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University Solutions: Problem Set 13 ECE 313 of Illinois Fall 1998
1. fX 1 (u) = fX 2 (u) = fX 3 (u) = 1 = 2 ; 0 u 2, and 0 elsewhere.
(a) There are 6 (3!) arrangements of X 1 ; X 2 , and X 3 among themselves, and each of these arrange- ments is equally likely since X 1 ; X 2 ; X 3 are i.i.d. Hence, each arrangement has probability 1 = 6 of o ccurring | is this reminiscent of Problem 6 in Homework 11? Of these 6 arrangements, X 2 o ccurs in the middle twice, (X 1 < X 2 < X 3 ) and (X 3 < X 2 < X 1 ). Hence the probability that X 2 lies b etween X 1 and X 3 is 2 = 6 = 1 =3. (b) Let N 1 ; N 2 and N 3 denote the smallest, middle and largest, resp ectively, of the three p oints X 1 ; X 2 and X 3 (note that X 1 do esn't have to b e the smallest). We derived the distribution of N 1 ; N 2 and N 3 in the previous homework.
FN 3 (u) = F (^) X^3 (u) =
0 ;^ u^ <^0
u 2
; 0 u < 2
1 ; u 2
Therefore P fN 3 > 1 g = 1 FN 3 (1) = 1 1 = 8 = 7 =8.
(c) Any of the three p oints can b e the largest. Therefore, the desired probability is
P fX 1 > X 2 + X 3 g + P fX 2 > X 1 + X 3 g + P fX 3 > X 1 + X 2 g = 3 P fX 1 > X 2 + X 3 g;
where the last equality follows b ecause the three RVs are iid. The only thing in calculating
P fX 1 > X 2 + X 3 g is in nding the correct integration limits. If u 1 ; u 2 and u 3 denote the
dummy variables for X 1 ; X 2 and X 3 , then we have u 3 2 [0; u 1 u 2 ]; u 2 2 [0; u 1 ] and u 1 2 [0; 2].
Therefore
P fX 1 > X 2 + X 3 g =
Z 2
u 1 =
Z u 1
u 2 =
Z u 1 u 2
u 3 =
du 3 du 2 du 1 = 1 6
and from this, we get the desired probability as 3 1 = 6 = 1 =2.
2. (a) Let pi = pfXYZ = ig denote the pmf of XYZ. The RV XYZ takes on 8 values (2 each for
each of X; Y and Z), 4 of which are distinct | 1 ; 2 ; 4, and 8.
p 1 = 18 ; p 2 = 38 ; p 4 = 38 ; p 8 = (^18)
(b) Let pi = pfXY + YZ + XZg denote the pmf of XY + YZ + XZ. This random variable takes
on 8 values again (why?), 4 of which are again distinct | 3 ; 5 ; 8 and 12.
p 3 = 1 8
; p 5 = 3 8
; p 8 = 3 8
; p 12 = 1 8
(c) Let pi = pfX^2 + Y 2 + Z^2 g denote the pmf of X^2 + Y 2 + Z^2. This random variable takes on 8
values (why?), of which again 4 are distinct | 3 ; 6 ; 9 and 12.
p 3 =
8 ;^ p^6 =^
8 ;^ p^9 =^
8 ;^ p^12 =^
- U is uniform on (0; 2 ) and Z, indep endent of U, is exp onential with parameter = 1.
(a) De ne two new random variables by
X =
p
2 Z cos U ; Y =
p
2 Z sin U
The system of equations that follows has a unique solution
a =
p
2 z cos u b =
p
2 z sin u )^
z 1 = a
(^2) + b 2 2 u 1 = tan ^1
b a
The ranges of a and b are: 1 < a < 1 ; 1 < b < 1. The Jacobian of the transformation is
J (z ; u) =
cos p u
2 z
p
2 z sin u
sin p u
2 z
p
2 z cos u
= 1 6 = 0 ; for all z ; u
Therefore
fX ;Y (a; b) =
fZ;U (z 1 ; u 1 )
jJ (z 1 ; u 1 )j =^
2 e
a^2 + 2 b^2 ; 1 < a < 1 ; 1 < b < 1
(b) From the form of fX ;Y (a; b) ab ove, we see that it can b e written as
fX ;Y (a; b) =
p^1
e ^
a 22 ^
p^1
e ^
b 22 ^
= fX (a) fY (b)
where X and Y are b oth distributed as N (0; 1). (c) De ne the new random variable R =
p
X^2 + Y 2. There are many ways to calculate the p df of R. The easiest is to realize that R =
p
2 Z. We know that Z exp (1), i.e.,
fZ (z ) =
e z^ ; z 0
0 ; z < 0
Therefore, with v = g (z ) =
p
2 z as our transformation, g 0 (z 1 ) = 1 =
p
2 z = 1 =v. Now we get
fR (v ) = fz^ (v^
p
2 z 1
v e ^ v^
2
2 ; v 0
0 ; v < 0
which is plotted in Figure 1.
