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University of Illinois Spring 2008
ECE 313: Solutions to Problem Set 13
- (a) fX 2 (v) = 1 2
v
[
fX (
v) + fX (−
v)
]
v
× 1
σ
2 π
[
exp(−v/ 2 σ^2 ) + exp(−v/ 2 σ^2 )
]
σ
2 v
× √^1
π
exp
− v 2 σ^2
= λ(λv)^
(^12) − 1
Γ
2
) exp(−λv) where λ = 1 2 σ^2
and Γ
π.
Hence, X 2 ∼ Gamma
2 σ^2
(b) The sum of independent Gamma(ti, λ) random variables is a Gamma(
ti, λ) random variable. Hence, W = X 2 + Y^2 + Z^2 is a Gamma
2 σ^2
random variable whose pdf
is fW (α) =
σ^3
√ (^) α 2 π exp^
− 2 ασ 2
, α ≥ 0 , 0 , α < 0. If σ^2 = 4, fW (5) = (1/8)
5 / 2 π exp(− 5 /8). (c) E[W] = E[X 2 + Y^2 + Z^2 ] = E[X 2 ] + E[Y^2 ] + E[Z^2 ] = 3σ^2 since W ∼ Gamma(3/ 2 , 1 / 2 σ^2 )), and its expected value is the ratio of the parameters, viz. 3 σ^2. (d) The pdf of H = 12 mW is fH(β) = (2/m)fW (2β/m). Since σ^2 = kTm , we get that the kinetic energy H has the Maxwell-Boltzmann pdf: fH(β) = √^2 π
(kT )−^
β exp
− β kT
for β ≥ 0. (e) FV (γ) = P {V ≤ γ} = P {W ≤ γ^2 } = FW (γ^2 ). Hence,
fV (γ) = 2γfW (γ^2 ) =
√^4
π
( (^) m 2 kT
γ^2 exp
mγ^2 2 kT
for γ ≥ 0.
(f) E[V] =
0
γ · √^4 π
( (^) m 2 kT
γ^2 exp
− mγ
2 2 kT
dγ =
0
2 kT mπ
x·exp(−x) dx = 2
2 kT mπ on substituting mγ^2 / 2 kT = x. Alternatively, E[V] = E[
W] =
0
α 1 σ^3
α 2 π
exp
− α 2 σ^2
dα =
0
√^4 σ 2 π
x · exp(−x) dx = √^4 σ 2 π
2 kT mπ on substituting^ α/^2 σ
(^2) = x and remembering that σ (^2) = kT m.
- E[X ] = 1, E[Y] = 4, var(X ) = 4, var(Y) = 9, and ρX ,Y = 0.1. (a) E[Z] = E[2(X +Y)(X −Y)] = 2E[X 2 −Y^2 ] = 2E[X 2 ]− 2 E[Y^2 ] = 2[4+1^2 ]−2[9+4^2 ] = −40. (b) cov(T , U) = cov(2X +Y, 2 X −Y) = 4·cov(X , X )+2·cov(Y, X )− 2 ·cov(X , Y)−cov(Y, Y) = 4 · var(X ) + 2 · cov(X , Y) − 2 · cov(X , Y) − var(Y) = 4 · var(X ) − var(Y) = 4 · 4 − 9 = 7. (c) E[W] = E[3X + Y + 2] = 3E[X ] + E[Y] + 2 = 9. var(W) = var(3X +Y +2) = 3^2 ·var(X )+var(Y)+2· 3 · 1 ·cov(X , Y) = 9·4+9+6· 2 · 3 · 0 .1 = 48 .6. (d) P {W > 0 } = 1 − Φ
√^0 −^9
− √^9
√^9
- (a) var(X + Y) = var(X ) + var(Y) + 2 · cov(X , Y) = 36. var(X − Y) = var(X ) + var(Y) − 2 · cov(X , Y) = 64. Hence, cov(X , Y) = −7. From the above, 2 · var(X ) + 2 · var(Y) = 8 · var(Y) = 100, giving var(Y) = 12. 5 , var(X ) = 37 .5 and ρX ,Y = cov(X , Y)/
var(X )var(Y) = − 7 / 12. 5
(b) var(X + Y) = var(X ) + var(Y) + 2 · cov(X , Y) equals var(X − Y) = var(X ) + var(Y) − 2 · cov(X , Y) if and only if cov(X , Y) = 0, that is, if and only if X and Y are uncorrelated. (c) No, whether var(X ) equals var(Y) or not has no bearing on the question of whether cov(X , Y) is zero or not.
