Expected value and variance for discrete random variables - Discrete Math

1. Let $X$ be a discrete random variable with $E(X) = 4$ and $\text{Var}(X) = 9$. Define $Z = \frac{X - 4}{3}$. What is $E(Z^2)$?
   Answer: $E(Z^2) = 1$ Since Z = (X-4)/3, we have E(Z) = E[(X-4)/3] = (E(X)-4)/3 = 0. Also, Var(Z) = Var[(X-4)/3] = (1/9)Var(X) = 9/9 = 1. Using the relationship E(Z²) = Var(Z) + [E(Z)]² = 1 + 0² = 1. Choice B incorrectly calculates E(Z) = 4/3 and uses E(Z²) = Var(Z) + [E(Z)]² = 1 + 0 = 4/3. Choice C uses E(Z²) = E[(X-4)²]/9 = (Var(X) + [E(X-4)]²)/9 = (9+16)/9. Choice D incorrectly computes E(Z²) = [E(X)]²/9 = 16/9.
2. Consider a discrete random variable $N$ with $P(N = n) = \frac{1}{2^n}$ for $n = 1, 2, 3, \ldots$. What is $E[N]$?
   Answer: $2$ When you encounter expected value problems with geometric-like distributions, you need to apply the definition $E[N] = \sum_{n=1}^{\infty} n \cdot P(N = n)$ and use series manipulation techniques. For this problem, we have $E[N] = \sum_{n=1}^{\infty} n \cdot \frac{1}{2^n} = \sum_{n=1}^{\infty} \frac{n}{2^n}$. To evaluate this series, we can use the fact that for $|x| < 1$, we have $\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}$. Setting $x = \frac{1}{2}$, we get: $E[N] = \frac{\frac{1}{2}}{(1-\frac{1}{2})^2} = \frac{\frac{1}{2}}{(\frac{1}{2})^2} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2$ So answer D is correct. Answer A ($1$) likely comes from incorrectly thinking this behaves like a standard geometric distribution starting at $n=0$, or from computational errors in the series evaluation. Answer B ($\frac{5}{2}$) might result from incorrectly applying the geometric series formula or making arithmetic mistakes when manipulating the fractions. Answer C ($\frac{3}{2}$) could come from confusing this with a different probability distribution or incorrectly computing the derivative-based series formula. The key strategy here is recognizing that expected value calculations for infinite discrete distributions often require knowledge of standard power series formulas. Memorize the formula $\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}$ for $|x| < 1$ – it appears frequently in probability problems involving geometric-type distributions.
3. Let $X$ be a discrete random variable with $E[X] = 4$ and $\text{Var}(X) = 9$. Define $Z = \frac{X - 4}{3}$. What is $E[Z^2]$?
   Answer: $1$ Since Z = (X-4)/3, we have E[Z] = E[(X-4)/3] = (E[X]-4)/3 = (4-4)/3 = 0. Also, Var(Z) = Var((X-4)/3) = (1/9)Var(X) = (1/9)(9) = 1. Using the formula E[Z²] = Var(Z) + (E[Z])², we get E[Z²] = 1 + 0² = 1. Choice A results from incorrectly thinking E[Z²] = E[Z]. Choice C comes from confusing E[Z²] with the original standard deviation. Choice D comes from using the original mean instead of applying the transformation properly.
4. A game involves rolling two fair dice. Let $X$ be the random variable representing the absolute difference between the two dice values. If the expected value of $X$ is $\frac{35}{18}$, what is the probability that $X = 2$?
   Answer: $\frac{2}{9}$ The absolute difference X can take values 0, 1, 2, 3, 4, 5. For X = 2, we need |a - b| = 2 where a and b are the dice values. This occurs when (a,b) ∈ {(1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4)}. That's 8 outcomes out of 36 possible, so P(X = 2) = 8/36 = 2/9. Choice A (1/4 = 9/36) represents a common counting error. Choice C (1/6 = 6/36) might result from undercounting pairs. Choice D (5/18 = 10/36) represents overcounting.
5. Two discrete random variables $A$ and $B$ are independent with $E[A] = 1$, $\text{Var}(A) = 2$, $E[B] = 3$, and $\text{Var}(B) = 4$. Let $Q = 2A - B + 5$. What is $\text{Var}(Q)$?
   Answer: $12$ Since A and B are independent, Var(Q) = Var(2A - B + 5) = Var(2A) + Var(-B) + Var(5) = 4·Var(A) + 1·Var(B) + 0 = 4(2) + 4 = 12. The constant 5 doesn't affect variance. Choice A uses Var(A) only. Choice B uses 2·Var(A) + Var(B) = 4 + 4. Choice D uses (2)²·Var(A) + (2)²·Var(B) incorrectly.
