Download Economics 765 – Assignment 2

Economics 765 – Assignment 2
2.5
Let (X, Y ) be a pair of random variables with joint density function
(
fX,Y (x, y) =
2|x|+y
√
2π
exp −
(2|x|+y)2
2
if y ≥ −|x|,
if y < −|x|.
0
Show that X and Y are standard normal variables and that they are uncorrelated but not
independent.
Fix x ≥ 0 and integrate the density with respect to y. In this case y varies from −|x| = −x
to ∞, and so the integral is
1
√
2π
Z
Z ∞
(2x + y)2
1
(2x + y)2 dy = − √
(2x + y) exp −
d exp −
2
2
2π −x
−x
2
1
1
1
= √ exp −−
(2x − x)2 = √ e−x /2 = φ(x).
2
2π
2π
∞
For fixed x < 0, y varies from −|x| = x to ∞, and so the integral is
1
√
2π
Z
∞
x
Z ∞
(y − 2x)2
1
(y − 2x)2 (y − 2x) exp −
dy = − √
d exp −
2
2
2π x
1
1
= √ exp −−
(x − 2x)2 = φ(x).
2
2π
Thus for all x, the marginal density of X is φ(x), the standard normal density.
Now fix y ≥ 0 and integrate with respect to x. We split the integral into two parts, from
0 to ∞, and from −∞ to 0. For positive x, the contribution is
1
√
2π
Z
0
∞
Z ∞
(2x + y)2
1
(2x + y)2 (2x + y) exp −
d exp −
dx = − √
2
2
2 2π 0
2
1
1
= √ e−y /2 = −
φ(y).
2
2 2π
For negative x, we have similarly
1
−√
2π
Z
(2x − y)2
1
(2x − y) exp −
dx = √
2
2 2π
−∞
0
Z
0
−∞
d exp −
(2x − y)2
1
dx = −
φ(y),
2
2
and so, adding up, we see that the total integral for y > 0 is φ(y).
1
For fixed y < 0, the integral over x splits into integrals over two disjoint intervals, namely
] − ∞, y] and [−y, ∞[. The second is
Z ∞
1
(2x + y)2
1
1
1
√
(2x + y) exp −
dx = √ exp −−
(−2y + y)2 = −
φ(y).
2
2
2
2π −y
2 2π
The integral over negative values of x is computed exactly similarly, and has the same
value. Thus the total integral is again φ(y), which means that the marginal distribution
of Y is standard normal.
The correlation is the expectation of XY , since the variables are centred and have variance 1. We have to compute the double integral of xy fX,Y (x, y) over the area in the
(x, y)-plane bounded below by y = x for x < 0 and y =√−x for x > 0. For the region
where x > 0, the integral is, ignoring the denominator of 2π,
Z ∞Z ∞
(2x + y)2
dy dx.
xy(2x + y) exp −
2
0
−x
For the region with x < 0, the integral is
Z 0 Z ∞
(−2x + y)2
dy dx.
xy(−2x + y) exp −
2
−∞ x
In this second integral, change the variable x by the relation x0 = −x. Then dx0 = −dx,
and x0 varies from ∞ to 0. Thus the integral becomes
Z ∞Z ∞
(2x0 + y)2
dy dx0 .
(−x0 )y(2x0 + y) exp −
2
0
0
−x
But this is exactly the negative of the integral over positive values of x, and so the whole
integral vanishes, showing that X and Y are uncorrelated.
The density fX,Y is manifestly not the product density φ(x)φ(y), which it would be if X
and Y were independent, and so we conclude that they are not.
Note that there are many other ways of arriving at the conclusions of this exercise. Some
of them may even be neater than what’s above!
2.7 Let Y be an integrable random variable on a probability space (Ω, F, P ), and let G
be a sub-σ-algebra of F. Based on the information in G, we can form the estimate E(Y |G)
of Y and define the error of the estimation Err = Y − E(Y |G). This is a random variable
with expectation zero and some variance Var(Err). Let X be some other G--measurable
random variable, which we can regard as another estimate of Y . Show that
Var(Err) ≤ Var(Y − X).
In other words, the estimate E(Y |G) minimises the variance of the error among all estimates
based on the information in G. (Hint: Let µ = E(Y − X). Compute the variance of Y − X
as
h
i
2
E[(Y − X − µ)2 ] = E ((Y − E(Y |G)) + (E(Y |G) − X − µ)) .
