probability8

 

 You should look at the questions first if you can complete them or not. Then decide to do it. 

   

  typed answer required. 

 

 detail procedures required. 

 

 Strict due. 

Math 464 – Fall

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3 – Homework 9

1. Let X and Y be independent random varaibles. Each has an exponential
distribution with parameter λ. Define two new random variables by

W = min{X, Y }

Z = max{X, Y }

Find the joint cdf of W and Z, and use it to find their joint pdf. Hint: We ca

n

split up the event {W ≤ w, Z ≤ z} as the union of {W ≤ w, Z ≤ z, X ≤ Y }
and {W ≤ w, Z ≤ z, X > Y }.

2

. Let X and Y be random variables with joint pdf

f(x, y) =

{

1

4
exp(−1

2
(x + y), if x ≥ 0, y ≥ 0

0 otherwise
(1)

Let

U =
1

2
(X − Y ), V = Y

(a) Show that the joint pdf for U, V is of the form

f(u, v) =

{
1

2
exp(−u − v), if (u, v) ∈ A

0 otherwise
(2)

where A is a subset of the (u, v) plane that you should determine.
(b) Find the marginal pdf of U.

3. Let a, b > 0. The random variables X and Y are independent and their
densities are

fX(x) =
1

Γ(a)
xa−1e−x, x ≥ 0

fY (y) =
1

Γ(b)
yb−1e−y, y ≥ 0

(These are gamma distributions.) Let

U = X + Y, V =
X

X + Y

1

Find the joint density of U and V and show they are independent.

4. Let X and Y be independent random variables, each of which is uniform
on [0, 1]. Define two new random variables by

U = exp(X + Y ), W = exp(X − Y )

Find the joint density of U, W . Be sure to specify the range of (U, W) in
your solution.

5. Let X1, X2, · · · , Xn be independent, indentically distributed random vari-
ables. Let µ be their common mean, and σ2 their common variance. Define
two new random variables by

Xn =
1

n

n
∑

i=1

Xi

Yn =
1

n − 1

n
∑
i=1

(Xi − X)
2

(a) Find the mean and variance of the sample mean Xn. (This is easy, you’ve
seen it before.)

(b) Find the mean of Yn.
In both parts your answer can contain n, µ and σ2, but nothing else.
Remark: Yn is usually called the sample variance. It is an estimator for the
population variance σ2. One may wonder why there is a 1/(n − 1) instead of
a 1/n in this estimator. Your answer should help explain why.

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