- The random point (X , Y) is uniformly distributed on the shaded region shown in the figure
below. Clearly, fX ,Y (u, v) =^4 3
on this region.
(a) fX (u) is the area of the cross-section of the pdf surface along the line u. There are two cases to be considered, as shown in the left-hand figure above. It is obvious almost by
inspection that fX (u) =
2 3 ,^0 ≤^ u^ ≤^
1 4 2 , 3 ,^
1 2 < u^ ≤^1 , 0 , elsewhere.
E[X ] =
−∞
u · fX (u) du =
0
u · 2 3
du +
1 / 2
u · 4 3
du = 7 12
E[X 2 ] =
−∞
u^2 · fX (u) du =
0
u^2 ·
3 du^ +
1 / 2
u^2 ·
3 du^ =^
var(X ) = E[X 2 ] − (E[X ])^2 = 5 12
(b) By symmetry, fX and fY are the same function: fY (v) =
2 3 ,^0 ≤^ v^ ≤^
1 4 2 , 3 ,^
1 2 < v^ ≤^1 , 0 , elsewhere. E[Y] =
12 ,^ var(Y) =^
(c) Given that X = α, the conditional pdf fY|X (v|α) is the cross-section of the pdf surface at u = α unitized to have area 1. For 0 < α < 12 : fY|X (v|α) ∼ Uniform[ 12 , 1] ⇒ E[Y|X = α] = 34 , var[Y|X = α] = 481 For 12 < α < 1: fY|X (v|α) ∼ Uniform[0, 1] ⇒ E[Y|X = α] = 12 , var[Y|X = α] = 121
(d) 0 ≤ v ≤ 1 2
: fY (v) =
0
fY|X (v|α)fX (α) dα =
0
dα +
1 / 2
dα =^2 3 1 2 ≤^ v^ ≤^ 1 :^ fY^ (v) =
0
fY|X (v|α)fX (α) dα =
0
(2) · 23 dα +
1 / 2
(1) · 43 dα =^43 Yes, we get the same answer as in part (b).
Since Z and W are zero-mean random variables, we get var(Z) = E[Z^2 ] = E[X 2 cos^2 θ + Y^2 sin^2 θ + 2X Y sin θ cos θ] = σ^21 cos^2 θ + σ^22 sin^2 θ + 2 ρσ 1 σ 2 sin θ cos θ, and var(W) = E[W^2 ] = E[X 2 sin^2 θ + Y^2 cos^2 θ − 2 X Y sin θ cos θ] = σ 12 sin^2 θ + σ^22 cos^2 θ − 2 ρσ 1 σ 2 sin θ cos θ. (b) Since Z and W are zero-mean random variables,
cov(Z, W) = E[ZW] = E[(X cos θ + Y sin θ)(Y cos θ − X sin θ)] = E[Y^2 ] sin θ cos θ − E[X 2 ] sin θ cos θ + E[X Y](cos^2 θ − sin^2 θ) = sin θ cos θ · (σ 22 − σ^21 ) + (cos^2 θ − sin^2 θ) · ρσ 1 σ 2 =
2 ·^ sin 2θ^ ·^ (σ
2 2 −^ σ
2 1 ) + cos 2θ^ ·^ ρσ^1 σ^2
(c) Z and W are jointly Gaussian random variables and thus they are independent if cov(Z, W) = 0. We get independent random variables if we choose θ =^12 arctan
2 ρ · σ 1 σ 2 σ^21 − σ 22
Note that if θ 0 is a solution to this equation, then θ 0 +π is also a solution, as are θ 0 ±π/2. That is, there are four different values of θ in the range [0, 2 π) that can be used to get independent Gaussian random variables from X and Y. In particular, if σ 1 = σ 2 , then θ can take on values ±π/4 and ± 3 π/4.
