Download Mathematics 4250 Midterm 2 with solutions

Mathematics 4250
Midterm 2 with solutions
Instructor: Hakima Bessaih
March 29, 2007
Partial credit will be awarded for your answers, so it is to your advantage to explain your
reasoning and what theorems you are using when you write your solutions. Please answer
the questions in the space provided and show your computations.
Good luck!
1
2
3
4
5
6
Total
Name:
1
I. Let
f (x, y) =
24xy 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ x + y ≤ 1
0
otherwise
1. Show that f (x, y) is a joint probability density function.
2. Find E[X].
3. Find E[Y ].
Solution:
1. We must show that
Z
∞
Z
R∞ R∞
−∞
−∞
∞
f (x, y)dxdy = 1.
Z
1
Z
1−y
1
Z
−∞
24xydxdy =
0
0
Z
Z 1
2
12y(1 − y) dy =
=
12x2 y|01−y dy
f (x, y)dxdy =
−∞
0
1
12(y − 2y 2 + y 3 )dy
0
0
2
3
= 12(1/2y − 2/3y + 1/4y 4 )|10 = 12(1/2 − 2/3 + 1/4) = 1.
2.
∞
Z
E[X] =
1
Z
xfX (x)dx =
1−x
Z
x
−∞
Z 1
24xydydx
0
0
12x2 (1 − x)2 dx
=
Z0 1
2
Z
2
1
(12x2 − 24x3 + 12x4 )dx
12x (1 − 2x + x )dx =
=
0
0
= (4x3 − 6x4 + 12/5x5 )|10 = 2/5
3.
Z
∞
E[Y ] =
1
Z
yfY (y)dy =
−∞
Z 1
Z
1−y
24xydxdy
y
0
0
12y 2 (1 − y)2 dy
=
Z0 1
2
2
Z
12y (1 − 2y + y )dy =
=
0
1
(12y 2 − 24y 3 + 12y 4 )dy
0
3
4
= (4y − 6y + 12/5y
5
)|10
= 2/5
PS: for (3), we could have found directly that E[Y ] = 2/5 by symmetry of f .
2
II. The joint density function of X and Y is
x + y 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
f (x, y) =
0
otherwise
1. Are X and Y independent?
2. Find P (X + Y < 1).
Solution:
1. For 0 < x < 1
Z
∞
1
Z
(x + y)dy
f (x, y)dy =
fX (x) =
−∞
0
= (xy + y 2 /2)|10 = x + 1/2.
And for 0 < y < 1
Z
∞
1
Z
f (x, y)dx =
fY (y) =
−∞
2
(x + y)dx
0
= (x /2 + xy)|10 = y + 1/2.
Hence
fX (x) · fY (y) = (x + 1/2)(y + 1/2) 6= (x + y) = f (x, y).
Thus, X and Y are not independent.
2. Let
C = (x, y) ∈ (0, 1)2 , x + y < 1 = {(x, y), 0 < x < 1, 0 < y < 1 − x} .
Z
1
Z
1−x
P {X + Y < 1} = P {(X, Y ) ∈ C} =
(x + y)dydx
0
0
Z 1
=
[x(1 − x) + (1 − x)2 /2]dx = 1/3.
0
3
III. Assume X and Y are independent exponential random variables with common parameter
λ.
1. Find the probability density function of Z = X + Y .
2. Find P (Y < t|X + Y > s), for some s > 0 and t > 0.
Solution:
1.
λe−λx
x≥0
0
otherwise
λe−λy
y≥0
0
otherwise
fX (x) =
and
fY (y) =
Since X are Y are independent, the joint density function of X and Y is given by their
product
2 −λ(x+y)
λe
x ≥ 0, y ≥ 0
fX,Y (x, y) =
0
otherwise
Let
C = {(x, y), x ≥ 0, y ≥ 0, x + y ≤ a}
= {(x, y), 0 ≤ x ≤ a − y, 0 ≤ y ≤ a} .
For a > 0
FX+Y (a) = P (X + Y ≤ a) = P [(X, Y ) ∈ C]
Z Z
Z Z
2
fX,Y (x, y)dxdy = λ
e−λ(x+y) dxdy
=
C
Z Ca Z a−y
= λ2
e−λx e−λy dxdy = 1 − e−λa (1 + λa).
