positive) or below (negative) the mean.
Intuition
The probabilities ride along. If b \ne 0, different values of x give different values of a + bx, so Y = a + bx happens on exactly the same outcomes as X = x:
The table of Y is the table of X with the top row rewritten and the bottom row copied. On a bar chart, every bar keeps its height and only the numbers under the bars change (Figure 1).
What the new labels do. Adding a slides every label by a: the balance point slides with them, but the gaps between values do not change, so neither does the spread. Multiplying by b makes every gap |b| times as long, so the balance point is multiplied by b and every distance from it by |b|. A negative b also reverses the order of the values, so the chart comes out as a mirror image. Distances have no sign, which is why the spread scales by |b| and not by b.
Standardizing first subtracts \mu_X, which moves the balance point to 0, then divides by \sigma_X, which makes one standard deviation one unit long. Nothing else happens. The bars keep their heights and their order, so the shape of the distribution is exactly the shape of X (Figure 2).
Results
Day 1: the rules for a + bX
For any random variable X and constants a and b, with Y = a + bX:
Two special cases. Adding a constant (b = 1): \mu_{X + a} = \mu_X + a and \sigma_{X + a} = \sigma_X, so the center shifts and the spread stays. Multiplying by a constant (a = 0): \mu_{bX} = b\,\mu_X and \sigma_{bX} = |b|\,\sigma_X, so both scale.
Probabilities carry over with the matching event. For b > 0, P(X \le x) = P(Y \le a + bx); for b < 0 the inequality turns around, P(X \le x) = P(Y \ge a + bx). For b > 0 the distribution of Y has the shape of the distribution of X; for b < 0 it has the mirror-image shape.
Day 2: standardizing
For any random variable with \sigma_X > 0, the standardized variable Z = \dfrac{X - \mu_X}{\sigma_X} has
and the same shape as X. Standardizing a variable that is not normal does not make it normal.
Why it is true
The mean. Weight each value of Y by its probability, which is the probability of the matching x, and split the sum:
The second line uses the distributive property and splits one sum into two; the third uses that the probabilities add to 1 and that \sum x\,P(x) is \mu_X.
The variance. Each deviation of Y from its mean is b times the matching deviation of X:
The a cancels, which is the whole reason adding a constant leaves the spread alone. Squaring and weighting by the same probabilities,
Taking square roots, \displaystyle \sigma_Y = \sqrt{b^2}\,\sigma_X = |b|\,\sigma_X, because the square root of b^2 is |b|: a square root is never negative.
Standardizing. With b = \dfrac{1}{\sigma_X} and a = -\dfrac{\mu_X}{\sigma_X}, the rules give \mu_Z = -\dfrac{\mu_X}{\sigma_X} + \dfrac{\mu_X}{\sigma_X} = 0 and \sigma_Z = \dfrac{1}{\sigma_X}\,\sigma_X = 1, since b > 0. The probabilities are unchanged and b > 0 keeps the order, so the shape is unchanged.
Worked examples
Day 1: from rentals to profit
Leire’s kayak rentals X on a weekday have this distribution (made-up probabilities). She charges 12 dollars per rental and pays 30 dollars a day for the dock spot, so her profit is Y = 12X - 30 dollars. Find \mu_X and \sigma_X, then \mu_Y and \sigma_Y two ways, and the probability that a day loses money.
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y | -18 | -6 | 6 | 18 | 30 |
| P | 0.1 | 0.1 | 0.4 | 0.3 | 0.1 |
For X, \mu_X = 1(0.1) + 2(0.1) + 3(0.4) + 4(0.3) + 5(0.1) = 3.2 rentals. The squared deviations from 3.2 are 4.84, 1.44, 0.04, 0.64, 3.24; weighted by the probabilities they are 0.484, 0.144, 0.016, 0.192, 0.324, so \operatorname{Var}(X) = 1.16 and \displaystyle \sigma_X = \sqrt{1.16} \approx 1.08 rentals.
Directly from the table of Y. Each profit in the middle row has the probability of the count above it, because each profit happens exactly on the days with the matching count.
