How to Use This Page
1. Solve each exercise problem on your own first — attempt every step before looking here. 2. Compare your method with the step-by-step solution shown. 3. If your answer differs, rework from the step where you diverged. 4. Pay special attention to the Common Mistakes section — these are the errors that cost marks in exams. 5. Use the Exam Checklist before your test to ensure you have covered every topic.
Key Concepts (Quick Recap)
Random Variable
- Random variable: A real-valued function defined on the sample space S of a random experiment
- Discrete random variable: Takes at most a countable number of values (e.g., number of heads in 3 tosses: 0, 1, 2, 3)
- Continuous random variable: Takes values in an interval (e.g., weight, height)
- Probability mass function (PMF): P(X = x_i) = p_i for discrete RV, where Σ p_i = 1 and p_i ≥ 0 for all i
- Probability density function (PDF): f(x) ≥ 0 and ∫ f(x) dx = 1 for continuous RV
Mathematical Expectation (Mean)
- E(X) = Σ x_i · P(x_i) — for discrete random variable
- E(X) = ∫ x · f(x) dx — for continuous random variable
- E(X) is also called the mean or expected value of X, denoted μ
Properties of Expectation
- E(c) = c — expectation of a constant is the constant itself
- E(aX) = a · E(X) — constant multiplier comes out
- E(aX + b) = a · E(X) + b — linear transformation
- E(X + Y) = E(X) + E(Y) — always true (no independence needed)
- E(XY) = E(X) · E(Y) — only when X and Y are independent
Variance and Standard Deviation
- Var(X) = E(X²) − [E(X)]² — the fundamental formula
- Var(X) = E[(X − μ)²] — equivalent definition
- σ = √Var(X) — standard deviation
- Var(c) = 0 — constant has zero variance
- Var(aX) = a² · Var(X) — multiplier is squared
- Var(X + c) = Var(X) — shifting by a constant does not change variance
- Var(X + Y) = Var(X) + Var(Y) — only when X and Y are independent
Moment Generating Function
- M_X(t) = E(e^{tX}) — definition
- E(Xⁿ) = M_X⁽ⁿ⁾(0) — nth derivative at t = 0 gives nth moment
I. Choose the Best Answer (MCQs)
1. If E(X) = 3 and E(Y) = 5, then E(X + Y) equals: (a) 8 (b) 15 (c) 2 (d) 3
Answer(a) 8 Explanation: By the property E(X + Y) = E(X) + E(Y) = 3 + 5 = 8. This property holds always, regardless of whether X and Y are independent.
2. If X is a random variable with E(X) = 2, then E(3X + 1) equals: (a) 6 (b) 7 (c) 9 (d) 3
Answer(b) 7 Explanation: Using E(aX + b) = a · E(X) + b, we get E(3X + 1) = 3 · E(X) + 1 = 3(2) + 1 = 7.
3. If Var(X) = 4, then Var(3X) equals: (a) 12 (b) 36 (c) 7 (d) 4
Answer(b) 36 Explanation: Var(aX) = a² · Var(X) = 3² · 4 = 9 · 4 = 36. Note the multiplier is squared, unlike expectation where it comes out as-is.
4. A random variable X has PMF P(X = x) = x/10 for x = 1, 2, 3, 4. Then E(X) equals: (a) 2 (b) 3 (c) 2.5 (d) 3.5
Answer(b) 3 Explanation: E(X) = Σ x · P(x) = 1(1/10) + 2(2/10) + 3(3/10) + 4(4/10) = 1/10 + 4/10 + 9/10 + 16/10 = 30/10 = 3.
5. If E(X) = 5 and E(X²) = 40, then Var(X) equals: (a) 15 (b) 35 (c) 200 (d) 10
Answer(a) 15 Explanation: Var(X) = E(X²) − [E(X)]² = 40 − 25 = 15.
