SOLUTIONS TO Final — ECE 313 – FALL 2007
1. (i) T, (ii) T, (iii) F, (iv) F, (v) F, (vi) T, (vii) F, (viii) T, (ix) T, (x) F.
Brief reasons:
(i) Xand Yare independent.
(ii) Xand Yare uncorrelated.
(iii) Xand Ymight not have the same variance.
(iv) Wand Zare uncorrelated but might not be independent.
(v) Same as (iv)
(vi) Xand Yare independent.
(vii) Wand Zmight not be independent.
(viii) cov(2W, Z −W) = cov(2W, Z )−cov(2W,W ) = −2Var(W)
(ix) Xand Yare independent.
(x) E[X2]6=E[X]2.
2. (i) T, (ii) T, (iii) F, (iv) T, (v) F, (vi) F, (vii) T, (viii) T
Brief reasons:
(ii) P(Y= 0) = 0.5 and therefore Yis mixed.
(v) ρ(X, Y ) = −1/2.
(vi) var(X−Y) = var(X) + var(Y).
3. (i) b, (ii) b, (iii) a, (iv) a, (v) b, (vi) a, (vii) b, (viii) a, (xi) a, (x) c.
Brief reasons:
(ii) By symmetry
(iv) E[(X−1/2)Y] = E[XY ] = EE[X Y |X=x]= 0.
(v) By symmetry
(vii) fZ,W (z, w) = 1,0< z ≤1,0< w ≤1 and zero otherwise.
(viii) E[X|Y] = 1/2 so variance is 0.
(x) Var(U) = Var(Z) = VarUnf[0,1]= 1/12.
4. (i) P(Y=−1) = P(Y= 1) = 1/2.
(ii) E[Y] = 0, var(Y) = 1 ×1/2+1×1/2 = 1
(iii) cov(X, Y ) = E[X Y ] = EX E[Y|X]=EX sgn(X)=E|X|= 2 R∞
0
1
√2πe−x2/2x dx =q2
π
(iv) ˆ
XLMM SE =cov(X,Y )
var(Y)Y=q2
πYand MSE = var(X)−cov(X,Y )
var(Y)= 1 −2
π
(v) fX|Y(x|1) = 2fX(x) for x > 0 and 0 otherwise.
Similarly, fX|Y(x| − 1) = 2fX(x) for x < 0 and 0 otherwise.
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