The deviations y - 8.4 are -26.4, -14.4, -2.4, 9.6, 21.6, each 12 times the matching deviation of X. Squared and weighted they are 69.696, 20.736, 2.304, 27.648, 46.656, which add to \operatorname{Var}(Y) = 167.04 square dollars, so \displaystyle \sigma_Y = \sqrt{167.04} \approx 12.92 dollars.
By the rules, with a = -30 and b = 12:
The two ways agree. Keep \sigma_X exact until the end: 12(1.08) = 12.96 is off in the second decimal.
In context: over many weekdays Leire’s profit averages 8.4 dollars a day, and a day’s profit is typically about 12.92 dollars away from that. The 30 dollar fee lowers every day’s profit by the same amount, so it moves the mean and not the spread; the 12 stretches every gap.
A day loses money when 12x - 30 < 0, that is x < 2.5, so on days with 1 or 2 rentals: P(Y < 0) = P(X \le 2) = 0.1 + 0.1 = 0.2.
Day 1: kayaks left on the dock
Leire starts each weekday with 8 kayaks, and each rental in Example 1 takes one out for the afternoon. Let L = 8 - X be the number left on the dock. Find \mu_L and \sigma_L, describe how the shape of L compares with the shape of X, and find P(L \le 4).
SolutionL = 8 + (-1)X, a linear transformation with a = 8 and b = -1. By the rules, \mu_L = 8 - 3.2 = 4.8 kayaks and \sigma_L = |-1|\,\sigma_X = \sigma_X \approx 1.08 kayaks.
The table of L rewrites the top row and keeps the probabilities:
| l | 7 | 6 | 5 | 4 | 3 |
|---|---|---|---|---|---|
| P | 0.1 | 0.1 | 0.4 | 0.3 | 0.1 |
As a check, 7(0.1) + 6(0.1) + 5(0.4) + 4(0.3) + 3(0.1) = 4.8. The values now run in reverse order. For X the tallest bar is at 3 and the next tallest, 0.3, is to its right; for L the tallest bar is at 5 and the 0.3 bar is to its left. The chart is a mirror image of the chart of X. The spread is unchanged because a mirror keeps every distance.
The event turns around with the order:
Day 1: minutes, hours and a trip charge
Haizea leads guided trips from the dock. A trip lasts T minutes, with T = 30, 60 or 90 with probabilities 0.5, 0.3 and 0.2 (made up). (a) Find \mu_T and \sigma_T. (b) Find the mean, standard deviation and variance of the length in hours, H = T/60. (c) Leire charges 5 dollars to launch plus 0.20 dollars a minute, so a trip costs C = 5 + 0.2T dollars. Find \mu_C, \sigma_C and \operatorname{Var}(C).
Solution(a) \mu_T = 30(0.5) + 60(0.3) + 90(0.2) = 15 + 18 + 18 = 51 minutes. The deviations are -21, 9, 39, so \operatorname{Var}(T) = 441(0.5) + 81(0.3) + 1521(0.2) = 220.5 + 24.3 + 304.2 = 549 square minutes and \displaystyle \sigma_T = \sqrt{549} \approx 23.43 minutes.
(b) \displaystyle H = \frac{1}{60}T, so \displaystyle b = \frac{1}{60} and a = 0. Then \displaystyle \mu_H = \frac{51}{60} = 0.85 hours and \displaystyle \sigma_H = \frac{\sqrt{549}}{60} \approx 0.39 hours. The variance divides by 60^2 = 3600, not by 60: \displaystyle \operatorname{Var}(H) = \frac{549}{3600} = 0.1525 square hours. A variance is in squared units, and a square minute is \displaystyle \frac{1}{3600} of a square hour.
(c) Here a = 5 and b = 0.2:
and \operatorname{Var}(C) = 0.2^2(549) = 21.96 square dollars. The launch fee adds 5 dollars to every trip, so it shows up only in the mean. Check: the trips cost 11, 17 and 23 dollars, and 11(0.5) + 17(0.3) + 23(0.2) is 5.5 + 5.1 + 4.6 = 15.2.
