ISYE 6644 ASY Probabilistic Models Exam 1
Simulation ACTUAL EXAM 2026/2027 |
Georgia Tech | Verified Q&A | Pass
Guaranteed - A+ Graded
Section A: Probability Review – Random Variables, Distributions, Expectation, Variance, MGFs (Q1–15)
Q1: Let X ~ Exponential(λ) with PDF f(x) = λe^,-λx- for x ≥ 0. The moment generating function (MGF) of X
is M_X(t) = E*e^,tX-+. For t < λ, this MGF equals:
A. λ / (λ + t)
B. λ / (λ - t) [CORRECT]
C. (λ - t) / λ
D. (λ + t) / λ
Correct Answer: B
Rationale: M_X(t) = ∫_0^∞ e^,tx- λ e^,-λx- dx = λ ∫_0^∞ e^,-(λ-t)x- dx = λ / (λ-t) for t < λ, which is the
standard exponential MGF.
Q2: A linear congruential generator has parameters m = 16, a = 5, c = 3, X₀ = 2. The next two random
numbers U₁ and U₂ (using U = X/m) are:
A. U₁ = 0.4375, U₂ = 0.5625
B. U₁ = 0.8125, U₂ = 0.4375
C. U₁ = 0.8125, U₂ = 0.5625 [CORRECT]
D. U₁ = 0.5625, U₂ = 0.8125
Correct Answer: C
Rationale: X₁ = (5×2 + 3) mod 16 = (10+3) mod 16 = 13, U₁ = 13/16 = 0.8125. X₂ = (5×13+3) mod 16 =
(65+3) mod 16 = 68 mod 16 = 4, U₂ = 4/16 = 0.5625.
Q3: Using the inverse transform method, derive a random variate X from the distribution with CDF F(x) =
1 - e^{-x^2- for x ≥ 0. Set U ~ Uniform(0,1). Then X equals:
,A. sqrt(-ln(1-U)) [CORRECT]
B. -ln(1-U)
C. sqrt(-ln(U))
D. -ln(U)
Correct Answer: A
Rationale: Set F(X) = 1 - e^{-X²} = U, then e^{-X²} = 1-U, so -X² = ln(1-U), thus X = sqrt(-ln(1-U)). Note that
1-U is also Uniform(0,1), so sqrt(-ln(U)) is equivalent in distribution.
Q4: Let X ~ Poisson(λ). The probability mass function is P(X = k) = (e^,-λ- λ^k) / k! for k = 0, 1, 2, ... The
expected value E[X] equals:
A. λ²
B. λ [CORRECT]
C. e^λ
D. 1/λ
Correct Answer: B
Rationale: For a Poisson random variable with parameter λ, the expected value E*X+ = λ. This follows
directly from the definition of the Poisson distribution as the limit of binomial distributions where np →
λ, or by summing k·P(X=k) over all k and simplifying using the Taylor series for e^λ.
Q5: Let X ~ Normal(μ, σ²). The moment generating function M_X(t) equals:
A. e^,μt-
B. e^,μt + σ²t²/2- [CORRECT]
C. e^,μt - σ²t²/2-
D. e^,μt + σt-
Correct Answer: B
Rationale: The MGF of a normal random variable X ~ N(μ, σ²) is derived by completing the square in the
exponent of the integral E*e^,tX-+. The result is M_X(t) = exp(μt + σ²t²/2), which exists for all real t. This
MGF uniquely characterizes the normal distribution and is fundamental for deriving distributions of
sums of independent normal random variables.
Q6: Let X ~ Binomial(n, p). The variance Var(X) equals:
A. np
B. np(1-p) [CORRECT]
, C. np²
D. n(1-p)
Correct Answer: B
Rationale: For a binomial random variable X ~ Binomial(n, p), the variance is Var(X) = np(1-p). This can
be derived by noting that X is the sum of n independent Bernoulli(p) random variables, each with
variance p(1-p), so the variance of the sum is the sum of the variances: n·p(1-p).
Q7: Let X and Y be independent random variables with MGFs M_X(t) and M_Y(t). The MGF of Z = X + Y is:
A. M_X(t) + M_Y(t)
B. M_X(t) · M_Y(t) [CORRECT]
C. M_X(t) / M_Y(t)
D. M_X(t) - M_Y(t)
Correct Answer: B
Rationale: For independent random variables X and Y, the MGF of their sum Z = X + Y is the product of
their individual MGFs: M_Z(t) = E[e^{t(X+Y)}] = E[e^{tX}]·E[e^{tY}] = M_X(t)·M_Y(t). This follows from the
independence assumption allowing the expectation of the product to factor into the product of
expectations.
Q8: Let X ~ Uniform(a, b). The expected value E[X] equals:
A. (b - a) / 2
B. (a + b) / 2 [CORRECT]
C. a + b
D. (b - a)² / 12
Correct Answer: B
Rationale: For a continuous uniform random variable X ~ Uniform(a, b), the PDF is constant at 1/(b-a)
over *a, b+. The expected value is the midpoint of the interval: E*X+ = ∫_a^b x·(1/(b-a)) dx = (a+b)/2. This
reflects the symmetry of the uniform distribution about its center.