6. For a random variable X, E(X − μ) equals: (a) σ² (b) 0 (c) μ (d) Var(X)
Answer(b) 0 Explanation: E(X − μ) = E(X) − E(μ) = μ − μ = 0. This is because the expected deviation from the mean is always zero.
7. If X and Y are independent with Var(X) = 3 and Var(Y) = 7, then Var(X + Y) equals: (a) 10 (b) 4 (c) 21 (d) 10
Answer(a) 10 Explanation: When X and Y are independent, Var(X + Y) = Var(X) + Var(Y) = 3 + 7 = 10.
8. Which of the following is always true? (a) E(XY) = E(X) · E(Y) (b) Var(X + Y) = Var(X) + Var(Y) (c) E(X + Y) = E(X) + E(Y) (d) Var(X + Y) = Var(X) · Var(Y)
Answer(c) E(X + Y) = E(X) + E(Y) Explanation: Option (a) and (b) require independence. Option (c) holds always. Option (d) is never correct.
9. The standard deviation of X is 3. Then Var(2X) is: (a) 18 (b) 36 (c) 6 (d) 12
Answer(b) 36 Explanation: σ = 3, so Var(X) = 9. Then Var(2X) = 4 · Var(X) = 4 · 9 = 36.
10. If X takes values 0, 1, 2 with probabilities 1/4, 1/2, 1/4, then E(X²) is: (a) 1 (b) 1.5 (c) 2 (d) 0.5
Answer(b) 1.5 Explanation: E(X²) = Σ x² · P(x) = 0²(1/4) + 1²(1/2) + 2²(1/4) = 0 + 1/2 + 1 = 1.5.
II. Fill in the Blanks
1. A function that assigns a real number to each outcome in the sample space is called a random variable.
2. If E(X) = 4, then E(5X) = 20. (Since E(aX) = a · E(X) = 5 · 4 = 20.)
3. The variance of a constant is always 0. (Since a constant does not vary at all.)
4. If E(X) = 3 and E(Y) = 7, then E(X + Y) = 10. (By additivity of expectation.)
5. Var(X) = E(X²) − [E(X)]². (This is the computational formula for variance.)
6. The positive square root of variance is called the standard deviation.
7. If X and Y are independent random variables, then E(XY) = E(X) · E(Y).
8. Var(X + 5) = Var(X) + 0, so Var(X + 5) = Var(X). (Adding a constant does not change variance.)
9. The moment generating function M_X(t) is defined as M_X(t) = E(e^{tX}).
10. If E(X) = 2 and E(X²) = 9, then Var(X) = 5. (Var(X) = 9 − 4 = 5.)
III. True or False
1. Statement: E(X + Y) = E(X) + E(Y) only when X and Y are independent.
False. Correction: E(X + Y) = E(X) + E(Y) holds always, regardless of independence. It is E(XY) = E(X) · E(Y) that requires independence.
2. Statement: Variance is always non-negative.
True. Variance = E[(X − μ)²], which is a sum of squares and hence ≥ 0.
3. Statement: Var(2X) = 2 · Var(X).
False. Correction: Var(aX) = a² · Var(X), so Var(2X) = 4 · Var(X), not 2 · Var(X).
4. Statement: E(X − μ) = 0, where μ = E(X).
True. E(X − μ) = E(X) − μ = μ − μ = 0.
5. Statement: A random variable can take only positive values.
False. Correction: A random variable can take any real value — positive, negative, or zero. For example, profit/loss can be positive or negative.
6. Statement: If Var(X) = 0, then X is a constant.
True. Zero variance means all values of X are the same, so X is essentially a constant.
7. Statement: For independent X and Y, Var(X − Y) = Var(X) + Var(Y).
True. For independent variables, Var(X − Y) = Var(X) + (−1)² · Var(Y) = Var(X) + Var(Y).
8. Statement: E(X²) = [E(X)]² for all random variables.
False. Correction: E(X²) ≥ [E(X)]² by Jensen's inequality (variance is non-negative). Equality holds only when Var(X) = 0.