Day 2: standardizing the rentals
Standardize Leire’s rentals X from Example 1. List the value of Z for each count, confirm that \mu_Z = 0 and \sigma_Z = 1, and interpret a 5-rental day. Then standardize the profit on that same day.
Solution\displaystyle Z = \dfrac{X - 3.2}{\sqrt{1.16}}. Dividing each deviation -2.2, -1.2, -0.2, 0.8, 1.8 by \displaystyle \sqrt{1.16} \approx 1.077 gives
| rentals x | z | P |
|---|---|---|
| 1 | -2.04 | 0.1 |
| 2 | -1.11 | 0.1 |
| 3 | -0.19 | 0.4 |
| 4 | 0.74 | 0.3 |
| 5 | 1.67 | 0.1 |
By the rules, \displaystyle Z = -\dfrac{3.2}{\sqrt{1.16}} + \dfrac{1}{\sqrt{1.16}}X, so \displaystyle \mu_Z = \dfrac{3.2 - 3.2}{\sqrt{1.16}} = 0 and \displaystyle \sigma_Z = \dfrac{\sqrt{1.16}}{\sqrt{1.16}} = 1. The table agrees: the weighted sum of the deviations is 0 (the balance property of the mean), so the weighted sum of the z values is 0; and the weighted squares of the z values add to \dfrac{1.16}{1.16} = 1.
A 5-rental day has z \approx 1.67: it sits about 1.67 standard deviations above Leire’s long-run average.
That day’s profit is 30 dollars, and standardizing the profit gives
the same number. Since Y = 12X - 30 with 12 > 0, each value of Y is exactly as many of its standard deviations from its mean as the matching x is from \mu_X. Standardizing strips away the units, the a and the b, and keeps only the position within the distribution.
Day 2: standardizing does not make it normal
The number W of kayaks that come back damaged in a week is 0, 1, 2 or 3 with probabilities 0.6, 0.25, 0.1 and 0.05 (made up). Standardize W, and decide whether the empirical rule from z-scores and the empirical rule describes Z.
Solution\mu_W = 0.25 + 0.2 + 0.15 = 0.6. By the shortcut, E(W^2) = 0.25 + 0.4 + 0.45 = 1.1, so \operatorname{Var}(W) = 1.1 - 0.6^2 = 0.74 and \displaystyle \sigma_W = \sqrt{0.74} \approx 0.86. The values of \displaystyle Z = \dfrac{W - 0.6}{\sqrt{0.74}} are about -0.70, 0.46, 1.63 and 2.79, with the same probabilities. By the rules \mu_Z = 0 and \sigma_Z = 1.
The shape has not changed: most of the probability sits at the left and a thin tail runs right. The empirical rule describes a normal distribution, where about 2.5\% lies more than 2 standard deviations above the mean and about 16\% more than 1 below. Here P(Z > 2) = P(W = 3) = 0.05, twice the normal figure, and P(Z < -1) = 0, because Z never goes below -0.70. So the empirical rule, and any normal table, would give wrong answers for Z. Standardizing changes the labels on the axis, never the shape.
Day 2: two deals for the dock
The park offers Leire a second deal for her weekdays. Plan A is the current one, Y_A = 12X - 30. Under plan B the park charges no daily fee but keeps all but 2.50 dollars of each rental, so Y_B = 2.5X. Customers pay the same either way, so X has the distribution of Example 1 under both. (a) Compare the two plans’ means, standard deviations and worst days. (b) What amount c per rental would give plan B the same mean as plan A, and what would its standard deviation be?
Solution(a) From Example 1, plan A has \mu = 8.4 dollars and \sigma \approx 12.92 dollars, with a worst day of -18 dollars and a loss on 20\% of days. For plan B, a = 0 and b = 2.5: \mu_B = 2.5(3.2) = 8 dollars and \displaystyle \sigma_B = 2.5\sqrt{1.16} \approx 2.69 dollars. Plan B’s worst day, one rental, still earns 2.5 dollars, so it never loses. Plan B gives up 0.40 dollars a day on average in exchange for a far steadier income.
