Download Third Assignment: Solutions 1. Since P(X(p) > n) = (1 − p) n, n = 0,1

Third Assignment: Solutions
1. Since P (X(p) > n) = (1 − p)n , n = 0, 1, 2, . . ., we have, for x ≥ 0,
P (X(p)/p > x) = P (X(p) > [px] + 1) = (1 − p)[px]+1 → e−x ,
as p → 0.
The conclusion follows from the fact that e−x is continuous for x > 0.
2. (i) We only give sufficient conditions. First assume that the probability measure Q
has density, with respect to the Lebesgue measure, f . Then
Z
1
f (x) =
e−iθ·x Φ(θ)dθ.
(2π)d Rd
Second, assume that f (x) is positive definite:
N
N X
X
j=1 k=1
cj ck f (xj − xk ) ≥ 0,
for all N ∈ N, all c1 , . . . , cN ∈ C, and all x1 , . . . , xN ∈ Rd . Then Φ(θ) ≥ 0, by
Bochner’s theorem.
(ii) If R is a rotation and if X is a random variable with values in Rd and law Q, then
T
Φ(θ) = Eeiθ·X . Hence Φ(Rθ) = Eei(Rθ)·X = Eeiθ·(R X) = Eeiθ·X = Φ(θ).
(iib) No: take, e.g., the probability measure Q on R2 with density
f (x, y) = c(x2 + y 2 )e−(x
2 +y 2 )
.
Then Q is invariant under rotations and has characteristic function
2
2
Φ(θ1 , θ2 ) = cπ(1 − (θ12 + θ22 )/4)e−(θ1 +θ2 )/4 ,
which takes both positive and negative values.
3. If there are two probability measures Q′ , Q′′ corresponding to the same characteristic
function, then the alternating sequence Q′ , Q′′ , Q′ , Q′′ , . . . satisfies the assumptions of
Lévy’s theorem and so it converges weakly to a probability measure. The only way
for this sequence to converge is if Q′ = Q′′ .
√
4. (i) Since (X1 +· · ·+Xn )/ n is a linear combination of (X1 , . . . , Xn ) [which has normal
√
law on Rn ], it follows immediately that Z := (X1 + · · · + Xn )/ n has normal law on
R1 . That it is standard is clear from the fact that E(Z) = 0, E(Z 2 ) = (1/n) × n = 1.
(ii) We have
Z ∞
1/π
eiθx
Φ(θ) =
dx = e−|θ| ,
1 + x2
−∞
as can be easily established by a contour integral. Therefore,
E[eiθXj ] = e−|θ| ,
for all j. Let W := (X1 + · · · + Xn )/n. Then
E[eiθW ] = E
n
Y
eiθXj /n =
n
Y
j=1
j=1
1
E[ei(θ/n)Xj ],
by independence, and so W has characteristic function
E[eiθW ] = e−|θ/n| )n = e−|θ| ,
which is identical to the characteristic function of X1 . Hence W and X1 have the same
law.
(iii) Indeed, the function
α
Φ(θ) := e−|θ|
is positive semidefinite and hence a characteristic function of some probability measure Qα . The stated property follows exactly in the same manner as above [part (i)
corresponds to α = 2, and part (ii) to α = 1]. If α > 2, the function above is not
positive semidefinite.
5. (i) As in the hint, let U be uniform in (−1, 1). Then, for all n,
n
X
ξj
U
+ n.
U =
j
2
2
(d)
j=1
P
ξj
The right-hand side converges weekly to ∞
j=1 2j . Hence the later has the law of U .
(ii) The left-hand side is the characteristic function of U . The characteristic function
P n ξj
P
ξj
of ∞
j=1 2j is the limit of the characteristic function of
j=1 2j , as n → ∞. The
Qn
Q
Pn ξ j
j
characteristic function of j=1 2j is j=1 cos(x/2 ). Hence limn→∞ nj=1 cos(x/2j )
exists and is equal to (sin x)/x.
(iii) Set x = π/2. Then (sin x)/x = 2/π. From the formula cos(2x) = 1 + 2 cos2 (x), it
follows that
q
p
√
2 + 2 + ··· + 2
j+1
cos(π/2 ) =
,
2
where the square root symbol appears j times. Hence
2
=
π
∞
Y
j=1
q
p
√
2 + 2 + ··· + 2
2
(j times)
.
6. If X(n), n = 1, 2, . . . is a tight family of random variables in Rd then it is obvious
that, for each j, Xj (n), n = 1, 2, . . . is a tight family of random variables in R.
