1

Assignment-3 Solutions

1. Consider MMSE estimation for the wireless sensor network (WSN) scenario as described in lectures with each observationy(k) = h+v(k), for 1 ≤ k ≤ N , i.e. number of observations N and IID Gaussian noise samples of variance σ2. The expressionfor the MSE (Mean Squared Error) of the ML estimate of the unknown Gaussian parameter h is σ2

NAns-(a)

2. Minimum Mean Squared Error(MMSE) in estimating h is defined as,

E{(h− h)2} = rhh − rhyR−1yy ryh (1)

Considering wireless sensor network,y = 1h+ v (2)

with h ∼ N (µh, σ2h)

rhh = E{(h− h)2} = σ2h

rhy = E{(h− µh)(y − µy)T = σ2

h1T = rTyh

Ryy = E{(y − µy)(y − µy)T = σ2

h11T + σ2I

E{(h− h)2} = σ2h − σ2

h1T (σ2

h11T + σ2I)−11σ2

h

= σ2h − σ4

hN

(σ2hN + σ2)

=σ2hσ

2

(σ2hN + σ2)

E{(h− h)2} =1

1σ2

N

+ 1σ2h

Ans-(b)

3. Given, N = 5, µh = 1/2, σ2h = 1/4, σ2 = −3dB = 1/2

MSE for ML estimate=

=σ2

N

=1/2

5

=1

10

Ans-(b)

4. Given, σ2 = −3dB = 1/2, σ2h = 1/4

P(∣∣∣h− h

∣∣∣ < σ

4

)> 0.999

P(∣∣∣h− h

∣∣∣ ≥ σ

4

)≤ 10−3

let w =∣∣∣h− h

∣∣∣

2

w = (h− h) ∼ N

0,1

1σ2

N

+ 1σ2h

σ2 =

11σ2

N

+ 1σ2h

(let)

P(|w| ≥ σ

4

)≤ 10−3

2P(w ≥ σ

4

)≤ 10−3

P(w ≥ σ

4

)≤ 5× 10−4

P

(w

σ≥ σ/4

σ

)≤ 5× 10−4

Q

(σ/4

σ

)≤ 5× 10−4(

σ/4

σ

)≥ Q−1

(5× 10−4

)σ1√

1σ2N

+ 1

σ2h

≥ 4Q−1(5× 10−4

)

σ

√(N

σ2+

1

σ2h

)≥ 4Q−1

(5× 10−4

)√(

N +σ2

σ2h

)≥ 4Q−1

(5× 10−4

)N +

σ2

σ2h

≥(4Q−1

(5× 10−4

))2N +

1/2

1/4≥(4Q−1

(5× 10−4

))2N ≥

(4Q−1

(5× 10−4

))2 − 2

N ≥ (4× 3.2905)2 − 2

N ≥ 171.2411Nmin = 172

Ans-(a)

5. For fading channel estimation problem, where the output symbol y(k) is y(k) = hx(k)+ v(k), with h, x(k), v(k) denotingthe real channel coefficient, pilot symbol and noise sample respectively, the expression for the MSE (Mean Squared Error) ofthe ML estimate h of the unknown parameter h is, σ2

∥x∥2

Ans-(d)

6. For fading channel estimation problem, MMSE is,

E{(h− h)2} = rhh − rhyR−1yy ryh

rhh = E{(h− h)2} = σ2h

rhy = E{(h− µh)(y − µy)T = σ2

hxT = rTyh

Ryy = E{(y − µy)(y − µy)T = σ2

h1xT + σ2I

E{(h− h)2} = σ2h − σ2

hxT (σ2

hxxT + σ2I)−1xσ2

h

we know that,σ2hx

T (σ2hxx

T + σ2I)−1 = (σ2hx

Tx+ σ2)−1σ2hx

T

3

substituting this in above expression gives,

E{(h− h)2} = σ2h − (σ2

hxTx+ σ2)−1σ2

hxTxσ2

h

= σ2h − σ2

hxTxσ2

h

(σ2hx

Tx+ σ2)

=σ2σ2

h

(σ2hx

Tx+ σ2)

=1

1σ2

xT x

+ 1σ2h

=1

1σ2

∥x∥+ 1

σ2h

Ans-(b)

7. We know that,MMSE =

11

MSE of ML + 1Prior Var

As the number of samples N becomes very large, the ML estimate becomes more accurate, and hence, MSE of ML<<<PriorVariance. which imply

1MSE of ML >>> 1

Prior Var

hence, MMSE becomes,

MMSE = MSE of ML =σ2

∥x∥2

Ans-(b)

8. Given x =

1/2213/2

, y =

−22−11

, σ2 = 3dB = 2, σ2h = 1/2, µh = 1

MSE for ML estimate =σ2

∥x∥

=2

(1/2)2 + (2)2 + (1)2 + (3/2)2

=4

15

MMSE =1

1σ2

∥x∥2+ 1

σ2h

=1

12

15/2

+ 11/2

=1

154 + 2

1

=4

23

Ans-(a)

4

9. True parameter h to be within σ/8 of the estimate h with probability greater than 0.9999,

P(∣∣∣h− h

∣∣∣ < σ

8

)> 0.9999

P(∣∣∣h− h

∣∣∣ ≥ σ

8

)≤ 10−4

let w =∣∣∣h− h

∣∣∣ also, we have |x(k)|2 = 4 for all k.

w = (h− h) ∼ N

0,1

1σ2

∥x∥2+ 1

σ2h

σ2 =

11σ2

∥x∥2+ 1

σ2h

(let)

P(|w| ≥ σ

8

)≤ 10−4

2P(w ≥ σ

8

)≤ 10−4

P(w ≥ σ

8

)≤ 5× 10−5

P

(w

σ≥ σ/8

σ

)≤ 5× 10−5

Q

(σ/8

σ

)≤ 5× 10−5(

σ/8

σ

)≥ Q−1

(5× 10−5

)σ1√

1σ2

∥x∥2+ 1

σ2h

≥ 8Q−1(5× 10−5

)

σ

√√√√(∥x∥2

σ2+

1

σ2h

)≥ 8Q−1

(5× 10−5

)√(

4N +σ2

σ2h

)≥ 8Q−1

(5× 10−5

)4N +

σ2

σ2h

≥(8Q−1

(5× 10−5

))24N ≥

(8Q−1

(5× 10−5

))2 − σ2

σ2h

N ≥ 1

4

((8Q−1

(5× 10−5

))2 − σ2

σ2h

)Nmin =

1

4

((8Q−1

(5× 10−5

))2 − σ2

σ2h

)Ans-(d)

10. Given x =

2 + j−1− j1− 2j−1 + j

, y =

2 + 2j−j

−2 + j1− j

, σ2 = 3dB = 2, σ2h = 1/2, µh = 1 + j

∥x∥2 = |2 + j|2 + |−1− j|2 + |1− 2j|2 + |−1 + j|2

= (5) + (2) + (5) + (2) = 14

5

MMSE Esimate =

xTyσ2 + µh

σ2h

1σ2

∥x∥2+ 1

σ2h

=

12 + 1+j

1/2

1214

+ 11/2

=52 + 2j71 + 2

1

=5

18+

2

9j

Ans-(c)