Density Function - Probability - Solved Exam, Exams of Probability and Statistics

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

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Probability: Midterm 2 Solutions
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 Xwith normal distri-
bution with expected value µand variance σ2. Express P(aXb) as a definite
integral of the standard normal density function.
Solution. (Xµ)/σ2has a standard normal distribution, so
P(aXb) = P(aµ
σ2Xµ
σ2bµ
σ2)
=1
2πZbµ
σ2
aµ
σ2
ex2/2dx.
(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
ndegrees of freedom. Explain how the chi squared distribution and Student’s
distribution arise from the normal distribution.
(4) (4 points) Suppose Xhas a binomial distribution based on ntrials 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 Xhas a beta distribution with parameters α=β= 1/2.
Find the cumulative distribution function FX(x) of X.
Solution. The density function for Xis given by
f(y) = Γ(1)
Γ(1/2)2y1/2(1 y)1/2
=1
π
1
y1y
pf3

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Probability: Midterm 2 Solutions

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) =

Γ(1/2)^2

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

0

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

] 1

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