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University of Illinois Fall 2003
ECE 313: Solutions to Problem Set
- Var(X 1 + X 2 + X 3 ) = Var(X 1 ) + Var(X 2 ) + Var(X 3 ). Expanding the right hand side and cancelling out identical terms, we have
Cov(X 1 , X 2 ) + Cov(X 2 , X 3 ) + Cov(X 1 , X 3 ) = 0.
This does not necessarily imply that each of the individual covariances Cov(Xi, Xj ) are 0. In fact, the equation can be rewritten as
Cov(X 2 , (X 1 + X 3 )) = −Cov(X 1 , X 3 ). (1)
Choose X 1 , X 3 such that Cov(X 1 , X 3 ) 6 = 0. Then for any function g(X 1 , X 3 )
Cov(g(X 1 , X 3 ), (X 1 + X 3 )) = −Cov(X 1 , X 3 ) ·
−Cov(g(X 1 , X 3 ), (X 1 + X 3 )) Cov(X 1 , X 3 ) ︸ ︷︷ ︸ =k = −k · Cov(X 1 , X 3 ), or,
Cov(
g(X 1 , X 3 ) k
, (X 1 + X 3 )) = −Cov(X 1 , X 3 ), ( ∵ Cov(kY, X) = kCov(Y, X) )
noting that for a given X 1 , X 3 , and g(X 1 , X 3 ), k is a constant. On comparing this result with Equation (1), it follows that we can always choose
X 2 =
g(X 1 , X 3 ) k
g(X 1 , X 3 )Cov(X 1 , X 3 ) Cov(g(X 1 , X 3 ), (X 1 + X 3 ))
to satisfy the given equation even when X 1 , X 3 are pairwise correlated, so long as Cov(g(X 1 , X 3 ), (X 1 + X 3 )) 6 = 0.
- (a) Since Z = 2(X + Y )(X − Y ) = 2(X^2 − Y 2 ), we have
E[Z] = 2(E[X^2 ] − E[Y 2 ]) = 2(Var(X) − Var(Y ) + E^2 [X] − E^2 [Y ]) = − 40.
(b) We have
Cov(T, U ) = Cov(2X + Y, 2 X − Y ) = 4Cov(X, X) − 2Cov(X, Y ) + 2Cov(Y, X) − Cov(Y, Y ) = 4Var(X) − Var(Y ) = 7.
(c) Here
E[W ] = E[3X + Y + 2] = 3E[X] + E[Y ] + 2 = 9. Var(W ) = Var(3X + Y + 2) = 9Var(X) + Var(Y ) + 6Cov(X, Y ) = 48. 6
since Cov(X, Y ) = ρ
Var(X)Var(Y ) = 3. 6.
- (a)
E[X] =
−∞
xfX (x)dx
= 0 (For any odd function g(x),
∫ (^) a
−a
g(x)dx = 0).
Since (x|x|fX (x)) is also an odd function E[X|X|] = 0 as well. Thus, Cov(X, |X|) = E[X|X|] − E[X]E[|X|] = 0 − 0 = 0, ie., |X| and X are uncorrelated.
(b) However, P (|X| ≤ y, X ≤ x) =
P (|X| ≤ y), x ≥ y ≥ 0 P (−y ≤ X ≤ x), y ≥ max(x, 0) 0 , otherwise. Clearly, P (|X| ≤ y, X ≤ x) 6 = P (|X| ≤ y)P (X ≤ x). Hence, X and |X| are not independent.
