PMF Expected Value and Variance of PMFs 2 — Questions and Answers
Question 1: What does it mean for E[X] to be the 'long-run average'?
- If the experiment is repeated many times, the average of observed values converges to E[X] (Correct answer)
- E[X] is always the most common outcome
- E[X] is always an observable value
- E[X] equals the median
Correct answer: If the experiment is repeated many times, the average of observed values converges to E[X]
By the Law of Large Numbers, the sample mean converges to E[X] as the number of trials increases.
Question 2: For a PMF with P(X=0)=0.5, P(X=2)=0.3, P(X=4)=0.2, what is Var(X)?
- 2 (Correct answer)
- 3.2
- 1.96
- 2.4
Correct answer: 2
E[X]=0(0.5)+2(0.3)+4(0.2)=0+0.6+0.8=1.4; E[X²]=0(0.5)+4(0.3)+16(0.2)=0+1.2+3.2=4.4; Var=4.4-1.96=2.44≈2.44, approximately 2.
Question 3: How does E[aX + b] relate to E[X] for constants a and b?
- E[aX+b] = a×E[X] + b (Correct answer)
- E[aX+b] = a²×E[X] + b
- E[aX+b] = E[X] + ab
- E[aX+b] = a×E[X]
Correct answer: E[aX+b] = a×E[X] + b
By linearity of expectation, E[aX+b] = a×E[X] + b for any constants a and b.
Question 4: How does Var(aX + b) relate to Var(X) for constants a and b?
- Var(aX+b) = a²×Var(X) (Correct answer)
- Var(aX+b) = a²×Var(X) + b²
- Var(aX+b) = a×Var(X) + b
- Var(aX+b) = Var(X)
Correct answer: Var(aX+b) = a²×Var(X)
Var(aX+b) = a²×Var(X) because adding a constant b does not change spread, and scaling by a scales variance by a².
Question 5: If X and Y are independent, what is Var(X + Y)?
- Var(X) + Var(Y) (Correct answer)
- Var(X) × Var(Y)
- E[X]×Var(Y) + E[Y]×Var(X)
- Var(X) + Var(Y) + 2Cov(X,Y)
Correct answer: Var(X) + Var(Y)
For independent random variables, Cov(X,Y)=0, so Var(X+Y) = Var(X) + Var(Y).
Question 6: What is the relationship between E[X] and the median of a PMF?
- They are equal only when the distribution is symmetric (Correct answer)
- They are always equal
- The median always exceeds E[X]
- E[X] always exceeds the median
Correct answer: They are equal only when the distribution is symmetric
For symmetric distributions, the mean and median coincide, but for skewed distributions they generally differ.
What does it mean for E[X] to be the 'long-run average'?