Download Math 4250 Spring 2007 Take-Home Assignment # 4 Due Date: In

Math 4250
Spring 2007
Take-Home Assignment # 4
Due Date: In-Class on March 6th 2007
Instructions: Your solutions must appear in an organized and legible format to be
given full consideration.
1. (ex4 pp247) The probability density function of X, the lifetime of a certain type of
electronic device (measured in hours), is given by
f (x) =
10
x2
x > 10
x ≤ 10
0
(a)Find P (X > 20).
(b) What is the cumulative distribution function of X?
(c) What is the probability that of 6 such of devices at least 3 will function for at
least 15 hours? What assumptions are you making?
Solution:
(a)
∞
Z
P (X > 20) =
20
= −
10
dx
x2
10 ∞ 1
| =
x 20 2
(b)If x < 10 then FX (x) = 0.
If x > 10 then
Z
x
FX (x) = P (X ≤ x) =
f (t)dt
−∞
Z
x
=
10
10
10
10
dx = − |x10 = 1 −
2
x
x
x
Hence
FX (x) =
1−
0
10
x
x > 10
x ≤ 10
(c) Let us denote by Yi the ith device. The lifetime of each device is given by the
rv X. The probability that a device will function for at least 15 hours is given by
P (X ≥ 15) = 1 − P (X ≤ 15) = 1 − FX (15) = 10
= 23 .
15
Moreover Yi ∼ B(1, 32 ). We can define the new rv Y =
of devices that will work for more than 15 hours.
P6
i=1
Yi being the number
2
Y ∼ B(6, ).
3
And we are looking for the following probability
6!
P (Y ≥ 3) = 1 − P (Y ≤ 2) = 1 −
4!2!
2 4
5
2
1
1
6! 2
−
.
3
3
5! 3
3
2. (ex7 pp248) The density function is given by
f (x) =
a + bx2 0 ≤ x ≤ 1
0
otherwise
If E [X] = 23 , Find a and b.
solution: f is a density function hence
Z inf ty
Z 1
f (x)dx = 1 =⇒
(bx2 + a)dx = 1 = bx3 /3 + ax|10 = b/3 + a.
−∞
0
on the other hand
Z
Z inf ty
xf (x)dx =
E[X] =
−∞
1
x(bx2 + a)dx = bx4 /4 + ax2 /2|10 = b/4 + a/2 = 2/3.
0
Hence, we have to solve the following system of 2 equations
b
+a=0
3
b a
2
+ =
4 2
3
1
=⇒ a = , b = 2.
3
3. (ex14 pp253) If X is an exponential random variable with parameter λ, and c > 0,
show that cX is exponential with parameter λc .
Solution:
X ∼ exp(λ) =⇒
f (x) =
λe−λx
x>0
0
otherwise
FcX (x) = P (cX ≤ x) = P (X ≤ x/c) = FX (x/c)
Hence
fcX (x) =
dFcX (x)
dFX (x/c)
1
=
= fX (x/c)
dx
dx
c
fcX (x) =
λ −λ xc
e
c
0
x>0
otherwise
=⇒ cX ∼ exp(λ/c).
4. (ex29 pp254) Let X have probability density fX . Find the probability density
function of the random variable Y , defined by Y = aX + b.
Solution:
FY (y) = P (aX + b ≤ y) = P (aX + b ≤ y) = P (aX ≤ y − b)
case1: If a > 0 then
y−b
y−b
) = FX (
)
a
a
dFY (y)
1
y−b
=⇒ fY (y) =
= fX (
)
dy
a
a
FY (y) = P (aX ≤ y − b) = P (X ≤
case2: If a < 0 then
y−b
y−b
y−b
) = 1 − P (X ≤
) = 1 − FX (
)
a
a
a
dFY (y)
−1
y−b
=⇒ fY (y) =
=
fX (
)
dy
a
a
FY (y) = P (aX ≤ y − b) = P (X ≥
5. (ex30 pp254) Find the probability density function of Y = eX when X is normally
distributed with parameters µ and σ 2 . The random variable Y is said to have a
lognormal distribution (since logY has a normal distribution with parameters µ
and σ 2 .
Solution:
FY (y) = P (eX ≤ y) =y>0 = P (X ≤ ln y) = FX (ln y)
=⇒ fY (y) =
FX (ln y)
1
dFY (y)
=
= fX (ln y).
dy
dy
y
6. (Bonus) (ex18 pp253) If X is an exponential random variable with mean λ1 , show
that
k!
E X k = k , k = 1, 2, . . . .
λ
Hint: Make use of the gamma density function to evaluate the preceding .
bf Solution:
Z
∞
xk λe−λx dx
Z ∞
−k
λe−λx (λx)−k dx
= λ
E Xk =
0
0
= λ−k Γ(k + 1) =
k!
λk