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(ELEC210)b4c353 - quiz2_refsol.pdf
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ELEC210 Solutions to Quiz 2 16/11/09

1.

Suppose that X and Y are jointly continuous random variables with joint density
0..if ..... ..
..
0otherwise .,..

(a) What is the value of c?
?.and ...
(b)What are the marginal densities.
. .
(c)What is the probability that
.., ..
,

?

Solutions:
(a) The integration of the joint density function over the entire plane should be 1
. .
.. .. ce.... dx dy .
..
. ..
. ..

.. ..
.

.x, y.
..

..
f

dx d

f

dx dy

.
.
ydy .c.1
..

x, y
, so c=1;
X,Ye.. dx
X,Y
..
.e..
c

.x.
..
f

.
f

dy .dx e.... dy . e.., for x. 0e.... dx . e..
x,y ...

.x,y. X,Y ...
. .
(c)The probability that
. ..
.
y. dxd....e ..

.. .
.
X
X,Y
f
(b)
;

fY.y. .
.
, for y. 0
...

;

.

is
.
.

e..dx.dy .
e...1 . e...dy . 1
2

.

...

...

e..
The number of mistakes made by a typesetter on any given page is a random
.
2.

, independent for each page.
.
variable having the Poisson distribution with parameter
A three-page article is prepared by one of four typesetters, for which the values of
are 1, 2, 3 and 4, respectively. The typesetter is selected randomly with uniform probability. What is the expected number of misprints throughout the article?
Solutions: The expected value of a Poisson distribution with parameter . is just .. So if the selected typesetter is typesetter 1, then the expected number of misprints is 1*3=3. And if the selected typesetter is the second one, the expected number of misprints is 2*3=6. If the selected typesetter is typesetter 3, the expected number of misprints will be 3*3=9. And if the 4th typesetter is selected, the expected number of misprints will be 4*3=12. As the typesetter is selected randomly with equal probability, so the expected number of misprints will be ... .3.1 4.6.1 4.9.1 4.12. 1 4.7.5
1 |2
3. Suppose that a continuous valued signal S is uniformly distributed on [0, 2], and that you measure the signal corrupted by noise twice:
V1 = S + N1
V2 = S + N2

where the noises N1 and N2 are independent of each other and of S. Assume that N1 and N2 are uniform on [-0.5, 0.5].
(a) Find the cross correlation between S and V1: E[SV1].
r
(b) Find the mean vector and covariance matrix of V = [V1 V2]T
Solutions:
(a) E.SV.. .E.S.S.N... .E.S. .SN.. .E.S.. .E.SN.. .E.S.. .E.S.E.N.. As S is uniformly distributed on [0,2], so E[S2]=4/3. And E[N1]=0, so E[SV1]=4/3.
(b) E[V1]=E[S]+E[N1]=1, E[V2]=E[S]+E[N2]=1, so the mean vector is .11.T. VAR[V1]=VAR[S]+VAR[N1]=5/12, VAR[V2]=5/12, E[V1V2]-E[V1]E[V2]=E[S2]-E[S]E[S]=4/3-1=1/3, 5/12 1/31/3
so the covariance matrix is .5/12..
4. Suppose X and Y are discrete random variables with joint pmf shown in the table below

(a)
Find the marginal probability mass function of X and Y

(b)
Find the probability of the following events
{X < Y}
{Y2 C 4Y + 3 < 0}



Solutions:
(a) The