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Midterm Solutions: Jacobian and Independent Random Variables in Probability Distributions, Exams of Probability and Statistics

Solutions to midterm problems related to finding the jacobian and checking the independence of random variables in probability distributions. Topics include the jacobian matrix, marginalization, and the characteristic function of a poisson distribution.

Typology: Exams

2012/2013

Uploaded on 04/01/2013

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Midterm 2 Solutions
EE126 - Fall 2000
Problem 1.
X Y
unif
(
;
1
1)
Z
=
XY
Let
U
=
X
and
V
=
XY
and form the jacobian for
f
UV
(
u v
).
f
UV
=
1
j
U
j
f
XY
(
u v=u
)
Marginalize with respect to u and employ the independence of X and Y.
f
V
(
v
)=
f
Z
(
z
)=
Z
+
1
;1
1
j
U
j
f
X
(
u
)
f
Y
(
z=u
)
du
f
Z
(
z
)=
Z
+
1
;1
1
2
j
U
j
f
Y
(
z=u
)
du
The domain for
f
Y
is bounded awayfrom 0bythe
z=u
term.
First consider
z >
0 for
f
Y
(
z=u
)
6
=0,1
>
j
u
j
>z
. Therefore, the integral
above can be completed for
z>
0by splitting it up into when
z<u<
1 and
;
1
<u<
;
z
.
so, for
z>
0
f
Z
(
z
)=
Z
1
z
1
4
U
du
+
Z
;
z
;
1
1
;
4
U
du
f
Z
(
z
)=
Z
1
z
1
2
U
du
f
Z
(
z
)=
;
1
2
ln
(
z
)
now, similarly,for
z<
0
f
Z
(
z
)=
Z
;
z
;
1
1
4
U
du
+
Z
1
z
1
;
4
U
du
f
Z
(
z
)=
Z
1
z
1
;
2
U
du
1
pf3
pf4

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Download Midterm Solutions: Jacobian and Independent Random Variables in Probability Distributions and more Exams Probability and Statistics in PDF only on Docsity!

Midterm  Solutions EE  Fall 

Problem 

X  Y  unif  

Z  X Y Let U  X and V  X Y and form the jacobian for fU V u v 

fU V 

jU j

fX Y u v u

Marginalize with resp ect to u and employ the indep endence of X and Y

fV v   fZ z  

Z 



jU j

fX ufY z udu

fZ z  

Z 



jU j

fY z udu

The domain for fY is b ounded away from  by the z u term First consider z   for fY z u     juj  z  Therefore the integral ab ove can b e completed for z   by splitting it up into when z  u   and   u  z  so for z  

fZ z  

Z 

z

U

du

Z z



 U

du

fZ z  

Z 

z

U

du

fZ z   

l nz 

now similarly for z  

fZ z  

Z z



U

du

Z 

z

 U

du

fZ z  

Z 

z

U

du

fZ z   

l nz  so

fZ z   

l njz j

it is unde ned for z but that s ok b ecause it happ ens with a probability of  The integral can b e evaluated in a closed form though b since X is indep endent of Y

E X Y   E X E Y   

Problem 

P T  t 

 t

F t 

t  t

fT t 

dFT dt

 t b

P atl eastbul bw or k ing att  g iv enw or k ing att  

   P w or k ing att  j w or k ing att  



P att   att   P  att  

P t   t   

t t 

 t

P t   

P w or k ing att  jal l w or k ing att    



P atl eastbul b       



c

Z  minT  T  T  T 

P Z  t 

P T  tP T  tP T  tP T  t

fr  r   r  fX Y r cos r sin

Problem  extra credit

The characteristic function of a Poission rv is

Gs  exps  

The sum of two indep endent rv s is the pro duct of their characteristic func tions H s  expas    expbs    expa bs   so converting back the characteristic function which is in standard form

fZ k   a bk k 

expa bk