Question
Asked Oct 25, 2019

Show that Y has a CDF that is U(a,b).

Let X be a uniform RV U(a, b) with CDF
{0
a
(r-a)/(b a) a <x <b
F(x)
for a <b
(a) Sketch the PDF of this distribution
(b) Let Z be a standard uniform random variable U(0, 1) and let Y be the
RV
Y =(b a)Z a.
Show that Y has a CDF that is U(a,b).
This means that a general uniform RV is just a linear transformation of
the standard one in the obvious way
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Let X be a uniform RV U(a, b) with CDF {0 a (r-a)/(b a) a <x <b F(x) for a <b (a) Sketch the PDF of this distribution (b) Let Z be a standard uniform random variable U(0, 1) and let Y be the RV Y =(b a)Z a. Show that Y has a CDF that is U(a,b). This means that a general uniform RV is just a linear transformation of the standard one in the obvious way

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check_circleExpert Solution
Step 1

a.

 

Show the graph of the probability density function of Uniform distribution.

 

From the given information, the random variable X be a uniform random variable, U(a,b) and its probability density function is  f(x)=1/(b-a)

 

The graph of the probability density function is,

f(X)
1/(Ь-а)
X
b
а
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f(X) 1/(Ь-а) X b а

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Step 2

b)

 

From the given information, Y=(b-a)Z+a and Z is a standard uniform random variable U(0,1).

The probability density function for Z is, f(Z)=1/(1-0)=1. The mean for Z is 1/2 and the variance for Z is 1/12.

 

The mean for Y=(b-a)Z+a is,

E(T) E[ (b-a)Z+a]
(b-a) E(Z)+E(a)
(6-a)4
b-a+2a
2
b+a
E(Y)=
2
(1)
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E(T) E[ (b-a)Z+a] (b-a) E(Z)+E(a) (6-a)4 b-a+2a 2 b+a E(Y)= 2 (1)

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Step 3

The variance for Y=(b-a)Z+a is,

...
V (r)=V[(b-a)Z+a]
=(b-a)v(Z)+V(a)
(constant) 0
(b-a)x
+0
12
(b-a)
V (r) =
(2)
12
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V (r)=V[(b-a)Z+a] =(b-a)v(Z)+V(a) (constant) 0 (b-a)x +0 12 (b-a) V (r) = (2) 12

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