EECS 501 SOLUTIONS TO PROBLEM SET #5 Fall 2001
1. We are given that fy|θ(Y|Θ) ∼ N(Θ, σ2)→E[y|θ= Θ] = Θ.
Then E[y] = Eθ[Ey[y|θ= Θ]] = RE[y|θ= Θ]fθ(Θ)dΘ = R2π
0Θ1
2πdΘ = π.
2. From pp. 138-139, we know that fz(Z) = Z
σ2e−Z2/(2σ2), Z ≥0 (Rayleigh pdf).
E[z] = R∞
0Z2e−Z2/2dZ =−Ze−Z2/2|∞
0+R∞
0e−Z2/2dZ = 0 + pπ/2 = pπ/2
using integration by parts: Ru dv =uv −Rv du where u=Zand v=−e−Z2/2
and R∞
−∞
1
√2πe−X2/2dX = 1 →R∞
0e−X2/2dX =1
2√2π=pπ/2 by symmetry.
E[z2] = R∞
0Z3e−Z2/2dZ =R∞
02Ue−UdU = 2 where U=Z2/2.
So σ2
z=E[z2]−(E[z])2= 2 −π/2 = 0.4292.
3a. Ey[Ex[g(x, y)|y]] = RdY fy(Y)(Rg(X, Y )fx|y(X|Y)dX)
=RdY RdX g(X, Y )fy(Y)fx|y(X|Y) = RdY RdX g(x, y)fx,y(X , Y ) = E[g(x, y)].
3b. g(x, y) = xand g(x, y) = x2→σ2
x=E[x2]−(E[x])2=Ey[E[x2|y]] −(Ey[E[x|y]])2.
Ey[σ2
x|y] = Ey[E[x2|y]]−Ey[(E[x|y])2] and V ary[E[x|y]] = Ey[(E[x|y])2]−(Ey[E[x|y]])2.
Adding, we get σ2
x=Ey[σ2
x|y] + V ary[E[x|y]] Q.E.D. What does this mean? See #4.
4a. E[y] = En[E[y|n]] = En[nE[x]] = E[n]E[x] since y=x1+x2+. . . +xn.
σ2
y=En[σ2
y|n] + V arn[E[y|n]] = En[nσ2
x] + V arn[nE[x]] = E[n]σ2
x+ (E[x])2σ2
n.
nconstant→σ2
y=nσ2
x. But xconstant→σ2
y=x2σ2
n. Makes sense since y=xn.
4b. Total number rof items sold=sum of random number (k) of random variables (n).
σ2
r=E[k]σ2
n+ (E[n])2σ2
k=µ1µ2+µ1µ2
2where E[k] = σ2
k=µ1and E[n] = σ2
n=µ2.
Increasing µ1, as opposed to µ2, leads to a smaller increase in σ2
r.
5. See overleaf.
6. Let An={ω:x(ω)>1
n}.A1⊂A2⊂ ··· ⊂ {ω:x(ω)>0}. Continuity of probability
→P r[x > 0] = P r[{ω:x(ω)>0}] = P r [LIM
n→∞An] = LIM
n→∞P r[An] = LIM
n→∞P r[x > 1
n]
≤LIM
n→∞nE[x] = 0 →P r[x= 0] = 1 using the Markov inequality and E[x] = 0.
