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Math 180A, Fall 2009
Homework 10 Solutions
4.5.4. If a > 0, then
FaX+b(x) = P (aX + b ≤ x) = P (X ≤ (x − b)/a) = F ((x − b)/a).
If a < 0, then
FaX+b(x) = P (aX + b ≤ x) = P (X ≥ (x − b)/a)
= 1 − P (X < (x − b)/a) = 1 − F
(x − b)
a
4.5.5. By the symmetry of fX , we have FX (−x) = 1 − FX (x), so it is enough to calculate FX (x)
for x ≥ 0. We have
FX (x) =
∫ (^) x
−∞
e−|t|^ dt
∫ (^) x
0
e
−t dt
(1 − e
−x )
e−x.
Therefore,
FX (x) =
2 e
x, x ≤ 0,
1 − 12 e−x, x ≥ 0.
4.5.6. (a) P (X ≥ 1 /2) = 1 − P (X < 1 /2) = 1 − (1/2)^3 = 7/8.
(b) f (x) = F ′(x) = 3x^2 for 0 < x < 1.
(c) E(X) =
0 x^ ·^3 x
(^3) dx = ∫^1 0 3 x
(^3) dx = 3/4.
(d) X is the largest of Y 1 , Y 2 , and Y 3. Therefore, for 0 ≤ x ≤ 1,
P (X ≤ x) = P (Y 1 ≤ x, Y 2 ≤ x, Y 3 ≤ x)
= P (Y 1 ≤ x) · P (Y 2 ≤ x) · P (Y 3 ≤ x)
= x · x · x = x^3 ,
as desired.
4.5.7. (a) Observe that Y is a one-to-one function of T. For y =
t we have t = y^2 and
dt/dy = 2y, so
fY (y) = fT (y
2 ) · 2 y = λe
−λy^2 2 y = 2λye
−λy^2 ,
for y ≥ 0.
(b) When λ = 3,
E(Y ) =
0
6 y^2 e−^3 y
2 dy = 3−^1 /^2
0
z^1 /^2 e−z^ dz,
where we have made the change of variables z = 3 y^2. This integral is equal to Γ(3/2) =
(1/2)Γ(1/2) =
π/2. Therefore, E(Y ) = (^12)
π/3 = 0.512.
Alternatively, we can evaluate E(Y ) by integrating by parts:
E(Y ) =
0
6 y
2 e
− 3 y^2 dy =
0
y · 6 ye
− 3 y^2 dy
= −ye
− 3 y^2
∞
0
0
e
− 3 y^2 dy
0
e
− 3 y^2 dy
Making the change of variables x =
6 y (so x^2 /2 = 3y^2 and dx =
6 dy), this last integral can be
rewritten as 1 √ 6
0
e
−x^2 / 2 dx.
But the symmetry of the standard normal density implies that
∫ (^) ∞
0
2 π
e
−x^2 / 2 dx = 1/ 2 ,
so that (^) ∫ (^) ∞
0
e−x
(^2) / 2 dx =
π/ 2.
It follows that
E(Y ) =
π/ 2 √ 6
π/ 3 ,
as before.
4.5.8. Let the lifetimes of the individual components be Ti and let L be the lifetime of the system.
Let us write λi = μ− i 1 for the rate of the exponential random variable Ti.
(c) The random variable min(T 1 , T 2 ) has the exponential distribution with parameter λ 1 + λ 2 ;
the random variable min(T 3 , T 4 ) has the exponential distribution with parameter λ 3 +λ 3. Therefore
P (L > t) = 1 − P (L ≤ t) = 1 − P (min(T 1 , T 2 ) ≤ t, min(T 3 , T 4 ) ≤ t)
= 1 − (1 − e
−t(λ 1 +λ 2 ) )(1 − e
−t(λ 3 +λ 4 ) ).
When μi = i this becomes
P (L > t) = 1 − (1 − e
− 2 t/ 3 )(1 − e
− 7 t/ 12 ) = e
− 2 t/ 3
− 7 t/ 12 − e
− 7 t/ 18 .
(d) We now have
P (L > t) = P (max(T 1 , T 2 ) > t, T 3 > t) = P (max(T 1 , T 2 ) > t) · P (T 3 > t)
[
1 − (1 − e
−λ 1 t )(1 − e
−λ 2 t )
]
e
−λ 3 t .
When μi = i this becomes
P (L > t) =
[
1 − (1 − e
−t )(1 − e
−t/ 2 )
]
e
−t/ 3 = e
− 4 t/ 3
− 5 t/ 6 − e
− 11 t/ 6 .
