ECON2303
Answer key 6: Joint Distributions and Covariance (advanced)
(To be covered in tutorial sessions in Week 9 from 17–20 March 2020)
1. Let (X, Y ) be any pair of random variables. From the two conditions
E[(aX + bY )2 ] = a2 E(X 2 ) + b2 E(Y 2 ) + 2abE(XY ) ≥ 0
E[(aX − bY )2 ] = a2 E(X 2 ) + b2 E(Y 2 ) − 2abE(XY ) ≥ 0,
we have
2a2 b2 + 2abE(XY ) ≥ 0
2a2 b2 − 2abE(XY ) ≥ 0.
Therefore, combining these two inequality, we obtain
−ab ≤ E(XY ) ≤ ab,
or
)| ≤ ab. This completes the proof because a =
p equivalently, |E(XY
p
2
E(Y ) and b = E(X 2 ). The inequality is called Cauchy-Schwarz inequality.
2. The result is straightforward from the direct application of Cauchy-Schwarz
inequality to X − E(X) and Y − E(Y ). Since the inequality holds for any
pair of random variables, it also holds for X − E(X) and Y − E(Y ).
3. (a) P (X ≤ 1 and Y ≤ 1) = FXY (1, 1) = 81 .
(b) P (1 ≤ X ≤ 2 and
FXY (1, 12 ) = 33
.
64
(c) P (1 ≤ X ≤ 2 and
1
2
1
2
≤ Y ≤ 2) = FXY (2, 2) − FXY (1, 2) − FXY (2, 21 ) +
≤ Y ≤ 5) = P (1 ≤ X ≤ 2 and
1
2
≤ Y ≤ 2) =
(d) P (X ≤ 10 and Y ≤ 100) = P (X ≤ 2 and Y ≤ 2) = 1.
1
33
.
64
(e)
FX (x) = FXY (x, ∞) = FXY (x, 2) =



1
x(x
8
0 when x < 0
+ 2) when 0 ≤ x ≤ 2
1 when 2 < x.
Y has the same distribution as X by symmetry- the cdf will be the same.
(f) From the answer in part (e),
1
x+
4
fX (x) =
1
4
when 0 ≤ x ≤ 2
0 otherwise.
Rx
With fX (x) above, you can see that −∞ fX (x̃)dx̃ = FX (x). Y has the
same distribution as X by symmetry- the pdf will be the same.
4. Let (X, Y ) be a pair of continuous random variables which take values on
[0, 1] × [0, 1], and the joint cdf is given by
FXY (x, y) = min(x, y) when 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1
(a) P (X ≤
1
2
1
2
and Y ≤ 13 ) = FXY ( 12 , 13 ) = min( 12 , 31 ) = 31 .
(b) P (X > and Y > 13 ) = FXY (1, 1)−FXY (1, 13 )−FXY ( 12 , 1)+FXY ( 12 , 13 ) =
min(1, 1) − min(1, 31 ) − min( 12 , 1) + min( 12 , 13 ) = 12 .
It is important to note that, unlike univariate case, P (X > 12 and Y >
1
) 6= 1 − P (X ≤ 12 and Y ≤ 13 ).
3
(c)
FXY (x2 , y2 ) − FXY (x1 , y2 ) − FXY (x2 , y1 ) + FXY (x1 , y1 )
= min(x2 , y2 ) − min(x1 , y2 ) − min(x2 , y1 ) + min(x1 , y1 )
There are three possible cases:
1) When x1 ≤ x2 ≤ y1 ≤ y2 , min(x2 , y2 ) − min(x1 , y2 ) − min(x2 , y1 ) +
min(x1 , y1 ) = 0. (similar when y1 ≤ y2 ≤ x1 ≤ x2 ).
2) When x1 ≤ y1 ≤ x2 ≤ y2 , min(x2 , y2 ) − min(x1 , y2 ) − min(x2 , y1 ) +
min(x1 , y1 ) = x2 − y1 ≥ 0. (similar when y1 ≤ x1 ≤ y2 ≤ x2 ).
3) When x1 ≤ y1 ≤ y2 ≤ x2 , min(x2 , y2 ) − min(x1 , y2 ) − min(x2 , y1 ) +
min(x1 , y1 ) = y2 − y1 ≥ 0. (similar when y1 ≤ x1 ≤ x2 ≤ y2 ).
In all these cases, FXY (x2 , y2 )− FXY (x1 , y2 )− FXY (x2 , y1 )+ FXY (x1 , y1 ) ≥
0 as it is supposed to be.
Recall that FXY (x2 , y2 )− FXY (x1 , y2 )− FXY (x2 , y1 )+ FXY (x1 , y1 ) = P (x1 ≤
X ≤ x2 and y1 ≤ Y ≤ y2 ).
2
(d)


0 when x < 0
x when 0 ≤ x ≤ 1
FX (x) = FXY (x, ∞) = FXY (x, 1) =

1 when 1 < x.
Therefore, X follows the uniform distribution: X˜U [0, 1]. By symmetry,
Y ˜U [0, 1].
(e) When Y = X,
FXY (x, y) = P (X ≤ x and Y ≤ y) = P (X ≤ x and X ≤ y) = P (X ≤ min(x, y)).
Since FX (x) = P (X ≤ x) = x -this is the result in (d)P (X ≤ min(x, y)) = min(x, y).
(f) Since Y = X, Corr(X, Y ) = 1.
3