Exponentially Distributed - 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: Exponentially Distributed, Continuous, Random Variable, Expected, Positive Integer, Geometric Distribution, Probability, Geometrically, Distributed, Value

Typology: Exams

2012/2013

Uploaded on 02/27/2013

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Probability: Problem Set 4 Solutions
Fall 2009
Instructor: W. D. Gillam
(1) Let Xbe an exponentially distributed (continuous) random variable with expected
value β. Define a (discrete!) random variable Yby setting
P(Y=k) := P(k1X < k)
for each positive integer k. Note
X
k=1
Y(k) =
X
k=1
P(k1X < k)
= 1.
Show that Yhas a geometric distribution with expected value
1
1e1 .
Solution. Set p:= 1 e1 ,q:= 1 p=e1 . Then
P(Y=k) = Zk
k1
ex/β
βdx
=hex/βik
k1
=ek/β +e(k1)
=e(k1)(1 e1 )
=pqk1,
which is exactly the probability P(Y=k) when Yis a geometrically distributed
discrete random variable with the indicated expected value.
(2) Fix a positive real number b. For CR, set
f(x) := ½Cx2, x b,
0,otherwise.
Find the value of C(this may depend on b) that makes fa density function. Find
the expected value and variance of the corresponding random variable. Find the
corresponding cumulative distribution function Fand graph fand Fon the same
axes.
Solution. Since
Z
b
x2dx =£x1¤
b
=1
b,
1
pf3

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Probability: Problem Set 4 Solutions

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

] 1

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?