IV. Match the Following
| Column A | Column B |
|---|
| 1. E(c) where c is constant | (a) E(X²) − [E(X)]² |
| 2. Var(X) | (b) c |
| 3. E(aX + b) | (c) a² · Var(X) |
| 4. Var(aX) | (d) 0 |
| 5. Var(X + c) | (e) a · E(X) + b |
Answer1 → (b): E(c) = c 2 → (a): Var(X) = E(X²) − [E(X)]² 3 → (e): E(aX + b) = a · E(X) + b 4 → (c): Var(aX) = a² · Var(X) 5 → (d): Var(X + c) = Var(X) + 0 = Var(X)
V. Answer Briefly
1. Define a random variable and give an example.
Answer
A random variable is a real-valued function defined on the sample space of a random experiment that assigns a numerical value to each outcome. For example, if two coins are tossed, the sample space is S = {HH, HT, TH, TT}. If X is the number of heads, then X(HH) = 2, X(HT) = 1, X(TH) = 1, X(TT) = 0. Here X is a discrete random variable taking values 0, 1, 2.
2. Distinguish between discrete and continuous random variables.
Answer
A discrete random variable takes at most a countable number of values. Examples: number of defective items in a batch (0, 1, 2, ...), number of heads in coin tosses. A continuous random variable takes values in an interval of real numbers. Examples: weight of a product, temperature reading, time to complete a task. Discrete RVs use PMF (probability mass function) while continuous RVs use PDF (probability density function).
3. State the properties of mathematical expectation.
Answer
The key properties are: 1. E(c) = c (expectation of a constant is the constant) 2. E(aX) = a · E(X) (constant multiplier comes out) 3. E(aX + b) = a · E(X) + b (linearity of expectation) 4. E(X + Y) = E(X) + E(Y) (always true, no independence needed) 5. E(XY) = E(X) · E(Y) (only when X and Y are independent)
4. Write the formula for variance and state its properties.
Answer
Var(X) = E(X²) − [E(X)]² Properties: 1. Var(c) = 0 2. Var(aX) = a² · Var(X) 3. Var(X + c) = Var(X) 4. Var(X + Y) = Var(X) + Var(Y) when X and Y are independent 5. Var(X) ≥ 0 always
5. What is the relationship between E(X²), E(X), and Var(X)?
Answer
Var(X) = E(X²) − [E(X)]². This means: - E(X²) = Var(X) + [E(X)]² - Since Var(X) ≥ 0, we always have E(X²) ≥ [E(X)]² - Equality holds only when Var(X) = 0, i.e., when X is a constant
6. Define moment generating function (MGF). What is its use?
Answer
The moment generating function of X is defined as M_X(t) = E(e^{tX}). Its uses: 1. The nth moment about the origin is obtained by differentiating n times and evaluating at t = 0: E(Xⁿ) = M_X⁽ⁿ⁾(0) 2. MGF uniquely determines the probability distribution 3. If X and Y are independent, M_{X+Y}(t) = M_X(t) · M_Y(t)
7. If E(X) = 4 and E(Y) = 6, and X and Y are independent, find E(XY) and E(X + Y).
Answer
Since X and Y are independent: E(XY) = E(X) · E(Y) = 4 × 6 = 24 E(X + Y) = E(X) + E(Y) = 4 + 6 = 10 (this holds even without independence)
8. If Var(X) = 9, find Var(3X − 2).
Answer
Var(3X − 2) = 3² · Var(X) = 9 × 9 = 81 (The constant −2 does not affect variance, and the multiplier 3 is squared.)
VI. Answer in Detail
1. A random variable X has the following probability distribution:
Answer
| x | 0 | 1 | 2 | 3 | 4 |
|---|
| P(x) | 1/16 | 4/16 | 6/16 | 4/16 | 1/16 |
Find E(X), E(X²), Var(X), and standard deviation.