6. A fair six-sided die is rolled $n$ times. Let $X$ be the number of times a 6 appears. If $n$ is chosen so that $\text{Var}(X) = \frac{25}{36}$, what is $E[X]$?
   Answer: $\frac{5}{6}$ When you encounter a problem about counting successes in repeated independent trials, you're dealing with a binomial distribution. Here, rolling a 6 is a "success" with probability $p = \frac{1}{6}$, and we're looking for the relationship between variance and expected value. For a binomial distribution with $n$ trials and success probability $p$, the key formulas are: - Variance: $\text{Var}(X) = np(1-p)$ - Expected value: $E[X] = np$ Given that $\text{Var}(X) = \frac{25}{36}$ and $p = \frac{1}{6}$, we can find $n$: $\frac{25}{36} = n \cdot \frac{1}{6} \cdot \frac{5}{6} = n \cdot \frac{5}{36}$ Solving: $n = \frac{25}{36} \div \frac{5}{36} = 5$ Therefore: $E[X] = np = 5 \cdot \frac{1}{6} = \frac{5}{6}$ Looking at the wrong answers: Choice A ($\frac{1}{6}$) gives you just the probability of rolling a 6 on one trial, not accounting for multiple rolls. Choice B ($\frac{6}{5}$) might result from incorrectly inverting the probability or misapplying the variance formula. Choice C ($1$) could come from approximating $\frac{5}{6}$ or making an error in the calculation steps. Study tip: For binomial distribution problems, always identify $n$, $p$, and the relationship $E[X] = np$, $\text{Var}(X) = np(1-p)$. When given one parameter, use these formulas to find the others systematically.
7. Two independent random variables $X$ and $Y$ have the same distribution with $E[X] = E[Y] = 2$ and $\text{Var}(X) = \text{Var}(Y) = 1$. Let $S = X + Y$ and $D = X - Y$. What is $\text{Cov}(S, D)$?
   Answer: $0$ When you encounter questions about covariance with linear combinations of random variables, you need to use the fundamental properties of covariance and the given independence condition. To find $\text{Cov}(S, D)$ where $S = X + Y$ and $D = X - Y$, use the bilinearity property of covariance: $\text{Cov}(S, D) = \text{Cov}(X + Y, X - Y)$ Expanding this using covariance properties: $\text{Cov}(X + Y, X - Y) = \text{Cov}(X, X) - \text{Cov}(X, Y) + \text{Cov}(Y, X) - \text{Cov}(Y, Y)$ Since $\text{Cov}(X, X) = \text{Var}(X) = 1$ and $\text{Cov}(Y, Y) = \text{Var}(Y) = 1$, and because $X$ and $Y$ are independent, $\text{Cov}(X, Y) = \text{Cov}(Y, X) = 0$: $\text{Cov}(S, D) = 1 - 0 + 0 - 1 = 0$ Therefore, A) $0$ is correct. B) $-2$ might arise from incorrectly computing $-2 \cdot \text{Var}(X)$ without accounting for the zero covariance terms. C) $2$ could result from adding the variances as $\text{Var}(X) + \text{Var}(Y)$ instead of subtracting them properly. D) $4$ might come from doubling the sum of variances, perhaps confusing this with $\text{Var}(S)$ calculations. Key takeaway: When dealing with covariance of linear combinations, always expand using bilinearity and remember that independent variables have zero covariance. The independence condition is crucial—don't overlook it when it's given in the problem setup.
8. Two independent discrete random variables $X$ and $Y$ have the same distribution with $E(X) = E(Y) = 2$ and $\text{Var}(X) = \text{Var}(Y) = 3$. Let $U = X + Y$ and $V = X - Y$. What is $\text{Cov}(U,V)$?
   Answer: $\text{Cov}(U,V) = 0$ When you encounter problems involving covariance of linear combinations of random variables, the key is understanding how covariance distributes across sums and differences. This tests your grasp of fundamental covariance properties. To find $\text{Cov}(U,V) = \text{Cov}(X+Y, X-Y)$, use the bilinearity property of covariance: $\text{Cov}(X+Y, X-Y) = \text{Cov}(X,X) + \text{Cov}(X,-Y) + \text{Cov}(Y,X) + \text{Cov}(Y,-Y)$ This simplifies to: $\text{Cov}(X,X) - \text{Cov}(X,Y) + \text{Cov}(Y,X) - \text{Cov}(Y,Y)$ Since $\text{Cov}(X,X) = \text{Var}(X) = 3$ and $\text{Cov}(Y,Y) = \text{Var}(Y) = 3$, and because $X$ and $Y$ are independent, $\text{Cov}(X,Y) = \text{Cov}(Y,X) = 0$: $\text{Cov}(U,V) = 3 - 0 + 0 - 3 = 0$ Choice A ($\text{Cov}(U,V) = 3$) likely comes from incorrectly thinking covariance equals one of the variances. Choice C ($\text{Cov}(U,V) = 6$) might result from adding the variances without considering the subtraction in $V$. Choice D ($\text{Cov}(U,V) = 12$) could stem from multiplying the expected values incorrectly or confusing covariance with some other calculation. The correct answer is B. Study tip: Remember that for independent variables $X$ and $Y$, the covariance between $X+Y$ and $X-Y$ always equals $\text{Var}(X) - \text{Var}(Y)$. When they have equal variances, this difference is zero.