For the converse, fix ε > 0 and pick Mj > 0 such that P (|Xj (n)| > Mj ) ≤ ε/d,
for
Pd all n. Let M := max(M1 , . . . , Md ). Then P (max(|X1 (n)|, . . . , |X(n)|) > M ) ≤
j=1 P (|Xj (n)| > M ) ≤ ε, for all n. Hence, for all n, the probability that X(n)
belongs to the compact set K := [−1, 1]d is at least 1 − ε, for all n, and hence the
sequence if tight.
7. Let f : Rd → R be continuous and bounded. Since g is continuous, it follows that
(d)
f ◦g : S → R is continuous and bounded. Therefore, Qn ◦(f ◦g)−1 −−→ Q◦(f ◦g)−1 . But
Qn ◦(f ◦g)−1 = (Qn ◦g −1 )◦f −1 , and Q◦(f ◦g)−1 = (Q◦g −1 )◦f −1 .
2
8. (i) If ϕ ∈ C[0, 1] then ϕ(g(ϕ)) = 0. Let τ := g(ϕ). If τ < 1 and there is ε > 0 such
that ϕ > 0 on (τ − ε, τ ) and ϕ < 0 on (τ, τ + ε) then g is continuous at ϕ. To see
this, let ϕn be a sequence of continuous functions such that ϕn → ϕ, uniformly. Then
ϕn will eventually be positive on (τ − ε, τ ) and negative on (τ, τ + ε), implying that
all the zeros of ϕn on (τ − ε, τ + ε) converge to τ . So, necessarily, any discontinuity
point ϕ of g must have the property that it has constant sign (or zero) on a small
neighborhood of its last zero τ . Intuitively, a Brownian motion cannot do this, almost
surely. The event that a Brownian motion B touches 0 at some point t0 but is positive
(respectively, negative) on a neighborhood of t0 implies that the infimum (respectively,
supremum) of the Brownian motion B on the same interval equals 0. By time scaling
and space translation, it suffices to prove that P (sup0≤t≤1 B(t) = x) = 0 for all x.
This follows because sup0≤t≤1 B(t) is a random variable with continuous distribution
function.
(ii) Since Sn = 0 if and only if n is even, it suffices to compute P (T2m = 2k). We have
P (T2m = 2k) = P (S2k = 0, S2k+1 6= 0, . . . , S2m 6= 0)
= P (S2k = 0)P (S2k+1 6= 0, . . . , S2m 6= 0 | S2k = 0)
= P (S2k = 0)P (S1 6= 0, . . . , S2m−2k 6= 0)
= P (S2k = 0)P (S2m−2k = 0).
2k −2k 2m − 2k −(2m−2k)
2
2
=
m−k
k
2k
2m − 2k −2m
=
2
.
k
m−k
(Alternatively, you can count the number of paths of length 2m, starting from 0 and
which hit 0 for the last time at time 2k, and find that this is equal to the product of
the binomial coefficients above.)
√
(iii) Let Zn (t) := B(nt)/ n if t ∈ (1/n)Z, and let Zn (t) to be obtained by linear
interpolation for other values of t. We know that the random sequence Zn converges
weakly to B. Since g is continuous at B, almost surely, we have that g(Zn ) converges
weakly to T = g(B) and so
P (g(Zn ) ≤ x) → P (g(B) ≤ x),
for all x such that P (g(B) = x) = 0. But
g(Zn ) = sup{t ≤ 1 : S[nt] = 1} =
1
Tn .
n
Therefore
P (Tn ≤ nx) → P (T ≤ x),
(*)
for all x such that P (T = x) = 0. From the formula for P (T2m = 2k) and Stirling’s
approximation we have
2/π
=: f (x),
2mP (T2m = 2km ) → p
x(1 − x)
if km /m → x,
In particular,
2mP (T2m = 2[mx]) → f (x),
3
as m → ∞.
as m → ∞.
By the dominated convergence theorem,
Z x
Z
2mP (T2m = 2[mt])dx →
0
x
f (t)dt,
0
as m → ∞.
But the left-hand side equals P (T2m ≤ 2mx) (here, we use the fact that T2m takes
only even values). Hence
Z x
P (T2m ≤ 2mx) →
f (t)dt.
(**)
0
Comparing (*) and (**), we have
Z
Z x
√
dt
2 x
2
p
f (t)dt =
P (T ≤ x) =
= arcsin x.
π 0
π
t(1 − t)
0
(Moreover, since P (T = x) = 0, for all x, convergence (*) holds for all x.)
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