Step 1: Verify ΣP(x) = 1 1/16 + 4/16 + 6/16 + 4/16 + 1/16 = 16/16 = 1 ✓ Step 2: Compute E(X) E(X) = Σ x · P(x) = 0(1/16) + 1(4/16) + 2(6/16) + 3(4/16) + 4(1/16) = 0 + 4/16 + 12/16 + 12/16 + 4/16 = 32/16 = 2 Step 3: Compute E(X²) E(X²) = Σ x² · P(x) = 0²(1/16) + 1²(4/16) + 2²(6/16) + 3²(4/16) + 4²(1/16) = 0 + 4/16 + 24/16 + 36/16 + 16/16 = 80/16 = 5 Step 4: Compute Var(X) Var(X) = E(X²) − [E(X)]² = 5 − 4 = 1 Step 5: Compute Standard Deviation σ = √Var(X) = √1 = 1
2. A random variable X has PMF P(X = x) = kx for x = 1, 2, 3, 4, and P(X = x) = 0 otherwise. Find the value of k, E(X), and Var(X).
Answer
Step 1: Find k using ΣP(x) = 1 Σ P(x) = k(1) + k(2) + k(3) + k(4) = k(1 + 2 + 3 + 4) = 10k = 1 Therefore k = 1/10 The PMF is: P(1) = 1/10, P(2) = 2/10, P(3) = 3/10, P(4) = 4/10 Step 2: Compute E(X) E(X) = Σ x · P(x) = 1(1/10) + 2(2/10) + 3(3/10) + 4(4/10) = 1/10 + 4/10 + 9/10 + 16/10 = 30/10 = 3 Step 3: Compute E(X²) E(X²) = Σ x² · P(x) = 1²(1/10) + 2²(2/10) + 3²(3/10) + 4²(4/10) = 1/10 + 8/10 + 27/10 + 64/10 = 100/10 = 10 Step 4: Compute Var(X) Var(X) = E(X²) − [E(X)]² = 10 − 9 = 1
3. Two unbiased dice are thrown. X is the sum of the numbers on the two dice. Find the probability distribution of X, E(X), and Var(X).
Answer
Step 1: Determine possible values and probabilities X takes values from 2 to 12. Total outcomes = 36. | x | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | |---|---|---|---|---|---|---|---|---|---|---|---| | P(x) | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 | Step 2: Compute E(X) E(X) = Σ x · P(x) = (2·1 + 3·2 + 4·3 + 5·4 + 6·5 + 7·6 + 8·5 + 9·4 + 10·3 + 11·2 + 12·1)/36 = (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)/36 = 252/36 = 7 Step 3: Compute E(X²) E(X²) = (4·1 + 9·2 + 16·3 + 25·4 + 36·5 + 49·6 + 64·5 + 81·4 + 100·3 + 121·2 + 144·1)/36 = (4 + 18 + 48 + 100 + 180 + 294 + 320 + 324 + 300 + 242 + 144)/36 = 1974/36 = 54.833... Step 4: Compute Var(X) Var(X) = E(X²) − [E(X)]² = 1974/36 − 49 = 1974/36 − 1764/36 = 210/36 = 35/6 ≈ 5.83
4. The probability distribution of a random variable X is given below. Find (i) the value of k, (ii) E(X), (iii) Var(X), and (iv) E(2X + 3).
Answer
(i) Find k: Σ P(x) = 1 → 0.1 + k + 0.3 + 0.4 = 1 → k = 0.2 (ii) Find E(X): E(X) = (−1)(0.1) + (0)(0.2) + (1)(0.3) + (2)(0.4) = −0.1 + 0 + 0.3 + 0.8 = 1.0 (iii) Find Var(X): E(X²) = (1)(0.1) + (0)(0.2) + (1)(0.3) + (4)(0.4) = 0.1 + 0 + 0.3 + 1.6 = 2.0 Var(X) = E(X²) − [E(X)]² = 2.0 − 1.0 = 1.0 (iv) Find E(2X + 3): E(2X + 3) = 2 · E(X) + 3 = 2(1) + 3 = 5
5. X and Y are two independent random variables with E(X) = 3, Var(X) = 2, E(Y) = 4, Var(Y) = 5. Find: (i) E(3X − 2Y), (ii) Var(3X − 2Y), (iii) E(XY), (iv) Var(X + Y).