- The densities of X 1 and X 2 are given by
fX 1 (u) =
2 ;^ ^1 ^ u^ ^1
0 ; elsewhere ;^ fX^2 (u)^ =
e u^ ; u 0
0 ; u < 0
(a) W = X 1 + X 2 and Z = X 1 =X 2.
(^0) −1 0 5
u −−>
pdf of W = X1 + X
−5^0 0
u −−>
pdf of W = X1/X
(a) (b) Figure 3: p dfs for Problem 4a.
(b) To nd the joint p df fW;Z (a; b), rst nd the solutions to the set of equations: a = u + v
b = u=v )^
u 1 = (^) bab+ v 1 = (^) b+1a Now calculate the Jacobian of the transformation at the unique solution p oint.
J (u; v ) = 11 =v^ u=v^1
u 1 ;v 1
v
u
v
u 1 ;v 1
= (b^ +^ 1)
2 a
The Jacobian is 0 for b = 1, i.e., X 1 = X 2 , but since P fX 1 = X 2 g = 0 (why?), we don't
have to worry ab out it. Now the joint density is
fW;Z (a; b) =
fX 1 ;X 2 (u 1 ; v 1 )
jJ (u 1 ; v 1 )j =^
jaj
2(b + 1)^2 exp
a b + 1
The ranges for a and b have to b e evaluated carefully. The uniform density imp oses the following
limits on a and b: 1 b ab+ 1 1, and the exp onential density imp oses b +a 1 0. These
two equations are a little tricky to solve: the second equation yields either a 0 ; b 1
OR a 0 ; b 1. The rst set together with 1 ab=(b + 1) 1 yields two regions:
1 a 0 ; 1 < b 1 =(1 + a) and 0 a < 1 ; 1 =(1 + a) b 0. The second
set together with 1 ab=(b + 1) 1 yields two regions: 0 a 1 ; 0 b < 1 and
1 a < 1 ; 0 b 1 =(a 1). These are graphed in Figure 4.
- For the rst part (X; Y ) is uniformly distributed on the interior of the square with vertices at (0; 0); (0; 1 ); (1; 1) and (1; 0). The marginals in this case are uniform on (0; 1), and the joint density is therefore a pro duct of the marginals. (Whenever the joint p df is constant over a rectangular region, you will b e able to write it as a pro duct of the marginals.)
(a) From the ab ove reasoning, X and Y are indep endent and hence uncorrelated.
(b) Now we determine if Z = X + Y and W = X Y are uncorrelated and/or indep endent. First
calculate the covariance of Z and W and check to see if it is zero. For starters,
E [Z] = E [X] + E [Y ] = 1 ; E [W ] = E [X] E [Y ] = 0 (why is this?)
Var(Z) = Var(X) + Var(Y ) = 1 = 6 ; Var(W ) = Var(X) + Var(Y ) = 1 = 6
-1 0 1 a
b
a = 1 + (^1) b
a = 1 ^1 b
Figure 4: Figure for Problem 4b.
So, Cov(Z; W ) = E [ZW ] E [Z]E [W ] = E [(X + Y )(X Y )] since W is zero mean. E [(X +
Y )(X Y )] = E [X^2 Y 2 ] = E [X^2 ] E [Y 2 ] = 0 b ecause E [X^2 ] = E [Y 2 ] = 1 =3; hence they
are uncorrelated. To determine if Z and W are indep endent, we have to evaluate their joint
density. First solve the equations a = u + v and b = u v , where a and b are particular values
for Z and W resp ectively. The unique solution is: u 1 = a^ +^ b 2
; v 1 = a^ ^ b
. The Jacobian of the transformation is
J (u; v ) = 11 ^11 = 2 6 = 0
Therefore
fZ;W (a; b) = fX^ ;Y^ (u;^ v^ )
jJ (u 1 ; v 1 )j
; 0 a^ +^ b
1 ; 0 a^ ^ b
and this is graphed in Figure 5(b) | we see that the supp ort of the joint p df is a rhombus, and hence Z and W cannot b e indep endent.
So X + Y and X Y are uncorrelated but not indep endent.
(c) (X; Y ) uniformly distributed on the interior of a square with corners at (2; 0); (1; 1), (0; 0), and
Clearly X and Y are not indep endent since the supp ort (or \fo otprint") of the p df is not rectangular. You can also determine this by calculating the marginals, which work out to b e as shown in Figure 5(a). To determine if they are correlated, we calculate the covariance. E [Y ] = 0
(b e careful, E [X] 6 = 0), and therefore E [X]E [Y] = 0.
Cov(X; Y ) = E [XY ] E [X]E [Y ] = E [XY ]
Z 1
u=
Z u
v = u
2 uv^ dv^ du^ +
Z 2
u=
Z 2 u
v =u 2
2 uv^ dv^ du^ =^0
Therefore X and Y are uncorrelated. While indep endent ) uncorrelated, the reverse do es not
necessarily hold (as this problem illustrates), i.e., uncorrelated do es not imply indep en- dent.