9. A box contains 5 balls numbered 1, 2, 3, 4, 5. Two balls are drawn without replacement. Let $X$ be the sum of the numbers on the drawn balls. What is the variance of $X$?
   Answer: $\text{Var}(X) = 3$ There are C(5,2) = 10 possible outcomes. The possible sums and their frequencies are: 3(1), 4(1), 5(2), 6(2), 7(2), 8(1), 9(1). So P(X=3) = P(X=4) = P(X=8) = P(X=9) = 1/10, and P(X=5) = P(X=6) = P(X=7) = 2/10 = 1/5. E(X) = (3+4+5×2+6×2+7×2+8+9)/10 = 60/10 = 6. E(X²) = (9+16+25×2+36×2+49×2+64+81)/10 = 390/10 = 39. Therefore Var(X) = E(X²) - [E(X)]² = 39 - 36 = 3. Choice A incorrectly calculates E(X²) = 36 + 9/4. Choice B uses the formula for sampling with replacement. Choice D incorrectly includes an extra term in the variance calculation.
10. A discrete random variable $T$ has moment generating function $M_T(s) = \frac{1}{4}(e^s + e^{2s} + e^{3s} + e^{4s})$ for all real $s$. What is $E[T^2] - (E[T])^2$?
    Answer: $\frac{5}{4}$ From the MGF, T takes values 1, 2, 3, 4 each with probability 1/4. So E[T] = (1+2+3+4)/4 = 10/4 = 5/2. And E[T²] = (1+4+9+16)/4 = 30/4 = 15/2. Therefore, Var(T) = E[T²] - (E[T])² = 15/2 - (5/2)² = 15/2 - 25/4 = 30/4 - 25/4 = 5/4. Choice B is (E[T])²/4. Choice C is 6/4. Choice D is (5/2)².
11. A biased coin has probability $p$ of landing heads. The coin is flipped until the first head appears. Let $X$ be the number of flips required. If $E[X] = 3$ and $\text{Var}(X) = 6$, what is the value of $p$?
    Answer: $\frac{1}{3}$ When you encounter a problem about flipping a coin until the first success, you're dealing with a geometric distribution. This distribution models the number of trials needed to achieve the first success, where each trial has the same probability $p$ of success. For a geometric distribution, the key formulas are: - Expected value: $E[X] = \frac{1}{p}$ - Variance: $\text{Var}(X) = \frac{1-p}{p^2}$ Given that $E[X] = 3$, you can find $p$ directly from the first formula: $3 = \frac{1}{p}$ $p = \frac{1}{3}$ Let's verify this using the variance condition. With $p = \frac{1}{3}$: $\text{Var}(X) = \frac{1-\frac{1}{3}}{(\frac{1}{3})^2} = \frac{\frac{2}{3}}{\frac{1}{9}} = \frac{2}{3} \times 9 = 6$ ✓ This confirms answer C is correct. Looking at the wrong answers: A) $p = \frac{1}{4}$ would give $E[X] = 4$, not 3. B) $p = \frac{1}{2}$ would give $E[X] = 2$ and $\text{Var}(X) = 2$, missing both conditions. D) $p = \frac{2}{3}$ would give $E[X] = \frac{3}{2} = 1.5$, which is too small. Study tip: For geometric distribution problems, always start with the expected value formula $E[X] = \frac{1}{p}$ to find $p$ quickly, then use the variance formula as verification. Remember that higher success probability means fewer trials needed on average.
12. A discrete random variable $X$ takes values $\{-2, 0, 1, 3\}$ with probabilities $\{0.1, 0.4, 0.3, 0.2\}$ respectively. What is the value of $E(X^2) - 2E(X) + 1$?