0
0
Hence a > 0
d
FX+Y (a)
da
d
=
1 − e−λa (1 + λa)
da
= λ2 ae−λa
fX+Y (a) =
Hence
fX+Y (a) =
λ2 ae−λa
a≥0
0
otherwise
4
2.
P (Y < t, X + Y > s)
P (X + Y > s)
P [(X, Y ) ∈ C1 ]
=
P [(X, Y ) ∈ C2 ]
P (Y < t|X + Y > s) =
Where
C1 = {(x, y), x ≥ 0, y ≥ 0, y < t, x + y > s}
= {(x, y), 0 < y < t, x > s − y} .
C2 = {(x, y), x ≥ 0, y ≥ 0, x + y > s}
= {(x, y), 0 < y, x > s − y} .
P [(X, Y ) ∈ C1 ] = λ
2
∞
Z tZ
e−λ(x+y) dxdy
0
s−y
−λs
= λte
.
and
P [(X, Y ) ∈ C2 ] = λ
2
Z
0
∞
Z
∞
e−λ(x+y) dxdy
s−y
= λ.
Hence
P (Y < t|X + Y > s) = te−λs .
5
VI. If X and Y are independent and identically distributed uniform random variables on
(0, 1), compute the joint density of U = Y − X and Y .
Solution:
Let fX (x) and fY (y) the associated densities of respectively X and Y .
1 0≤x≤1
0 otherwise
1 0≤y≤1
0 otherwise
fX (x) =
fY (y) =
Then using the independence of X and Y , we have that the joint density function of X and
Y is given by
f (x, y) =
1 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
0
otherwise
Let g1 (x, y) = y − x and g2 (x, y) = y Hence the Jacobian of g1 and g2 : Jac(g1 , g2 ) = −1.
On the other side if we let u = y − x and y = y then 0 < x < 1, 0 < y < 1 implies that
−1 < u < 1, 0 < y < 1. Hence using the formula for the change of variables we get that
fU,Y (u, y) = fX,Y (x, y) · |jac(g1 , g2 )|−1 .
Hence
fU,Y (u, y) =
1 −1 ≤ u ≤ 1, 0 ≤ y ≤ 1
0
otherwise
6
V. The joint probability mass function of the random variable X, Y , Z is given by
1
p(1, 2, 3) = p(2, 2, 1) = p(2, 3, 2) = .
3
1. Find E[XY Z].
Solution:
E[XY Z] =
X
xyzp(x, y, z)
x,y,z
1
(1 · 2 · 3 + 2 · 2 · 1 + 2 · 3 · 2)
3
1
(6 + 4 + 12) = 22/3.
=
3
=
7
VI. (Bonus) If X1 , X2 , X3 , X4 and X5 are independent and identically distributed exponential random variables with the parameter λ, compute
1. P (min(X1 , . . . , X5 ) ≤ a).
2. P (max(X1 , . . . , X5 ) ≤ a)
Solution:
Since the Xi ’s are exponential random variables with parameter λ
P (Xi > a) = e−λa i = 1, . . . , 5.
and
P (Xi ≤ a) = 1 − e−λa i = 1, . . . , 5.
1. We know that Xi ≥ min(X1 , . . . , X5 ) i = 1, . . . , 5.
On the other hand
P (min(X1 , . . . , X5 ) ≤ a) = 1 − P (min(X1 , . . . , X5 ) > a)
= 1 − P (X1 > a, . . . , X5 > a)
5
5
Y
Y
= 1−
P (Xi > a) = 1 −
e−λa
i=1
−5λa
i=1
= 1−e
2. We know that Xi ≤ max(X1 , . . . , X5 ) i = 1, . . . , 5. Hence
P (max(X1 , . . . , X5 ) ≤ a) = P (X1 ≤ a, . . . , X5 ≤ a)
5
5
Y
Y
=
P (Xi ≤ a) =
(1 − e−λa )
i=1
i=1
−λa 5
= (1 − e
8
)