Multiply out the right-hand side and use iterated conditioning to show that the cross term
is zero.)
2
The hint does most of the work, because, if the cross-term is indeed zero, we see that
Var(Y − X) = E (Y − X − µ)2 = E (Y − E(Y |G))2 + E (E(Y |G) − X − µ)2 ,
and so, since the last term above is nonnegative, we see that
Var(Y − X) ≥ E (Y − E(Y |G))2 = Var(Err),
as required.
To show that the cross-term is zero, we look at the expectation of the product
Y − E(Y |G) E(Y |G) − X − µ .
The factor E(Y |G) − X − µ is G-measurable, and so the expectation conditional on G of
the product is this variable unchanged times the conditional expectation of Y − E(Y |G),
which is clearly zero. The unconditional expectation is therefore also zero.
2.8 Let X and Y be integrable random variables on a probability space (Ω, F , P ). Then
Y = Y1 + Y2 , where Y1 = E(Y |X) is σ(X)--measurable and Y2 = Y − E(Y |X). Show
that Y2 and X are uncorrelated. More generally, show that Y2 is uncorrelated with every
σ(X)--measurable random variable.
We show only the second part, about any random variable Z that is σ(X)--measurable,
since the first part is a special case. Note first that E(Y2 ) = 0, because E(Y2 |X) =
E(Y |X) − E(Y |X) = 0. The covariance of Z and Y2 is thus the expectation of the product
ZY2 . Conditional on σ(X), we have
E(ZY2 | σ(X)) = E Z(Y − E(Y |X)) | σ(X) = ZE(Y |X) − ZE(Y |X) = 0,
since the notations E(·|X) and E(·|σ(X)) mean the same thing, and since we assume
that Z is σ(X)--measurable. The vanishing of the conditional expectation means that the
unconditional expectation is also zero, and so Z and Y2 are uncorrelated.
(i) Let X be a random variable on a probability space (Ω, F , P ), and let W be a
nonnegative σ(X)--measurable random variable. Show that there exists a function g
such that W = g(X). (Hint: Recall that every set in σ(X) is of the form {X ∈ B} for
some Borel set B ∈ R. Suppose first that W is the indicator of such a set, and then
use the standard machine.)
(ii) Let X be a random variable on a probability space (Ω, F, P ), and let Y be a nonnegative
random variable on this space. We do not assume that X and Y have a joint density.
Nonetheless, show that there is a function g such that E(Y |X) = g(X).
2.11
(i) Let the set A be in σ(X). Then A = X −1 (B) for some Borel set B. It follows that IA
is a σ(X)--measurable random variable, because, for any Borel set C,
A if 1 ∈ C,
−1
IA (C) = {ω | IA (ω) ∈ C} =
Ac if 1 ∈
/ C,
and both A and its complement Ac are in σ(X). Conversely, any indicator random variable
ID , D ∈ F, is σ(X)--measurable if D ∈ σ(X), that is, if there exists a Borel set E such
that D = X −1 (E). Thus an indicator random variable IA is σ(X)--measurable if and only
if A ∈ σ(X).
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Consider such a σ(X)--measurable set A with A = X −1 (B). Define the function g : R → R
as the indicator function IB , that is, g(x) = 1 if x ∈ B and g(x) = 0 otherwise. It is easy to
see that g ◦ X, considered as a mapping from Ω to R, is equal to the indicator function IA ,
as follows:
(g ◦ X)(ω) = g(X(ω)) = IB (X(ω)) = I(X(ω) ∈ B) = I(ω ∈ X −1 (B))
= I(ω ∈ A) = IA (ω).
This proves the result in (i) for any W that is a σ(X)--measurable indicator function.
P
Now let W = i ci IAi be a simple function, that is, a linear combination of σ(X)--measurable indicator functions IAi , with each Ai being σ(X)--measurable,Pand ci > 0. For
each Ai , let the indicator function gi be defined as above, and let g = i ci gi . Then, also
as above,
X
X
g(X(ω)) =
ci gi (X(ω)) =
ci IAi (ω) = W (ω).
i
i
This completes the second step of the standard machine. The third follows the usual
pattern with no special conditions needed, since we construct monotone convergence
up
P
to an arbitrary nonnegative function, and so obtain a limit to the sequence i ci gi (x) for
all real x. The fourth step is not asked for in this exercise.
(ii) This follows immediately from (i) on noting that the conditional expectation E(Y |X)
is a nonnegative σ(X)--measurable random variable.
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