

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
This is the Solved Exam of Probability which includes Exponentially Distributed, Continuous, Random Variable, Expected, Positive Integer, Geometric Distribution, Probability, Geometrically, Distributed, Value etc. Key important points are: Density Function, Definitions, Functional Equation, Gamma Function, Normal, Expected Value, Standard Normal Density Function, Standard Normal Distribution, Student Distributions, Function Specializes
Typology: Exams
1 / 3
This page cannot be seen from the preview
Don't miss anything!


Fall, 2009 Instructor: W. D. Gillam
(1) (3 points) Give the definitions of the beta and gamma functions, the relationship between them, and the functional equation for the gamma function. (No proofs required.) (2) (4 points) Give the density function for a random variable X with normal distri- bution with expected value μ and variance σ^2. Express P (a ≤ X ≤ b) as a definite integral of the standard normal density function.
Solution. (X − μ)/
σ^2 has a standard normal distribution, so
P (a ≤ X ≤ b) = P (
a − μ √ σ^2
X − μ √ σ^2
b − μ √ σ^2
2 π
∫ (^) √b−μ σ^2 √^ a−μ σ^2
e−x
(^2) / 2 dx.
(3) (6 points) Give the density functions and expected values for the gamma, beta, and Student distributions. Indicate the choices of the parameters α, β for which the gamma density function specializes to the chi squared density function with n degrees of freedom. Explain how the chi squared distribution and Student’s distribution arise from the normal distribution. (4) (4 points) Suppose X has a binomial distribution based on n trials with success probability p. Calculate the moment generating function gX (t) of X. Hint: You can use multiplicativity of MGFs under sums of independent random variables to reduce to the case n = 1.
Solution. When n = 1, this is gX (t) = E(etX^ ) = pet·^1 + qet·^0 = pet^ + q, so in general it is gX (t) = (pet^ + q)n.
(5) (4 points) Suppose X has a beta distribution with parameters α = β = 1/2. Find the cumulative distribution function FX (x) of X.
Solution. The density function for X is given by
f (y) =
y−^1 /^2 (1 − y)−^1 /^2
π
y
1 − y
2
(supported on [0, 1]). For x with 0 ≤ x ≤ 1, we compute 1 π
∫ (^) x
0
y
1 − y
dy =
π
arcsin
y
]x
π
arcsin
x, so
FX (x) =
0 , x ≤ 0 (2/π) arcsin
x, 0 ≤ x ≤ 1 1 , x ≥ 1
(6) (4 points) Suppose X and Y are IIDRV, each with density function
f (x) =
2 x, 0 ≤ x ≤ 1 0 , otherwise. Calculate the density function of X + Y.
Solution. The desired density is given by
g(x) =
−∞
f (y)f (x − y)dy.
The integrand will vanish unless 0 ≤ y ≤ 1 and 0 ≤ x − y ≤ 1. The second inequality is equivalent to x − 1 ≤ y ≤ x. So the integrand is zero if x < 0 or if x > 2. If 0 ≤ x ≤ 1 it is given by
g(x) =
∫ (^) x
0
(2y)(2(x − y))dy
∫ (^) x
0
4 xy − 4 y^2 dy
=
2 xy^2 − (4/3)y^3
]x 0 = (2/3)x^3. When 1 ≤ x ≤ 2, it is given by
g(x) =
x− 1
4 xy − 4 y^2
2 xy^2 − (4/3)y^3
x− 1 = 2 x − 4 / 3 − 2 x(x − 1)^2 + (4/3)(x − 1)^3.
(7) (4 points) A point P is chosen randomly (uniformly) along a thin rod (a coat hanger, say). The rod is then bent to form a right angle at P , so that the bent rod forms the two shorter sides of a right triangle. Let θ denote the smallest angle in this triangle. Find the expected value of tan θ.
Solution. By using the appropriate units, we may assume the length of the rod is 1, so that P is uniformly distributed on [0, 1]. Then the two shorter sides of