Partial solutions to assignment 3
Math 5080-1, Spring 2011
p. 283, #5. We know that T1 , T2 , · · · ∼ EXP(100) are independent. Recall
that EXP(100) is the same distribution as GAMMA(100 , 1).
P10
(a) i=1 Ti ∼ GAMMA(100 , 10).
P10
(b) “Successful operation for at least 1.5 years” is the same sentence as “ i=1 Ti >
1.5 × 365 = 547,” assuming that : (1) There are 365 days per year; and (2) The
Ti ’s are measured in days. Therefore we are interested in the total area, under
the GAMMA(100 , 10) density curve, to the right of 547. Since
)
( 10
)
(
10
X
2 X
Ti > 10.74 ,
P
Ti > 547 = P
100 i=1
i=1
Theorem 8.3.3 [page 269] tells us that we want the probability that if X ∼
χ2 (20), then
( 10
)
X
P
Ti > 547 = P {X > 10.74}.
i=1
According to Table 4 [page 604] of your text, P {X ≤ 9.59} = 0.025} and
P {X ≤ 10.85} = 0.05. Therefore,
( 10
)
X
0.95 = P {X > 10.85} ≤ P
Ti > 547 ≤ P {X > 9.59} = 0.975.
i=1
Therefore, the probability of interest is somewhere between 0.95 and 0.975.
[Note that the problem is stated incompletely and has other interpretations.
ForP
instance, if the Ti ’s are measured in years, then we should seek to compute
10
P { i=1 Ti > 1.5} = P {X > 0.03} ≈ 1.
(c) If we continue to assume that the Ti ’s are measured in days and there are
365 days per year, then 2 years is 2 × 365 = 730 days. And the question is
stating the following: Find the smallest integer N such that
(N
)
X
P
Ti > 730 ≈ 0.95.
i=1
The number
of N .
PN of needed spared is simply N − 1, for this valueP
N
Now i=1 Ti ∼ GAMMA(100 , N ); therefore, XN := 0.02 I=1 Ti ∼ χ2 (2N ).
In other words, we need to find N such that
P {XN ≤ 0.02 × 730} = P {XN ≤ 0.02 × 14.6} ≈ 0.05.
Use Table 4 on page 604 to see that P {X25 ≤ 14.61} ≈ 0.05. Therefore, we
can choose N ≤ 25/2 = 12.5. But we cannot choose N ≤ 24/2 = 12 because
1
P {X24 ≤ 14.6} > P {X24 ≤ 13.85} ≈ 0.05 Therefore, the answer is N = 13; i.e.,
13 − 1 = 12 spares are needed.
p. 283, #8. We consider MGF’s:
MY −X (t) = MY (t)M−X (t) = MY (t)MX (−t).
This holds simply because MX (−t) = Ee−tX = M−X (t). Now, we use the MGF
table [behind the front cover of the text, for example] to find that if |t| < 1/2,
then
n/2
m/2
1
1
,
MX (−t) =
.
MY (t) =
1 − 2t
1 + 2t
[Otherwise, one the preceding two is infinity.] Therefore,

1
1
1


 (1 − 2t)n/2 (1 + 2t)m/2 if − 2 < t < 2 ,
MY −X (t) =



∞,
otherwise.
And this is not equal to the MGF of a χ2 (n − m), which is

(n−m)/2

1


if −∞ < t < 12 ,

1 − 2t




∞
otherwise.
p. 283, #9. We compute the MGF once again: Because X and Y are independent, and since S − X = Y ,
MS (t) = MX (t)MS−X (t).
Now,
MS (t) =









And
MX (t) =
1
1 − 2t
∞
MY (t) =
1
1 − 2t
m/2
∞





if t < 12 ,
otherwise.









It follows that
(n+m)/2
if t < 21 ,
otherwise.
1
1 − 2t
nm/2




∞
if t < 21 ,
otherwise.
2
This is the MGF of a χ2 (n).
p. 283, #10. Let X1 , . . . , X15 be i.i.d. EXP(θ)-distributed random variables.
We are asked to find c such that
( 15
)
X
15θ
P
Xi <
= 0.95.
c
i=1
P15
P15
Because i=1 Xi ∼ GAMMA(θ , 15), we know that 2 i=1 Xi /θ ∼ χ2 (30) [Theorem 8.3.3, page 269]. Therefore, we can write our questions as follows: Find c
such that
( 15
)
2X
30
P
Xi <
= 0.95.
θ i=1
c
Table 4 [page 604] tells us that
)
( 15
2X
Xi < 43.77 = 0.95.
P
θ i=1
Therefore, we set (30/c) = 43.77 and solve to obtain c = 30/43.77 ≈ 0.685401.
p. 283, #14. We can realize T as
Z
,
T =p
V /ν
where Z ∼ N (0 , 1) and V ∼ χ2 (ν) are independent. Therefore,
T2 =
Z2
V /ν
has the form {χ21 (1)/1}/{χ22 (ν)/ν} for independent χ2 variables χ21 and χ22 . It
follows that T 2 ∼ F1,ν .
3