

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: Exponentially Distributed, Continuous, Random Variable, Expected, Positive Integer, Geometric Distribution, Probability, Geometrically, Distributed, Value
Typology: Exams
1 / 3
This page cannot be seen from the preview
Don't miss anything!


Fall 2009 Instructor: W. D. Gillam
(1) Let X be an exponentially distributed (continuous) random variable with expected value β. Define a (discrete!) random variable Y by setting
P (Y = k) := P (k − 1 ≤ X < k) for each positive integer k. Note ∑^ ∞
k=
Y (k) =
k=
P (k − 1 ≤ X < k)
= 1. Show that Y has a geometric distribution with expected value 1 1 − e−^1 /β^
Solution. Set p := 1 − e−^1 /β^ , q := 1 − p = e−^1 /β^. Then
P (Y = k) =
∫ (^) k
k− 1
e−x/β β
dx
−e−x/β^
]k
k− 1 = −e−k/β^ + e−(k−1)/β = e−(k−1)/β^ (1 − e−^1 /β^ ) = pqk−^1 , which is exactly the probability P (Y = k) when Y is a geometrically distributed discrete random variable with the indicated expected value.
(2) Fix a positive real number b. For C ∈ R, set
f (x) :=
Cx−^2 , x ≥ b, 0 , otherwise. Find the value of C (this may depend on b) that makes f a density function. Find the expected value and variance of the corresponding random variable. Find the corresponding cumulative distribution function F and graph f and F on the same axes.
Solution. Since ∫ (^) ∞
b
x−^2 dx =
−x−^1
b
=
b
1
2
we should set C = b. Neither the expected value nor the variance exist because ∫ (^) ∞
b
x(bx−^2 ) = ∞.
The cumulative distribution function is given by
F (x) =
∫ (^) x
b
by−^2 dy
=
−by−^1
]x b
=
1 − (^) xb , x ≥ b 0 , x < b.
(3) For C ∈ R, set
f (x) :=
Cx^2 (1 − x)^3 , 0 ≤ x ≤ 1 , 0 , otherwise. Find the value of C that makes f a density function. Find the expected value and variance of the corresponding random variable.
Solution. Since ∫ (^1)
0
x^2 (1 − x)^3 dx =
0
−x^5 + 3x^4 − 3 x^3 + x^2
dx
x^6 6
3 x^5 5
3 x^4 4
x^3 3
0 = −(1/6) + (3/5) − (3/4) + (1/3) = 1 / 60 , we should set C = 60. The expected value of X is given by
E(X) =
0
x^3 (1 − x)^3 dx
and the variance of X is given by
V (X) =
0
x^4 (1 − x)^2 dx − E(X)^2 ,
which can be calculated by elementary calculus as above.
(4) Let X be a continuous random variable uniformly distributed on the interval [a, b]. Show that X is “memoryless” in the sense that P (X ≥ c + d|X ≥ c) = P (X ≥ a + d|X ≥ a) as long as c ≥ a and c + d ≤ b. (5) The lifespan of a certain kind of fuse has an exponential distribution with an expected lifespan of 3 years. What is the probability that it will fail in the first year? What is Chebyshev’s upper bound for the probability that the fuse will last at least n years? How does this compare with the actual probability that the fuse will last at least n years?