Prof. H. K. Hsieh

Stat 515 Final Exam Practice Problems + Solution

Students Name: Section: .

NOTE: Write main steps of your work clearly and circle your answers.

1. Suppose that the joint p.d.f. of X and Y is given by

f(x, y) = 8xy, 0 < x < y < 1= 0 elsewhere

(a) Verify that the f(x, y) given above is indeed a p.d.f.

Answer: 10

y08xydxdy =

104yx2]y0dy =

104[y3]dy = y4]10 = 1.

So, f(x, y) is a pdf.

(b) Find the marginal probability density of X, f1(x).

Answer:

f1(x) = 1x8xydy = 4xy2]1x = 4x(1 x2), 0 < x < 1.

(c) Find the marginal probability density of Y , f2(y).

Answer:f2(y) =

y08xydy = 4yx2]y0 = 4y

3, 0 < y < 1.

(d) Are X and Y independent?

Answer: X and Y are not independent as f(x, y) 6= f1(x)f2(y). We can alsoargue that X and Y can not be independent as x and y in the support of f(x, y)depend each other.

(e) Find the expected value of X and variance of X.

Answer:

E(X) = 10xf1(x)dx =

10x4x(1x2)dx = 4[x3/3x5/5]10 = 4(1/31/5) = 8/15.

V (X) = E(X2) [E(X)]2 = (work out by yourself).(f) Find the conditional density fY |x(y) of Y given X = x.

Answer:

f(y|x) = 8xy/[4x(1 x2)] = 2x/(1 x2), 0 < x < y < 1

2. Suppose that X and Y are independent and each has p.d.f.

f(t) = 2t 0 < t < 1

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(a) Write down the joint p.d.f. of X and Y , f(x, y) =?, including the support of thefunction.

f(x, y) = (2x)(2y) = 4xy, 0 < x < 1, 0 < y < 1.

(b) Draw the line x + y = 12, indicating the origin and coordinates x and y. Then

indicate the area of x+ y < 12in the support of f(x, y).

(Draw by yourself.)

(c) Find Pr{X + Y < 12}.

Pr{X + Y < 12} =

12

0

12y

04xydxdy =

12

02y[x2]

12y

0 dy

= 1

2

02y[(

1

2 y)2]dy =

12

02y[1/4 y + y2]dy

= 2

[y2

8 y

3

3+y4

4

] 12

0

=1

96

3. Consider gains X1, X2 (in dollars) of two investiments. Assume that the two gains areindependent and normally distributed with

X1 N(100, 25), X2 N(50, 16)noting that the second mean is a negative number. Denote the total gain by T =X1 +X2.

(a) What is the distribution of the total gain? That is, give the name, mean andvariance of the distribution of T .

Since T is the sum of two normal variables, T is also a normal variable with

E(T ) = 100 + (50) = 50; V (T ) = 25 + 16 = 41

(b) What is the chance that the total gain will be greater than $60?

P [T > 60] = P [Z > (60 50)/41] = P [Z > 1.56] = 0.0594

(c) What are the name, mean, variance, and standard deviation of the distributionof Y = 2X1 3X2, respectively?Y has a normal distribution as it is a linear function of normal variavles. Themean and variance of Y are

E(Y ) = E[2X1 3X2] = 2(100) 3(50) = 200 + 150 = 350V (Y ) = V [2X1 3X2] = 22(25) + (3)2(16) = 244

4. Suppose X and Y are independent and each has a uniform distribution in the interval(0, 1).

(Omit this problem)

(a) Write down the joint pdf of g(x, y) of X and Y .

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(b) Find the m.g.f. m(t) of X.

(c) Compute the m.g.f. of U , where U = X 2Y .

5. Suppose X and Y are independent and each has a distribution with p.d.f.:

f(t) = 2e2t, for t > 0.

(a) Find the pdf of W = X2. (Noting that X has positive support.)

w = x2 which is one-to-one transformation as X is non-negative. The inverse isx = w

12 , and

J =dx

dw=

1

2w

12

So the pdf of W is

g(w) = 2e2w 1

2w

12 = w

12 e2

w, for w > 0.

(b) Write down the joint pdf of g(x, y) of X and Y .

f(x, y) = 2e2x2e2y = 4e2(x+y), for x > 0, y > 0.

(c) Define U = X + 3Y , and V = Y , then find the joint p.d.f. of U and V .

u = x+ 3y, v = y; x = u 3v, y = v, J = 1

u 3v > 0, v > 0 u > 3v > 0h(u, v) = 4e2[(u3v)+v]] = 4e2(u2v) u > 3v > 0

(d) Find the p.d.f. of U .

h(u) = 4 1

3u

0e2(u2v)dv

= 4e2u 1

3u

0e4vdv

= e2u[e4v]13u

0

= e2u[e43u 1], u > 0

6. Suppose X1, X2, X3, X4 are independent and each has a distribution with p.d.f.:

f(x) = 2x, for 0 < x < 1.

Let Y1, Y2, Y3, Y4 be their order statistics (Y1 Y2 Y3 Y4).(a) i. Pr[X1 < t] =

0 t2xdx = t

2, 0 < x < 1

ii. Pr[t1 < X1 < t2] = t22 t21, 0 < t1 t2 < 1

iii. Pr[X1 > t] = 1 t2

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(b) Find pdf of Y1.

g1(y1) =4!

1!3!f(y1)[P (X > y1]

3 = 4(2y1)[1 y21]3 = 8y1[1 y21]3, 0 < y1 < 1

(c) Find pdf of Y4.

g4(y4) =4!

3!1![F (y4)]

3f(y4) = 4[y24]

3(2y4) = 8y74, 0 < y4 < 1

(d) Find joint pdf of Y1 and Y3.

g13(y1, y3) =4!

1!1!1!1!(2y1)[y

23 y21](2y3)[1 y23], 0 < y1 < y3 < 1

7. Problems in HW#7. (See solution for HW#7)

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