The two models show where each part of plan A’s risk comes from. The large multiplier 12 makes the spread large; the fixed 30 dollars changes no distance at all but slides the whole distribution down, past zero on slow days.
(b) The mean of cX is 3.2c. Setting 3.2c = 8.4 gives c = 2.625 dollars, and then \displaystyle \sigma = 2.625\sqrt{1.16} \approx 2.83 dollars, still under a quarter of plan A’s.
Boundaries
- The rules hold only for a + bX with constants a and b. For any other rule, such as X^2, \displaystyle \sqrt{X} or \displaystyle \frac{1}{X}, the mean of the new variable is not the rule applied to \mu_X (Exercise 21), and its distribution must be worked out from its own table.
- a and b must be the same on every outcome. A price that changes with how busy the day is is not a constant b.
- If b = 0, Y is the constant a: the rules still give \mu_Y = a and \sigma_Y = 0, but the table of Y collapses to one column.
- Standardizing needs \sigma_X > 0; a variable that never varies cannot be standardized.
- Standardizing keeps the shape. Z is a normal random variable only when X was normal to begin with, so the empirical rule and normal tables apply to Z only then.
- 2X means one day’s count doubled, not the total over two separate days. A sum of two random variables follows its own rules, in Combining random variables.
Where this goes next
The rules here turn any change of units, fee, price or score scale into a one-line calculation, and standardizing puts every random variable on one common scale: mean 0, standard deviation 1, its own shape kept. Combining random variables combines two random variables instead of transforming one, and there the variances, not the standard deviations, add. For a normal random variable a linear transformation stays normal, which is why one table, the The standard normal distribution, serves every normal model. The same algebra on a graph is a Shifting a graph up, down, left and right followed by a vertical stretch.
Exercises
Day 1: the rules for a + bX
All probabilities are made up. Give standard deviations to two decimal places unless exact.
- A random variable has \mu_X = 6 and \sigma_X = 2. Find the mean and standard deviation of (a) X + 10, (b) 3X, (c) 3X + 10, (d) 5 - 2X.
- A random variable has \operatorname{Var}(X) = 9. Find \operatorname{Var}(4X - 1) and the standard deviation of 4X - 1.
- Leire also rents dry bags for 4 dollars each. The number D rented in a day is 0, 1 or 2 with probabilities 0.3, 0.5 and 0.2. Find \mu_D and \sigma_D, then use the rules for the mean and standard deviation of the dry-bag revenue R = 4D. Check \mu_R from the table of R.
- On a busy day Ekaitz waits M minutes for a kayak, with \mu_M = 12 and \sigma_M = 5 minutes. Find the mean, standard deviation and variance of his wait in seconds, S = 60M, with units.
- Haizea is paid 15 dollars an hour plus 10 dollars a day for lunch. Her hours H in a day have \mu_H = 5.5 and \sigma_H = 1.5 hours. Find the mean and standard deviation of her daily pay.
- X takes the values 0, 2, 4 with probabilities 0.5, 0.3, 0.2, and Y = 10 - 2X. (a) Write the table of Y and find \mu_Y from it, then by the rule. (b) Find \sigma_X and \sigma_Y. (c) Write P(Y \ge 6) as a probability about X and find it.
- A random variable has \operatorname{Var}(X) = 4. Ekaitz writes \operatorname{Var}(3X + 2) = 3(4) + 2 = 14. Find the correct variance and standard deviation, and name both errors.
- (a) Y = 5 + 2X has \mu_Y = 17 and \sigma_Y = 6. Find \mu_X and \sigma_X. (b) Answer the same question for Y = 5 - 2X with the same \mu_Y and \sigma_Y.
Day 2: standardizing and money models
- A random variable has \mu_X = 40 and \sigma_X = 8. Find the value of Z when X = 52 and when X = 30, and the value of X that gives Z = -0.5.
- Standardize the dry-bag count D of Exercise 3: give each value of Z as a fraction and to two decimal places, and confirm from the table that \mu_Z = 0 and \sigma_Z = 1.
