Download Probability II - Continuous Random Variables (1) - Solutions and more Exercises Statistics in PDF only on Docsity!
STAT/MATH 395 A - PROBABILITY II – UW
Winter Quarter 2017 Néhémy Lim
HW1 : Continuous Random Variables (1) – Solutions
Problem 1. Let X be a continuous random variable whose probability density
function is:
fX ( x ) = 3 x
2 (^1) 0 , 1
(a) Verify that fX is a valid probability density function.
Answer. First, fX ( x ) is clearly nonnegative since 3 x 2 ≥ 0 for 0 ≤ x ≤ 1
and 0 otherwise. Second, let us show that
R
fX = 1 :
−∞
fX ( x ) dx =
0
3 x
2 dx = x
3
1
0
(b) What is P( X = 1 / 2)?
Answer. X is a continuous random variable. Hence, the probability that
X takes on a particular value is zero : P( X = 1 / 2) = 0.
(c) What is P(1 / 2 < X < 1)?
Answer.
P(1 / 2 < X < 1) =
1 / 2
fX ( x ) dx
1 / 2
3 x
2 dx
= x
3
1
1 / 2
(d) Give the cumulative distribution function FX of X.
Answer. The cumulative distribution function FX of X is piecewise like
its probability density function :
- If x < 0 , then fX ( t ) = 0 for all t ≤ x.
FX ( x ) =
∫ (^) x
−∞
fX ( t ) dt =
∫ (^) x
−∞
0 dt = 0
- If 0 ≤ x ≤ 1 , then fX ( t ) = 3 t 2 for all 0 ≤ t ≤ x.
FX ( x ) = FX (0) +
x
0
fX ( t ) dt
x
0
3 t
2 dt
= t
3
x
0
= x
3
- If x > 1 , then fX ( t ) = 0 for all 1 ≤ t ≤ x.
FX ( x ) = FX (1) +
x
1
fX ( t ) dt
3
In a nutshell, FX is given by :
FX ( x ) =
0 if x < 0
x
3 if 0 ≤ x ≤ 1
1 if x > 1
(e) What is the expected value of X?
Answer.
E[ X ] =
∞
−∞
xfX ( x ) dx
0
3 x
3 dx
x
4
1
0
(f) What is the variance of X?
x
fX ( x )
(b) Verify that fX is a valid probability density function.
Answer. It is obvious that fX is nonnegative. Let us show that
R
fX = 1
:
∞
−∞
fX ( x ) dx =
− 1
−∞
fX ( x ) dx +
0
− 1
fX ( x ) dx +
1
0
fX ( x ) dx +
∞
1
fX ( x ) dx
− 1
( x + 1) dx +
0
(1 − x ) dx + 0
[
x 2
] 0
− 1
[
x −
x 2
] 1
0
(c) Give the cumulative distribution function FX of X.
Answer. The cumulative distribution function FX of X is piecewise like
its probability density function :
- If x < − 1 , then fX ( t ) = 0 for all t ≤ x.
FX ( x ) =
∫ (^) x
−∞
fX ( t ) dt =
∫ (^) x
−∞
0 dt = 0
- If − 1 ≤ x ≤ 0 , then fX ( t ) = t + 1 for all − 1 ≤ t ≤ x.
FX ( x ) = FX (−1) +
∫ (^) x
− 1
fX ( t ) dt
∫ (^) x
− 1
( t + 1) dt
t 2
x
− 1
x
2
x 2
( x + 1)
2
- If 0 < x ≤ 1 , then fX ( t ) = 1 − t for all 0 < t ≤ x.
FX ( x ) = FX (0) +
∫ (^) x
0
fX ( t ) dt
2
x
0
(1 − t ) dt
[
t −
t 2
] x
0
x −
x 2
x
2
- If x > 1 , then fX ( t ) = 0 for all 1 ≤ t ≤ x.
FX ( x ) = FX (1) +
x
1
fX ( t ) dt
2
In a nutshell, FX is given by :
FX ( x ) =
0 if x < − 1 1 2 ( x + 1)
2 if − 1 ≤ x ≤ 0
x 2
2
1 2
if 0 < x ≤ 1
1 if x > 1
