1MA01: Mathematical Methods Tutorial Sheet
1
Verify that the following are probability density functions
• f (x) = 4x3 on [0, 1].
• f (x) =
3 2
x
26
on [1, 3].
For f (x) = 4x3 on [0, 1], determine the probability P (1/4 ≤ X ≤ 3/4).
3 2
x on [1, 3], determine the probability P (1/4 ≤ X ≤ 3/4).
For f (x) = 26
In each case showing it’s a PDF and solving for the probability:
• f (x) = 4x3 on [0, 1]:
(i) x is positive or zero on the interval [0,1] so f (x) = 4x3 is
nonnegative on that
interval.
1
R1
1 4
3
(ii) 0 4x = 4 x = 1.
0
(iii) P (1/4 ≤ X ≤ 3/4) =
R 3/4
1/4
4x3 dx = 4
3/4
x4 4 1/4
= ( 34 )4 − ( 14 )4 =
120
.
400
3 2
• f (x) = 26
x on [1, 3].
(i) x is positive or zero on the interval [1,3] so f (x) = 3/26x2 is
nonnegative on that
interval.
3
R3 3 2
3
3 x
(ii) 1 26 x = 26
= 1.
3
1
(iii)P (1/4 ≤ X ≤ 3/4) =
R 3/4 3
1/4
x2 dx =
26
3/4
3 x3 26 3 1/4
=
1 27
.
26 64
In the following, find k such that the function is a probability density
function over the given interval.
• f (x) = kx on [1, 3]
• f (x) = kx2 on [−1, 1]
1
Sinéad Ryan, ryan@maths.tcd.ie, see also http://www.maths.tcd.ie/ ryan/123.html
For the first one we want, k
R3
1
xdx = 1 so
1 2 3
x
2 1
1
=
(9 − 1)
2
1
⇒k =
4
1
=
k
The PDF is f (x) = 14 x on [1, 3].
For the second,
Z 1
1
x2 dx
=
k
−1
1 3 1
=
x
3 −1
1
(1 − (−1))
=
3
3
⇒k =
2
Compute the probability using the given distribution
• P (3 ≤ X ≤ 4), X is uniformly distributed over [2, 4]
• P (200 ≤ X ≤ 350), X is uniformly distributed over [100, 500]
In each case the probability density is 1/(b − a) so 1/(4 − 2) = 1/2 for
the first case and 1/(500 − 100) = 1/400 for the second. Therefore:
•
P (3 ≤ X ≤ 4) =
Z 4
3
•
P (200 ≤ X ≤ 350) =
Z 350
200
1
1
dx =
2
2
1
350 − 200
3
=
= .
400
400
8
For each probability density function, find E(X), V ar(X) and SD(X).
• f (x) = 92 x, [0, 3].
• f (x) =
• f (x) =
4
(x + x3 ), [1, 2]
21
1
sin(x), [0, π]
2
• for f (x) = 29 x, [0, 3]:
2
2Z 3 2
2 1 3 3
E(X) =
x dx =
x = (27 − 0) = 2
9 0
9 3 0 27
2Z 3 3
x dx − (2)2
9 0
2 1 4 3
1
1
= x − 4 = (81) − 4 =
94 0
18
2
V ar(X) = E(X 2 ) − (E(X))2 =
And
SD(X) =
• for f (x) =
4
(x
21
q
7
V ar(X) = √
3 2
+ x3 ), [1, 2]:
4 1 3 1 5 2
4 Z2 2
4
x + x dx =
x + x E(X) =
21 1
21 3
5
1
4 7 31
4 128
=
+
.
=
21 3
5
21 15
V ar(X) = E(X 2 ) − (E(X))2
"
#
4 x4 x6
4 Z2 3
5
2
=
(x + x )dx − (E(X)) =
+
− (E(X))2
21 1
21 4
6
2
57
4 128
−
.
=
21
21 15
And SD(x) =
q
V ar(X).
• for f (x) = 12 sin(x), [0, π]:
Note that this question uses integration by parts. This would not
be asked in an exam question by me.
E(X) =
1Z π
π
x sin(x)dx =
2 0
2
Note that here I used integration by parts ie udv = uv − vdu
with u = x ⇒ du = dx a nd dv = sin(x)dx ⇒ v = − cos(x).
Now,
R
Z π
x2 sin(x)dx − (E(X))2 = −π 2 (−1) + 2
0
π
= π2 − 4 − .
4
V ar(X) =
Again there is another integration by parts here.
R
Z π
0
(−2x cos(x)dx −
π
4