EE306001 Probability, Fall 2012
Quiz #10, Problems and Solutions
Some probability density function you may need.
(a) Normal with parameters µ, σ 2 :
1 2 2
f (x) = √ e−(x−µ) /2σ , ∀ x ∈ R.
2πσ
(b) Exponential with parameter λ:
(
λe−λx , x ≥ 0,
f (x) =
0, x < 0.
Prob. 1: The number of years a radio functions is exponentially distributed random
1
variable with parameter λ = . If Jones buys a used radio, what is the probability that
8
it will be working after an additional 8 years?
Solution: Let X be the life time of radio. According to the problem, X is exponential
1
with parameter λ = . Assume that the radio have been used for y years when Jones
8
buys it. The probability that it will be working after an additional 8 years is
P (X > t + 8|X > t) =P (X > 8), since exponentially distributed r.v. is memoryless.
Z ∞
1 −(1/8)x
= e dx
8 8
= − e−(1/8)x |∞ −1
8 = e .
Prob. 2: The median of a continuous random variable having distribution function F is
1
that value m such that F (m) = . Find the median of X if X is
2
(a) uniformly distributed over (a, b);
(b) normal with parameters µ, σ 2 ;
(c) exponential with rate λ.
Solution: Let f (x) be the p.d.f. of X.
(a) For uniformly distributed X over (a, b), we have
1 , a<x<b
f (x) = b − a
0, otherwise.
Since F (a) = 0, F (b) = 1 and F (x) is monotonic increasing function, We have
m > a. So
Z m Z m
1 m−a 1
F (m) = f (x)dx = dx = =
−∞ a b−a b−a 2
b−a
⇒m = a + .
2
(b) Since Z µ Z ∞
1 2 2 1 2 2
F (µ) = √ e−(x−µ) /2σ dx = √ e−(y−µ) /2σ dy,
−∞ 2πσ µ 2πσ
where y = 2µ − x, and
Z µ Z ∞ Z ∞
1 −(x−µ)2 /2σ 2 1 −(y−µ)2 /2σ 2 1 2 2
√ e dx+ √ e dy = √ e−(x−µ) /2σ dx = 1.
−∞ 2πσ µ 2πσ −∞ 2πσ
We have
1
F (µ) = .
2
Therefore m = µ.
1
(c)Since F (m) = > 0, we have m > 0.
2
Z m Z m
1
F (m) = f (x)dx = λe−λx dx = 1 − e−λm = .
−∞ 0 2
ln 2
⇒m = .
λ
Prob. 3: Let X be a continuous random variable having cumulative distribution function
F . Define the random variable Y by Y = F (X). Show that Y is uniformly distributed
over (0, 1).
Solution: Since FX (x) is a distribution function, FX (x) ∈ [0, 1] for all x ∈ R so that
Y = FX (X) takes one of the values in [0, 1]. Thus for y ≥ 1, FY (y) = P (Y ≤ y) = 1 and
for y < 0, FY (y) = P (Y ≤ y) = 0. Now for 0 ≤ y < 1,
FY (y) = P (Y ≤ y) = P (0 ≤ Y ≤ y) = P (FX (X) ∈ [0, y]) = P (X ∈ FX−1 ([0, y])).
Note that
FX−1 ([0, y]) = {x ∈ R|FX (x) ≤ y} = (−∞, z]
for some z ∈ R, since FX (x) is a monotone continuous function on R. Note that FX (z) = y.
Thus
FY (y) = P (X ∈ F −1 ([0, y])) = P (X ≤ z) = FX (z) = y
and then Y is uniformly distributed over (0, 1).