- Z is a standardized random variable. Find the mean and standard deviation of 3Z + 7 and of -Z.
- A random variable has \mu_X = 20 and \sigma_X = 4. Find constants a and b > 0 so that Y = a + bX has mean 100 and standard deviation 15.
- At the dock’s open day, a free raffle ticket wins V = 10 dollars with probability 0.1 and 0 dollars otherwise. Find \mu_V and \sigma_V, the two values of the standardized Z, and P(Z \ge 3). The empirical rule puts about 0.15\% of a normal distribution more than 3 standard deviations above the mean. What does the comparison show?
- Ekaitz’s snack cart sells S sandwiches a day, with \mu_S = 14 and \sigma_S = 4. Each sells for 6 dollars and its ingredients cost 2.50 dollars, and the cart’s permit costs 20 dollars a day. Write his daily profit G as a linear transformation of S and find \mu_G and \sigma_G.
- Rentals N at a second dock have \mu_N = 5 and \sigma_N = 2. Plan A pays the owner 10N - 25 dollars a day; plan B pays 4N dollars. Find the mean and standard deviation under each plan and describe the choice.
- Rentals X at a third dock have \mu_X = 6 and \sigma_X = 1.5, and the daily profit is Y = pX - 30 dollars for a price p per rental. Find the p that makes \mu_Y = 45 dollars, and \sigma_Y at that price, assuming the price does not change X.
Reasoning
- Explain from the weighted sums why adding a constant to a random variable leaves its standard deviation unchanged.
- Explain why \sigma_Y = |b|\,\sigma_X and not b\,\sigma_X.
- Use the rules to show that a standardized variable has mean 0 and standard deviation 1, and explain why standardizing cannot change the shape of a distribution.
- For b < 0, explain why P(a + bX \le a + bx) = P(X \ge x).
- X is 0 or 2, each with probability 0.5. Compare E(X^2) with \mu_X^2, and explain why the rule for the mean of a + bX says nothing about X^2.
Answers
- (a) Mean 16, standard deviation 2. (b) Mean 18, standard deviation 6. (c) Mean 28, standard deviation 6. (d) Mean 5 - 2(6) = -7, standard deviation |-2|(2) = 4.
- \operatorname{Var}(4X - 1) = 4^2(9) = 144, and the standard deviation is \displaystyle \sqrt{144} = 12. The -1 changes neither.
- \mu_D = 0.5 + 0.4 = 0.9 bags. E(D^2) = 0.5 + 0.8 = 1.3, so \operatorname{Var}(D) = 1.3 - 0.81 = 0.49 and \sigma_D = 0.7 bags. Then \mu_R = 4(0.9) = 3.6 dollars and \sigma_R = 4(0.7) = 2.8 dollars. Check: R is 0, 4, 8 with the same probabilities, and 0 + 2 + 1.6 = 3.6.
- \mu_S = 60(12) = 720 seconds, \sigma_S = 60(5) = 300 seconds, and \operatorname{Var}(S) = 60^2(25) = 90{,}000 square seconds.
- Pay = 10 + 15H. Mean 10 + 15(5.5) = 92.5 dollars, standard deviation 15(1.5) = 22.5 dollars; the 10 dollars for lunch moves only the mean.
- (a) Y is 10, 6, 2 with probabilities 0.5, 0.3, 0.2, so \mu_Y = 5 + 1.8 + 0.4 = 7.2. By the rule, \mu_X = 0.6 + 0.8 = 1.4 and 10 - 2(1.4) = 7.2. (b) E(X^2) = 1.2 + 3.2 = 4.4, \operatorname{Var}(X) = 4.4 - 1.96 = 2.44, \sigma_X \approx 1.56; \displaystyle \sigma_Y = 2\sqrt{2.44} \approx 3.12. (c) 10 - 2X \ge 6 means X \le 2, so P(Y \ge 6) = P(X \le 2) = 0.8.
- \operatorname{Var}(3X + 2) = 3^2(4) = 36 and the standard deviation is 6. The multiplier must be squared, and the added 2 does not belong in a variance at all.
