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1 Theory of Probability I: Final Exam
Exercise 1.1. (20 points) Let h : R → [0, ∞) a measurable function, and
H(x) =
∫ (^) x
−∞
h(y)dy, −∞ < x < ∞.
If X is a random variable, show that
E[H(X)] =
−∞
h(x)P(X ≥ x)dx.
Using this, compute E[eθX^ ], θ > 0.
Solution: Actually, this is a homework problem, namely from Homework 4. The solution is just a usual application of Fubini. To be precise,
E[H(X)] = E
[∫ ∞
−∞
(^1) {y≤X}h(y)dy
]
−∞
h(y)E[1{y≤X}] =
−∞
h(y)P(X ≥ y)dy.
We can apply this for h(y) = θeθy^ and get
E[eθX^ ] =
−∞
θeθxP(X ≥ x)dx.
Exercise 1.2. (weak law for correlated random variables, 20 points) ). Let X 1 , X 2 ,... , Xn,... a sequence of random variables. Assume that E[Xn] = 0 for each n and that there exists a function r : { 0 , 1 , 2 ,... } → [0, ∞),
such that r(k) → 0 , k → ∞,
and E[XnXm] ≤ r(n − m), for each m ≤ n. Show that
X 1 +... Xn n
where the convergence above takes place in L^2 and in probability.
Solution: We just estimate
E
[(
X 1 +... Xn n
) 2 ]
n^2
∑^ n
i,j=
E[XiXj ] ≤
n^2
∑^ n
i,j=
r(|i − j|)
The Sum above is actually equal to (if we separate the n diagonal terms from the rest)
1 n^2
× nr(0) + 2 ×
n^2
(n − 1) × r(1) + (n − 2) × r(2) + .. 1 × r(n − 1)
Sn
The first (isolated term) above, obviously converges to zero. In order to finish the proof, we use a customary trick, namely, we separate the above (Sn) sum into 2 kind of terms: those for small k, and those for large k. More precisely, fix ε > 0. For k ≥ k(ε), we have r(k) ≤ ε).
Therefore, we have
Sn =
n^2
∑^ k(ε)
i=k
(n − k) ︸ ︷︷ ︸ ≤n
r(k) +
n^2
∑^ n−^1
k=k(ε)
(n − k) ︸ ︷︷ ︸ ≤n
r(k) ︸︷︷︸ ≤ε
n
∑^ k(ε)
i=k
r(k) + ε
Therefore, for each ε > 0 we have lim sup Sn ≤ ε, so Sn → 0. This means that X 1 +... Xn n
→^ L^2 .
Convergence in probability follows from Chebyshevs’s inequality.
Exercise 1.3. (15 points) Let Xn be an i.i.d sequence of random variables. Show that
P(|Xn| ≥ en^ i.o.) = 0
if and only if E[(log |Xn|)+] < ∞.
Solution: This is, clearly, just an application of Borel-Cantelli Lemmas. More precisely, since
{|Xn| ≥ en^ i.o.} = {log |Xn| ≥ en^ i.o.} = {(log |Xn|)+^ ≥ en^ i.o.}
we can apply both Borel-Cantelli lemmas to conclude that, P(|Xn| ≥ en^ i.o.) = 0 if and only if ∑∞
n=
P((log |Xn|)+^ ≥ n) =
∑^ ∞
n=
P((log |X|)+^ ≥ n) < ∞.
However, we have used repeatedly that, for a positive Y ,
n
P(Y ≥ n) < ∞ if and only if
E[Y ] < ∞. Applying this to Y = (log |X|)+, we complete the solution.
Exercise 1.4. (20 points) Let (Xn)n and X be random variables with a common uniform bound, i.e., such that |Xn|, |X| ≤ M < ∞ a.s. Show that the following two statements are equivalent: Xn converges in distribution to X if and only if E[Xnk ] → E[Xk], n → ∞
for each k = 0, 1 , 2 ,...
Solution: recalling the Theorems about weak convergence and characteristic function, we have that φXn (t) → φ(t) implies weak convergence. Therefore, it is enough to show that, if we have (point-wise)convergence of ALL moments, then we also have convergence of char- acteristic functions.
ϕXn (t) = E[eitX^ ] =
∑^ ∞
k=
E[
(itXn)k k!
] =
∑^ ∞
k=
E[
(itXn)k k!
] =
∑^ L
k=
(it)kE[
(Xn)k k!
] +
∑^ ∞
k=L+
E[
(itXn)k k!
].
Now, since all Xn’s have a uniform bound (the limiting r.v. included)., we conclude that,
∑^ ∞
k=L+
E[
(itXn)k k!
]| ≤
∑^ ∞
k=L+
M k k!
→ 0 , L → ∞.
Pick L large enough such that the above tail series of the exponential is smaller then ε. Using what we have above, we have