- The joint density of X and Y is
fX ;Y (u; v ) = c(u^2 v 2 )e u^ ; 0 u < 1 ; u v u
The conditional density of Y given that X = u is given by fY jX (v j X = u) = fX ;Y (u; v )=fX (u). The
interesting thing here is that we do not need to evaluate the constant c, since it cancels out of the numerator and denominator. The marginal density of X is
fX (u) =
Z u
v = u
c(u^2 v 2 )e u^ dv =
4 cu^3 e u
3 ;^0 ^ u^ <^1
Therefore,
fY jX (v j X = u) =
3 c(u^2 v 2 )e u
4 cu^3 e u^ =^
3(u^2 v 2 )
4 u^3 ;^ u^ ^ v^ ^ u
Rememb er here that u is now a sp eci c value of the random variable X.
- X; Y and Z are indep endent random variables having identical density functions f (u) = e u^ ;
0 < u < 1. The new RVs are de ned by
U = X + Y ; V = X + Z; W = Y + Z
If we denote by x; y , and z , the sp eci c values that X; Y and Z can take resp ectively, and by a; b; c, the sp eci c values that U; V ; W can take, then we have a = x + y ; b = x + z ; c = y + z , and
a; b; c 0 since x; y ; z 0. The unique solution to this system of equations is
x 1 = a^ +^2 b^ ^ c; y 1 = a^ ^2 b^ +^ c; z 1 = a^ + 2 b^ +^ c
and the Jacobian of the transformation (note that all the functions involved are linear, and hence continuously di erentiable) is
J (x; y ; z ) =
Therefore
fU;V ;W (a; b; c) =
fX ;Y ;Z
a + b c
; a^ ^ b^ +^ c
; a^ +^ b^ +^ c
jJ (x 1 ; y 1 ; z 1 )j =^
2 exp
a^ +^2 b^ +^ c
; a; b; c 0
Thus, U; V and W are also indep endent RVs.
8. X and Y are Gaussian random variables with E [X] = 1 ; E [Y ] = 1 ; Var[X] = 2 ; Var[Y ] = 1 and
X ;Y = 0 :5. (X and Y are not indep endent.) From this information, we calculate:
E [X^2 ] = Var(X) + (E [X])^2 = 3; E [Y 2 ] = Var(Y ) + (E [Y ])^2 = 2; Cov(X; Y ) = p^1
(a) E [2(X + Y )(X Y )] =a 2 E [X^2 Y 2 ] =b 2(E [X^2 ] E [Y 2 ]) = 2. In equalities \a" and \b", we
have made use of the linearity of exp ectation.
(b) Cov(U; W ) = E [UW ] E [U] E [W ]. Now E [U] = E [2X 3 Y ] = 2 E [X] 3 E [Y ] = 5, again,
by linearity of the exp ectation. Similarly, E [W ] = E [2X + 3 Y ] = 1. Also, E [UW ] =
E [(2X 3 Y )(2X + 3 Y )] = E [4X^2 9 Y 2 ] = 6. Therefore Cov(U; W ) = 6 5 ( 1) = 1.
We can also calculate the variances of U and W :
Var(U) = 4Var(X) + 9Var(Y ) 2(2)(3)Cov(X; Y ) = 4(2) + 9(1) 2(6)( p^1
p
Var(W ) = 4Var(X) + 9Var(Y ) + 2(2)(3)Cov(X; Y ) = 4(2) + 9(1) + 2(6)( p^1
p
(c) The trick here is to realize that b oth U and W are Gaussian random variables b ecause they are obtained by p erforming linear op erations on X and Y , each of which is Gaussian. We have calculated the means of U and W , their variances and their covariance. All we need to do is calculate the correlation co ecient. = Cov(U; W )=
p
Var(U)Var(W ) = 1 =
p
- Plugging into the general formula b elow we get
fU;W (a; b) =
2 U W
p
1 ^2
e
a U
U
b W
W
a U
U
b W
W
fU;W (a; b) =
p
e
(^12)
17 a 65 p^2 ^2 +^ ^ 17+6b+1p^2 ^2 +^ 2(a ^217 5)(b+1)
- [Extra Credit:] A; B; C are indep endent, and chosen uniformly in the interval [0; 1]. The ro ots
of the equation Ax^2 + Bx + C = 0 are real if and only if B^2 4 AC 0, i.e., B^2 = 4 AC. Let
us rst nd the distribution of the random variable Y = AC. We will then nd the probability
P fB^2 = 4 Y g. We see that Y 2 [0; 1]. Therefore FY ( ) can b e calculated as
0 < 1 : P fY > g = 1 FY ( ) =
Z 1
u=
Z 1
v = =u
1 dv du
= (1 + ln )
Therefore,
FY ( ) =
ln ; 0 < 1
and the probability density is given by fY ( ) =
ln ; 0 < 1
Now, since A; B; C are indep endent, so are B and AC (why?).
P fB^2 = 4 Y g =
Z 1
u=
Z u 2 = 4
=
ln d du
Z 1
u=
u^2 4
2 4
ln 4 u^2
du;
which can b e evaluated to yield
P fB^2 = 4 Y g =
36 +^
ln 2 6 :
