Download Jointly Distributed Continuous Random Variables - Lecture Notes | MATH 5010 and more Study notes Probability and Statistics in PDF only on Docsity!
Lecture 27
1. Two important properties
Theorem 27. 1 (Uniqueness). If X and Y are two random variables— discrete or continuous—with moment generating functions MX and MY , and if there exists δ > 0 such that MX (t) = MY (t) for all t ∈ (−δ , δ), then MX = MY and X and Y have the same distribution. More precisely:
(1) X is discrete if and only if Y is, in which case their mass functions are the same; (2) X is continuous if and only if Y is, in which case their density functions are the same.
Theorem 27.2 (L´evy’s continuity theorem). Let Xn be a random variables— discrete or continuous—with moment generating functions Mn. Also, let X be a random variable with moment generating function M. Suppose there exists δ > 0 such that:
(1) If −δ < t < δ, then Mn(t), M (t) < ∞ for all n ≥ 1 ; and (2) limn→∞ Mn(t) = M (t) for all t ∈ (−δ , δ), then
lim n→∞ FXn (a) = lim n→∞ P {Xn ≤ a} = P{X ≤ a} = FX (a),
for all numbers a at which FX is continuous.
Example 27. 3 (Law of rare events). Suppose Xn = binomal(n , λ/n), where λ > 0 is fixed, and n ≥ λ. Then, recall that
MXn (t) =
q + pe−t
)n
λ n
λe−t n
)n → exp
−λ + λe−t
Note that the right-most term is M (^) X (t), where X = Poisson(λ). Therefore, by L´evy’s continuity theorem,
nlim→∞ P^ {X^ n^ ≤^ a}^ = P^ {X^ ≤^ a}^ ,^ (20)
at all a where F (^) X is continuous. But X is discrete and integer-valued. Therefore, F (^) X is continuous at a if and only if a is not a nonnegative integer. If a is a nonnegative integer, then we can choose a non-integer b ∈ (a , a + 1) to find that
lim n→∞ P{X (^) n ≤ b} = P{X ≤ b}.
Because X (^) n and X are both non-negative integers, X (^) n ≤ b if and only if X (^) n ≤ a, and X ≤ b if and only if X ≤ a. Therefore, (20) holds for all a.
Example 27.4 (The de Moivre–Laplace central limit theorem). Suppose X (^) n = binomial(n , p), where p ∈ (0 , 1) is fixed, and define Yn to be its standardization. That is, Yn = (X (^) n − EX (^) n )/
VarX (^) n. Alternatively,
Yn = X (^) n − np √ npq
We know that for all real numbers t,
M (^) Xn (t) =
q + pe−t^
) (^) n .
We can use this to compute M (^) Yn as follows:
M (^) Yn (t) = E
[
exp
t · X (^) n − np √ npq
)]
Recall that X (^) n = I 1 + · · · + In , where Ij is one if the jth trial succeeds; else, I∑j = 0. Then, I 1 ,... , In are independent binomial(1 , p)’s, and X (^) n − np = n j=1 (Ij^ −^ p). Therefore,
E
[
exp
t · X (^) n − np √ npq
)]
= E
(^) √t npq
∑^ n
j=
(Ij − p)
E
[
exp
t √ npq (I 1 − p)
)])n
p exp
t √ npq
(1 − p)
t √ npq
(0 − p)
}) (^) n
p exp
t
q np
−t
p nq
}) (^) n .
E
A
Figure 1. Region of integration in Example 27.
Example 27.6 (Uniform joint density). Suppose A is a subset of the plane that has a well-defined finite area |A| > 0. Define
f (x , y) =
|A|
if (x , y) ∈ A,
0 otherwise.
Then, f is a joint density function, and the corresponding random vector (X, Y ) is said to be distributed uniformly on A. Moreover, for all planar sets E with well-defined areas,
P{(X, Y ) ∈ E} =
E∩A
|A|
dx dy =
|E ∩ A|
|A|
See Figure 1.
Example 27.7. Suppose (X, Y ) has joint density
f (x , y) =
Cxy if 0 < y < x < 1 , 0 otherwise.
- Jointly distributed continuous random variables 99
y=x/
y=x
x
y
1
1
0
Figure 2. Region of integration in Example 27.
Let us first find C, and then P{X ≤ 2 Y }. To find C:
1 =
−∞
−∞
f (x , y) dx dy =
0
∫ (^) x
0
Cxy dy dx
= C
0
x
(∫ (^) x
0
y dy
(^12) x 2
dx =
C
0
x^3 dx =
C
Therefore, C = 8, and hence
f (x , y) =
8 xy if 0 < y < x < 1 , 0 otherwise. Now P{X ≤ 2 Y } = P{(X, Y ) ∈ A} =
A
f (x , y) dx dy,
where A denotes the collection of all points (a , b) in the plane such that a ≤ 2 b. Therefore,
P{X ≤ 2 Y } =
0
∫ (^) x
x/ 2
8 xy dy dx =
See Figure 2.