Q9: Let X ~ Gamma(α, β) with shape parameter α and rate parameter β. The expected value E*X+ equals:
A. α / β [CORRECT]
B. α · β
C. β / α
D. α + β
Simulation ACTUAL EXAM 2026/2027 |
Georgia Tech | Verified Q&A | Pass
Guaranteed - A+ Graded
Section A: Probability Review – Random Variables, Distributions, Expectation, Variance, MGFs (Q1–15)
Q1: Let X ~ Exponential(λ) with PDF f(x) = λe^,-λx- for x ≥ 0. The moment generating function (MGF) of X
is M_X(t) = E*e^,tX-+. For t < λ, this MGF equals:
A. λ / (λ + t)
B. λ / (λ - t) [CORRECT]
C. (λ - t) / λ
D. (λ + t) / λ
Correct Answer: B
Rationale: M_X(t) = ∫_0^∞ e^,tx- λ e^,-λx- dx = λ ∫_0^∞ e^,-(λ-t)x- dx = λ / (λ-t) for t < λ, which is the
standard exponential MGF.
Q2: A linear congruential generator has parameters m = 16, a = 5, c = 3, X₀ = 2. The next two random
numbers U₁ and U₂ (using U = X/m) are:
A. U₁ = 0.4375, U₂ = 0.5625
B. U₁ = 0.8125, U₂ = 0.4375
C. U₁ = 0.8125, U₂ = 0.5625 [CORRECT]
D. U₁ = 0.5625, U₂ = 0.8125
Correct Answer: C
Rationale: X₁ = (5×2 + 3) mod 16 = (10+3) mod 16 = 13, U₁ = 13/16 = 0.8125. X₂ = (5×13+3) mod 16 =
(65+3) mod 16 = 68 mod 16 = 4, U₂ = 4/16 = 0.5625.
Q3: Using the inverse transform method, derive a random variate X from the distribution with CDF F(x) =
1 - e^{-x^2- for x ≥ 0. Set U ~ Uniform(0,1). Then X equals:
,A. sqrt(-ln(1-U)) [CORRECT]
B. -ln(1-U)
C. sqrt(-ln(U))
D. -ln(U)
Correct Answer: A
Rationale: Set F(X) = 1 - e^{-X²} = U, then e^{-X²} = 1-U, so -X² = ln(1-U), thus X = sqrt(-ln(1-U)). Note that
1-U is also Uniform(0,1), so sqrt(-ln(U)) is equivalent in distribution.
Q4: Let X ~ Poisson(λ). The probability mass function is P(X = k) = (e^,-λ- λ^k) / k! for k = 0, 1, 2, ... The
expected value E[X] equals:
A. λ²
B. λ [CORRECT]
C. e^λ
D. 1/λ
Correct Answer: B
Rationale: For a Poisson random variable with parameter λ, the expected value E*X+ = λ. This follows
directly from the definition of the Poisson distribution as the limit of binomial distributions where np →
λ, or by summing k·P(X=k) over all k and simplifying using the Taylor series for e^λ.
Q5: Let X ~ Normal(μ, σ²). The moment generating function M_X(t) equals:
A. e^,μt-
B. e^,μt + σ²t²/2- [CORRECT]
C. e^,μt - σ²t²/2-
D. e^,μt + σt-
Correct Answer: B
Rationale: The MGF of a normal random variable X ~ N(μ, σ²) is derived by completing the square in the
exponent of the integral E*e^,tX-+. The result is M_X(t) = exp(μt + σ²t²/2), which exists for all real t. This
MGF uniquely characterizes the normal distribution and is fundamental for deriving distributions of
sums of independent normal random variables.
Q6: Let X ~ Binomial(n, p). The variance Var(X) equals:
A. np
B. np(1-p) [CORRECT]
, C. np²
D. n(1-p)
Correct Answer: B
Rationale: For a binomial random variable X ~ Binomial(n, p), the variance is Var(X) = np(1-p). This can
be derived by noting that X is the sum of n independent Bernoulli(p) random variables, each with
variance p(1-p), so the variance of the sum is the sum of the variances: n·p(1-p).
Q7: Let X and Y be independent random variables with MGFs M_X(t) and M_Y(t). The MGF of Z = X + Y is:
A. M_X(t) + M_Y(t)
B. M_X(t) · M_Y(t) [CORRECT]
C. M_X(t) / M_Y(t)
D. M_X(t) - M_Y(t)
Correct Answer: B
Rationale: For independent random variables X and Y, the MGF of their sum Z = X + Y is the product of
their individual MGFs: M_Z(t) = E[e^{t(X+Y)}] = E[e^{tX}]·E[e^{tY}] = M_X(t)·M_Y(t). This follows from the
independence assumption allowing the expectation of the product to factor into the product of
expectations.
Q8: Let X ~ Uniform(a, b). The expected value E[X] equals:
A. (b - a) / 2
B. (a + b) / 2 [CORRECT]
C. a + b
D. (b - a)² / 12
Correct Answer: B
Rationale: For a continuous uniform random variable X ~ Uniform(a, b), the PDF is constant at 1/(b-a)
over *a, b+. The expected value is the midpoint of the interval: E*X+ = ∫_a^b x·(1/(b-a)) dx = (a+b)/2. This
reflects the symmetry of the uniform distribution about its center.
Q9: Let X ~ Gamma(α, β) with shape parameter α and rate parameter β. The expected value E*X+ equals:
A. α / β [CORRECT]
B. α · β
C. β / α
D. α + β
