6. we writedavar/math5010/summer2008/hw9... · 2008. 7. 23. · 2a a (2a-z)g(z)dz = z a 0...
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(oro)
V=r*x(x'&=-r)
Figure 1: The region: y+ x > 1, x > 0, y > 0
Chapter 8 Problems
3. (a) First, we find the value of c:
c =1∫∞
0
∫∞0 exp(−x− y)dxdy
=1∫1
0 e−x dx ·
∫10 e
−y dy= 1.
(b) Use Figure 1 to find that
P{X+ Y > 1} =∫10
(∫∞1−x
e−x−y dy
)dx+
∫∞1
(∫∞0
e−x−y dy
)dx
=
∫10
e−x(∫∞
1−x
e−y dy
)dx+
∫∞1
e−x(∫∞
0
e−y dy
)dx
=
∫10
e−x · e−(1−x) dx+∫∞1
e−x · 1dx
= 2e−1 ≈ 0.7358.
(c) Without integration: P{Y > 1 + X} = 1, therefore P{X < Y} = 1.
4. First we calculate FZ: FZ(a) = 0 if a < 0 since X, Y > 0 with probabilityone (why?). And if a > 0, then
FZ(a) = P{X+ Y 6 a}
=
∫a0
(∫a0
f(x ,y)dx)dy (plot the region)
=
∫a0
(∫a0
g(x+ y)dx
)dy.
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G-
\f,n'$w
Figure 2: The region between: 0 < y < a and y < z < a+ y
In order to calculate the inner integral: Fix y between zero and a andchange variables [z := x+ y] to find that∫a
0
g(x+ y)dx =
∫a+yy
g(z)dz.
Therefore [see Figure 2],
FZ(a) =
∫a0
∫a+yy
g(z)dzdy
=
∫a0
g(z)
(∫z0
dy
)dz+
∫2aa
g(z)
(∫az−a
dy
)dz
=
∫a0
zg(z)dz+
∫2aa
(2a− z)g(z)dz
=
∫a0
zg(z)dz+ 2a∫2aa
g(z)dz−
∫2aa
zg(z)dz.
Differentiate [d/da] to find that if a < 0, then fZ(a) = 0; else if a > 0,then
fZ(a) = ag(a) + 2∫2aa
g(z)dz+ 2a {g(2a) − g(a)} − 2ag(2a) − ag(a)
= 2∫2aa
g(z)dz− 2ag(a).
[The fundamental theorem of calculus] We can slightly simplify the latter:fZ(a) = 2
∫2aa {g(z) − g(a)}dz.
6. We writew(x ,y) = x2 + y2 and z(x ,y) = y/x.
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Then r :=√w and θ := arctan(z) define (x ,y) in polar coordinates. That
is,x =√w cos(arctan z) and y =
√w sin(arctan z).
Then, the Jacobian of this transformation is
J(w , z) =∂x
∂w
∂y
∂z−∂x
∂z
∂y
∂w.
But
∂x
∂w=
12√w
cos(arctan z)
∂x
∂z= −√w
sin(arctan z)1 + z2
∂y
∂w=
12√w
sin(arctan z)
∂y
∂z=√w
cos(arctan z)1 + z2
.
Therefore,
J(w , z) =1
2(1 + z2).
Consequently,
fW,Z(w , z) = fX,Y(x(w , z) ,y(w , z))× |J(w , z)|.
Because x2(w , z) + y2(w , z) = w2,
fX,Y(x(w , z) ,y(w , z)) =c
(1 +w2)3/2,
and sofW,Z(w , z) =
c
2(1 +w2)3/2(1 + z2),
for w > 0 and −∞ < z < ∞. Else, fW,Z(w , z) = 0. We can writefW,Z(w , z) =
cπ
2(1 +w2)3/2× 1π(1 + z2)
, w > 0, −∞ < z < ∞.Therefore,W and Z are independent with
fZ(z) =1
π(1 + z2)(i.e., Z is Cauchy),
andfW(w) =
cπ
2(1 +w2)3/2, w > 0.
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It remains to find c [this is a part of #5]. Clearly,
c =2
π∫∞0 (1 +w
2)−3/2 dw.
In order to compute the integral, change variables [a = 1/(1 + w2)] tofind that ∫∞
0
dw
(1 +w2)3/2=
12
∫10
da√1 − a
= 1.
Therefore, c = 2/π.
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