- (a) 17 = 5 + 2\mu_X gives \mu_X = 6; 6 = 2\sigma_X gives \sigma_X = 3. (b) 17 = 5 - 2\mu_X gives \mu_X = -6; 6 = |-2|\sigma_X gives \sigma_X = 3 again.
- \displaystyle z = \frac{52 - 40}{8} = 1.5 and \displaystyle z = \frac{30 - 40}{8} = -1.25. From \displaystyle -0.5 = \frac{x - 40}{8}, x = 40 - 4 = 36.
- With \mu_D = 0.9 and \sigma_D = 0.7: \displaystyle z = -\frac{9}{7} \approx -1.29, \displaystyle \frac{1}{7} \approx 0.14, \displaystyle \frac{11}{7} \approx 1.57. Weighted sum \displaystyle \frac{-2.7 + 0.5 + 2.2}{7} = 0. Weighted squares \displaystyle \frac{0.3(81) + 0.5(1) + 0.2(121)}{49} = \frac{49}{49} = 1, so \sigma_Z = 1.
- 3Z + 7: mean 3(0) + 7 = 7, standard deviation 3(1) = 3. -Z: mean 0, standard deviation |-1|(1) = 1.
- b(4) = 15 gives b = 3.75; a + 3.75(20) = 100 gives a = 25. So Y = 25 + 3.75X.
- \mu_V = 1 dollar; \operatorname{Var}(V) = 1^2(0.9) + 9^2(0.1) = 9, \sigma_V = 3 dollars. Z is \displaystyle \frac{0 - 1}{3} \approx -0.33 or \displaystyle \frac{10 - 1}{3} = 3, so P(Z \ge 3) = 0.1, far more than a normal model’s 0.15\%. Z has mean 0 and standard deviation 1 but keeps the two-value shape of V; it is not normal.
- Each sandwich earns 6 - 2.5 = 3.5 dollars, so G = 3.5S - 20. \mu_G = 3.5(14) - 20 = 29 dollars, \sigma_G = 3.5(4) = 14 dollars.
- Plan A: mean 10(5) - 25 = 25 dollars, standard deviation 10(2) = 20 dollars. Plan B: mean 4(5) = 20 dollars, standard deviation 4(2) = 8 dollars. Plan A earns 5 dollars more a day on average but swings much more widely; plan B is steadier.
- 6p - 30 = 45 gives p = 12.5 dollars, and then \sigma_Y = 12.5(1.5) = 18.75 dollars.
- The mean of X + a is \mu_X + a, so each deviation is (x + a) - (\mu_X + a) = x - \mu_X, the same as before. The probabilities are the same too, so the weighted sum of squared deviations, the variance, is the same, and so is its square root.
- Each deviation is multiplied by b, so each squared deviation by b^2, and the variance by b^2. The standard deviation is \displaystyle \sqrt{b^2}\,\sigma_X = |b|\,\sigma_X. A standard deviation is a square root and is never negative, while b\,\sigma_X would be negative for b < 0.
- Z = a + bX with \displaystyle b = \frac{1}{\sigma_X} and \displaystyle a = -\frac{\mu_X}{\sigma_X}, so \displaystyle \mu_Z = -\frac{\mu_X}{\sigma_X} + \frac{\mu_X}{\sigma_X} = 0 and \displaystyle \sigma_Z = \frac{1}{\sigma_X}\sigma_X = 1. Each value keeps its probability, b > 0 keeps the order, and every gap is divided by the same \sigma_X, so the bars sit in the same pattern with the same heights: the shape is the same.
- Subtracting a from both sides, a + bX \le a + bx is the same as bX \le bx, and dividing by the negative b reverses the inequality, giving X \ge x. The two statements are true on exactly the same outcomes, so they have the same probability.
- E(X^2) = 0(0.5) + 4(0.5) = 2 but \mu_X^2 = 1^2 = 1. The rule splits \sum (a + bx)P(x) because each value is a constant plus a constant times x. Squaring is not of that form: it stretches the large value much more than the small one, so the weighted sum does not reduce to a function of \mu_X.