STAT/MATH 395 A - Winter Quarter 2017 - Quiz 1
Name:
Problem. Let X be a continuous random variable whose probability density function is:
fX (x) = 4xe−2x 1[0,∞) (x)
(a) [2 points] Verify that fX is a valid probability density function.
Answer.
• Nonnegativity of fX : For any x ∈ [0, ∞), 4xe−2x ≥ 0, otherwise fX (x) = 0 ≥ 0.
R
• Let us now verify that R fX = 1 :
Z ∞
Z ∞
fX (x) dx =
4xe−2x dx
−∞
0
u = 4x
dv = e−2x dx
du = 4 dx v = −1/2 e−2x
∞ Z ∞
Z ∞
fX (x) dx = −2xe−2x −
−2e−2x dx
−∞
0
0
∞
−2x
−2x = −2xe
−e
Let us perform the following integration by parts :
0
=1
(b) [2 points] Compute P(2 ≤ X < 3).
Answer.
3
Z
P(2 ≤ X < 3) =
fX (x) dx
2
3
Z
4xe−2x dx
=
2
3
= −(2x + 1)e−2x 2
= 5e−4 − 7e−6
(c) [2 points] Give FX (1) where FX is the cumulative distribution function of X.
Answer.
Z
FX (1) = FX (0) +
1
fX (x) dx
0
Z
=0+
1
4xe−2x dx
0
1
= −(2x + 1)e−2x 0
= 1 − 3e−2
1
(d) [2 points] Compute the expected value of X.
Answer. The expected value of X is given by :
Z ∞
E[X] =
xfX (x) dx
−∞
Z ∞
4x2 e−2x dx
=
0
Let us perform the following integration by parts :
∞
E[X] = −2x2 e−2x +
u = x2
dv = 4e−2x dx
du = 2x dx v = −2e−2x
Z
0
0
|
∞
4xe−2x dx
{z
}
=1 (cf. previous question)
=0+1
=1
(e) [2 points] You are given that E[X 2 ] = 3/2. Find the variance of X.
Answer. The variance of X is given by :
Var(X) = E[X 2 ] − (E[X])2
3
= − 12
2
1
=
2
2