E
[∑n ∑i=1^ Xi n i=1 Xi
]
−∞
−∞
−∞
∑n ∑i=1^ xi n i=1 xi
fX 1 (x 1 )fX 2 (x 2 ) · · · fXn (xn)dx 1 dx 2 · · · dxn
∑^ n
i=
−∞
−∞
−∞
xi ∑n i=1 xi
fX 1 (x 1 )fX 2 (x 2 ) · · · fXn (xn)dx 1 dx 2 · · · dxn
(a)
∑^ n
i=
I = nI
Equality (a) follows from the fact that the integrals are all identical. This is because the pdfs for all the variables are identical, and the integration is over the entire support, so order of integration does not matter. Also, the LHS of the equation is simply E[1] = 1, which yields I = 1/n. Therefore,
E
[∑k ∑i=1^ Xi n i=1 Xi
]
∑^ k
i=
−∞
−∞
−∞
xi ∑n i=1 xi
fX 1 (x 1 )fX 2 (x 2 ) · · · fXn (xn)dx 1 dx 2 · · · dxn
∑^ k
i=
I =
k n
- The key point to observe here is that X 1 , X 2 , · · · are independent random variables, and therefore Yn and Yn+k are also independent if k ≥ 3. Thus Cov(Yn, Yn+k) = 0 if k ≥ 3. Now, for k = 0, we have:
Cov(Yn, Yn) = Var(Yn) = Var(Xn+Xn+1+Xn+2) = Var(Xn)+Var(Xn+1)+Var(Xn+2) = 3σ^2 ,
where we have used the independence of Xn, Xn+1, Xn+2 once again. For k = 1, we have:
Cov(Yn, Yn+1) = Cov(Xn + Xn+1 + Xn+2, Xn+1 + Xn+2 + Xn+3) = Cov(Xn, Xn+1 + Xn+2 + Xn+3) + Cov(Xn+1 + Xn+2, Xn+1 + Xn+2 + Xn+3) = 0 + Cov(Xn+1 + Xn+2, Xn+1 + Xn+2) + Cov(Xn+1 + Xn+2, Xn+3) = Var(Xn+1 + Xn+2) + 0 = Var(Xn+1) + Var(Xn+2) = 2σ^2 ,
where we have used the bilinearity of the covariance function, along with the independence of Xn, Xn+1, Xn+2. Finally for k = 2, we have:
Cov(Yn, Yn+2) = Cov(Xn + Xn+1 + Xn+2, Xn+2 + Xn+3 + Xn+4) = Cov(Xn + Xn+1, Xn+2 + Xn+3 + Xn+4) + Cov(Xn+2, Xn+2 + Xn+3 + Xn+4) = 0 + Cov(Xn+2, Xn+2) + Cov(Xn+2, Xn+3 + Xn+4) = Var(Xn+2) + 0 = σ^2 ,
where the reasoning is similar to that used for the case k = 1. Summarizing all of the above, we conclude that Cov(Yn, Yn+k) = (3 − k)σ^2 for k = 0, 1 , 2 and Cov(Yn, Yn+k) = 0 for k ≥ 3.
that is Q(λ) = E[(λ(X − X) − (Y − Y ))^2 ] = 0, for λ = BA. But the expectation of a non-negative expression can be 0 only when the expression equals 0 almost everywhere (a.e.):
[λ(X − X) − (Y − Y )]^2 = 0 , or, Y = λX + (Y − λX),
which is of the form Y = a + bX. To prove the converse, note that, for Y = a + bX,
|Cov(Y, X)|^2 = Cov^2 (a + bX, X) = b^2 Var^2 (X) = b^2 Var(X)Var(X) = Var(Y )Var(X).
E[max{X 1 , X 2 }] =
−∞
−∞
max{x, y}fX,Y (x, y) dx dy
−∞
−∞
max{x, y}fX (x)fY (y) dx dy
−∞
y
xfX (x)fY (y) dx dy +
−∞
∫ (^) y
−∞
yfX (x)fY (y) dx dy
(a) = μ +
−∞
y
(x − μ)fX (x)fY (y) dx dy +
−∞
∫ (^) y
−∞
(y − μ)fX (x)fY (y) dx dy
= μ +
−∞
fY (y)
[∫ ∞
y
(x − μ)fX (x) dx
]
dy +
−∞
fX (x)
[∫ ∞
x
(y − μ)fY (y) dy
]
dx
(b) = μ + 2
−∞
fY (y)
[ ∫ ∞
y
(x − μ)fX (x) dx
]
dy
= μ + 2
−∞
fY (y)
[ ∫ ∞
y
2 πσ
(x − μ)e−^
(x−μ)^2 2 σ^2 dx
]
dy
= μ + 2
−∞
σ √ 2 π
fY (y)
[ ∫ ∞
(y−μ)^2 2 σ^2
e−z^ dz
]
dy
substituting z =
(x − μ)^2 2 σ^2
= μ + 2
−∞
σ √ 2 π
2 πσ
e−^
(y−μ)^2 2 σ^2 , dy
= μ +
π
−∞
e−^
(y−μ)^2 σ^2 dy
= μ +
σ √ π
−∞
2 π(σ/
e−^
(y−μ)^2 2(σ^2 /2) (^) dy
= μ +
σ √ π
(since the integral of the pdf of an RV ∼ N (μ, σ/
- is 1.)
Equality (a) follows because μ =
−∞
−∞ μfX,Y^ (x, y)^ dx dy, and Equality (b) holds since fX (·) = fY (·), therefore, the 2 integrals are identical.
