Consider 100 i.i.d. random variables from an exponential distribution with
𝜆
=
1
3
. What is the variance of the sample sum
𝑆
100
?

Question

Consider 100 i.i.d. random variables from an exponential distribution with 
𝜆
=
1
3
.  What is the variance of the sample sum 
𝑆
100
?

BlackTom AI · Accepted Answer

For an exponential(λ) random variable, Var(X) = 1/λ^2, so first compute the variance of one observation and then scale by 100 for the sum.

1. Relevant knowledge points and concepts
- Exponential distribution: mean = 1/λ, variance = 1/λ^2.
- Independence and additivity of variance: for independent variables X1,...,Xn, Var(Sn) = Var(X1 + ... + Xn) = n Var(X1).

2. Step-by-step reasoning
- Given λ = 1/3.
- Compute λ^2: (1/3)^2 = 1/9.
- Compute variance of one exponential: Var(X) = 1/λ^2 = 1 / (1/9) = 9.
- For 100 i.i.d. observations, the sample sum S100 = X1 + ... + X100 has variance Var(S100) = 100 × Var(X) = 100 × 9.

3. Intermediate calculation steps
- λ^2 = (1/3)^2 = 1/9.
- 1/λ^2 = 1 ÷ (1/9) = 9.
- Var(S100) = 100 × 9 = 900.

4. Why the correct answer is correct
Because each exponential has variance 9 and variances add for independent random variables, the total variance of the sum of 100 such variables is 100×9 = 900. This directly follows from the property Var(aX+bY+...) = a^2Var(X)+b^2Var(Y)+... for independent variables with coefficients 1.

5. Why the other options are incorrect
- 1/9: equals Var(X) if λ = 3, not our λ = 1/3.
- 9: equals Var(X) for a single observation, not the sum of 100.
- 100/9: would be 100 × (1/9), which is the wrong reciprocal interpretation.
- 9/100: equals Var(sample mean) = Var(S100)/100^2 = 900/10000 = 0.09, not Var(S100).
- 1/900: unrelated; not produced by correct variance formulas here.

6. Conclusion
Therefore Var(S100) = 900, which matches the provided correct answer 900.

Consider 100 i.i.d. random variables from an exponential distribution with 𝜆 = 1 3 . What is the variance of the sample sum 𝑆 100 ?Single choice

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