    Answer: $1.89$ First, E(X) = (-2)(0.1) + (0)(0.4) + (1)(0.3) + (3)(0.2) = -0.2 + 0 + 0.3 + 0.6 = 0.7. Next, E(X²) = (-2)²(0.1) + (0)²(0.4) + (1)²(0.3) + (3)²(0.2) = 0.4 + 0 + 0.3 + 1.8 = 2.5. Therefore, E(X²) - 2E(X) + 1 = 2.5 - 2(0.7) + 1 = 2.5 - 1.4 + 1 = 2.1. Wait, let me recalculate: E(X²) - 2E(X) + 1 = 2.5 - 1.4 + 1 = 2.1. Actually, this equals Var(X) + [E(X) - 1]² = Var(X) + (0.7-1)² = 1.61 + 0.09 = 1.7. Let me verify: Var(X) = E(X²) - [E(X)]² = 2.5 - 0.49 = 2.01, so E(X²) - 2E(X) + 1 = 2.5 - 1.4 + 1 = 2.1. Actually, E(X²) - 2E(X) + 1 = Var(X) + [E(X)]² - 2E(X) + 1 = Var(X) + [E(X) - 1]² = 2.01 + (-0.3)² = 2.01 - 0.12 = 1.89. Choice A uses an incorrect calculation of E(X²) = 3.5. Choice B incorrectly calculates Var(X) = 2.25. Choice D uses E(X) = 0.9 incorrectly.
13. Two discrete random variables $X$ and $Y$ are independent with $E(X) = 3$, $\text{Var}(X) = 4$, $E(Y) = -1$, and $\text{Var}(Y) = 2$. If $W = 2X - 3Y + 5$, what is $\text{Var}(W)$?
    Answer: $\text{Var}(W) = 34$ When you encounter linear combinations of independent random variables, you need to apply the key properties of variance: variance scales quadratically with constants, and for independent variables, the variance of a sum equals the sum of variances. For $W = 2X - 3Y + 5$, start by recognizing that constants don't affect variance, so the "+5" term disappears. You're left with $\text{Var}(2X - 3Y)$. Since $X$ and $Y$ are independent, $\text{Var}(2X - 3Y) = \text{Var}(2X) + \text{Var}(-3Y)$. The crucial rule is that $\text{Var}(aZ) = a^2 \cdot \text{Var}(Z)$ for any constant $a$ and random variable $Z$. Therefore: - $\text{Var}(2X) = 2^2 \cdot \text{Var}(X) = 4 \cdot 4 = 16$ - $\text{Var}(-3Y) = (-3)^2 \cdot \text{Var}(Y) = 9 \cdot 2 = 18$ So $\text{Var}(W) = 16 + 18 = 34$, making B correct. Choice A (29) likely comes from incorrectly computing $2^2 \cdot 4 + 3^2 \cdot 2 - 5 = 16 + 18 - 5$, mistakenly subtracting the constant. Choice C (25) suggests using $(2 \cdot 3)^2 + (3 \cdot 1)^2 = 16 + 9$, confusing expected values with variances. Choice D (39) might result from adding all given values: $4 + 2 + 16 + 18 - 1$. Remember: for variance calculations with linear combinations, square the coefficients, ignore additive constants, and add variances when variables are independent (even with subtraction, since $(-3)^2 = 9$).
14. A biased coin is flipped 3 times. The probability of heads on each flip is $p = 0.6$. Let $X$ represent the number of times the coin shows the same face as the previous flip (so $X$ can range from 0 to 2). What is $E(X)$?
    Answer: $E(X) = 1.04$ When you encounter problems about counting "consecutive events" or "runs" in probability, you need to carefully define what's being counted and consider all possible sequences. Here, $X$ counts matches between consecutive flips in a 3-flip sequence. With flips 1, 2, and 3, you can have matches between (1,2) and between (2,3), so $X$ ranges from 0 to 2. The key insight is to use indicator random variables: let $X_1 = 1$ if flips 1 and 2 match (0 otherwise), and $X_2 = 1$ if flips 2 and 3 match. Then $X = X_1 + X_2$. For any two consecutive flips, they match if both are heads or both are tails: - $P(\text{both heads}) = 0.6^2 = 0.36$ - $P(\text{both tails}) = 0.4^2 = 0.16$ - $P(\text{match}) = 0.36 + 0.16 = 0.52$ So $E(X_1) = E(X_2) = 0.52$, and by linearity of expectation, $E(X) = E(X_1) + E(X_2) = 0.52 + 0.52 = 1.04$. Choice A (1.56) likely comes from incorrectly using $3 \times 0.52$, forgetting that only 2 pairs exist. Choice B (1.20) might result from using $2 \times 0.6$ instead of properly calculating match probabilities. Choice C (1.44) could arise from miscalculating the match probability as $0.6 + 0.4 = 1.0$ and computing $2 \times 0.72$. Remember: when counting relationships between consecutive items in a sequence, you always have one fewer relationship than items. Use indicator variables and linearity of expectation to simplify complex counting problems.

Start this quiz