Pre-Sessional Econometrics: Solutions to Exercises 1

R∞
1. To be a valid PDF we require −∞
f (x)dx = 1:
Z 2 2
2cx2 
2cxdx = 2 
0 0
= 4c
1
= 1 iff c = 4
1
We also require f (x) ≥ 0 ∀x. Given c = 4
we have f (x) = 21 x, 0 ≤ x ≤ 2, so this second
condition is also satisfied.
2. We obtain:
Z 2
P (1 ≤ X ≤ 2) = f (x)dx
1
Z 2
1
= 2
xdx
1
2
x2 
= 4 
1
4 1 3
= − =
4 4 4

3. We obtain:
Z 1 Z 0.5 
3 x y
P (0 ≤ X ≤ 0.5, 0.5 ≤ Y ≤ 1) = − − dxdy
0.5 0 2 2 2
Z 1  x=0.5 !
3 x2 xy 
= x− −  dy
0.5 2 4 2 x=0
Z 1 
3 1 y
= − − dy
0.5 4 16 4
Z 1 
11 y
= − dy
0.5 16 4
 1
11 1 2 
= y− y 
16 8 0.5
11 1 11 1
= ( − )−( − )
16 8 32 32
9 5 1
= − =
16 16 4

4. Let’s first calculate the marginal PDF of X:

Z ∞
f (x) = f (x, y)dy
−∞
Z 2
x2 + 31 xy dy


=
0

y=2
= x2 y + 61 xy 2 y=0
= 2x2 + 23 x

1
 Then we obtain:
Z ∞
P (X ≥ 0.5) = f (x)dx
Z0.51
2x2 + 23 x dx


=
0.5
2 3

1
= x + 1 x2 
3 3 0.5
= ( 23 + 13 ) − ( 12
1
+ 1
12
)
5
= 6 ≈ 0.83

5. We first calculate the conditional density function of Y given X = x:

f (x, y)
f (y|x) =
f (x)
x2 + 13 xy
=
2x2 + 23 x

We can then calculate the conditional density of Y at X = 0.5:

1
4
+ 16 y
f (y|x = 0.5) = 1
2
+ 13
3
= 10
+ 51 y

and
Z 0.5
3
+ 51 y dy


P (0 ≤ Y ≤ 0.5|X = 0.5) = 10
0
1 2 0.5
3

= 10
y + 10
y 0
3 1
= 20
+ 40
7
= 40
= 0.175