Answer
(i) E(3X − 2Y) = 3 · E(X) − 2 · E(Y) = 3(3) − 2(4) = 9 − 8 = 1 (ii) Var(3X − 2Y) = 3² · Var(X) + (−2)² · Var(Y) = 9(2) + 4(5) = 18 + 20 = 38 (Since X and Y are independent, variance of linear combination: Var(aX + bY) = a²Var(X) + b²Var(Y)) (iii) E(XY) = E(X) · E(Y) = 3 × 4 = 12 (since X and Y are independent) (iv) Var(X + Y) = Var(X) + Var(Y) = 2 + 5 = 7 (since X and Y are independent)
6. Find the moment generating function of X and use it to find E(X) and Var(X), given that X has the following distribution: P(X = 0) = 1/4, P(X = 1) = 1/2, P(X = 2) = 1/4.
Answer
Step 1: MGF M_X(t) = E(e^{tX}) = Σ e^{tx} · P(x) = e^{0} · (1/4) + e^{t} · (1/2) + e^{2t} · (1/4) = 1/4 + (1/2)e^{t} + (1/4)e^{2t} Step 2: Find E(X) M'_X(t) = (1/2)e^{t} + (1/2)e^{2t} M'_X(0) = 1/2 + 1/2 = 1 → E(X) = 1 Step 3: Find E(X²) M''_X(t) = (1/2)e^{t} + e^{2t} M''_X(0) = 1/2 + 1 = 3/2 → E(X²) = 3/2 Step 4: Find Var(X) Var(X) = E(X²) − [E(X)]² = 3/2 − 1 = 1/2
Common Mistakes to Avoid
1. Forgetting to verify that probabilities sum to 1 — Always check ΣP(x) = 1 before computing E(X) or Var(X). If probabilities don't sum to 1, there is an error in the PMF.
2. Confusing E(X²) with [E(X)]² — These are different. E(X²) is the expectation of X², computed as Σ x² P(x). [E(X)]² is the square of the mean. Var(X) = E(X²) − [E(X)]², not the other way around.
3. Applying independence-only properties without checking independence — E(XY) = E(X) · E(Y) and Var(X + Y) = Var(X) + Var(Y) only hold when X and Y are independent. E(X + Y) = E(X) + E(Y) always holds.
4. Incorrect variance of a linear transformation — Var(aX + b) = a²Var(X), NOT aVar(X). The multiplier is squared for variance. Also, adding a constant b does not change variance.
5. Computing E(X) incorrectly — Use E(X) = Σ x · P(x), not Σ x² · P(x). A common exam error is mixing up E(X) and E(X²).
6. Sign errors in variance of difference — Var(X − Y) = Var(X) + Var(Y) for independent variables, NOT Var(X) − Var(Y). Variance is always additive for independent variables regardless of whether you add or subtract.
Exam Checklist
- [ ] I can define a random variable and distinguish discrete from continuous types
- [ ] I can verify that a PMF sums to 1 before computing expectations
- [ ] I can compute E(X), E(X²), Var(X), and standard deviation from a probability distribution table
- [ ] I can apply the properties of expectation: E(c), E(aX + b), E(X + Y), E(XY) with independence
- [ ] I can apply the properties of variance: Var(c), Var(aX + b), Var(X + Y) with independence
- [ ] I can find the constant k when P(X = x) = k · f(x) using the condition ΣP(x) = 1
- [ ] I can use the moment generating function to find moments E(Xⁿ)
- [ ] I can solve complete problems: find k → find E(X) → find Var(X) → find E(aX + b) and Var(aX + b)
