Probability Problems with Solutions | STP 421, Assignments of Probability and Statistics

Material Type: Assignment; Class: Probability; Subject: Statistics and Probability; University: Arizona State University - Tempe; Term: Fall 1996;

Typology: Assignments

Pre 2010

Uploaded on 09/02/2009

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SOLUTION TO PROBLEM 31 OF CHAPTER 3
We are assuming that Xand Yare independent random variables both having a
discrete uniform density on the first Npositive integers. So, in particular,
fX(x)=
1
N,x=1,2,...,N;
0,otherwise.
and similarly for fY.
We want to determine the density of X+Y. For convenience, set Z=X+Y.
By (25) on page 72,
fZ(z)=
z
X
x=0
fX(x)fY(zx).()
Note that the range of Zis {2,3,..., 2N}.
Now, a summand in () is nonzero if and only if
1xN
1zxNor 1xN
zNxz1
or
max{1,zN}≤xmin{N,z 1}.
But,
max{1,zN}=(1,zN;
zN, z N+1
and
min{N,z 1}=(z1,zN;
N, z N+1
Thus, if z=2,3,...,N,
f
Z(z)=
z1
X
x=1
1
N·1
N=z1
N2,
if z=N+1, N+2,...,2N,
f
Z
(z)=
N
X
x=zN
1
N·1
N=2N+1z
N2,
and fZ(z) = 0 otherwise.

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SOLUTION TO PROBLEM 31 OF CHAPTER 3

We are assuming that X and Y are independent random variables both having a discrete uniform density on the first N positive integers. So, in particular,

fX (x) =

1 N ,^ x^ = 1,^2 ,... , N^ ;

0 , otherwise.

and similarly for fY.

We want to determine the density of X + Y. For convenience, set Z = X + Y. By (25) on page 72,

fZ (z) =

∑^ z

x=

fX (x)fY (z − x). (∗)

Note that the range of Z is { 2 , 3 ,... , 2 N }.

Now, a summand in (∗) is nonzero if and only if

1 ≤ x ≤ N 1 ≤ z − x ≤ N

or

1 ≤ x ≤ N z − N ≤ x ≤ z − 1

or max{ 1 , z − N } ≤ x ≤ min{N, z − 1 }.

But,

max{ 1 , z − N } =

{ (^1) , z ≤ N ;

z − N, z ≥ N + 1

and

min{N, z − 1 } =

{ (^) z − 1 , z ≤ N ;

N, z ≥ N + 1

Thus, if z = 2, 3,... , N ,

fZ (z) =

z∑− 1

x=

N

N

z − 1 N 2

if z = N + 1, N + 2,... , 2N ,

fZ (z) =

∑^ N

x=z−N

N

N

2 N + 1 − z N 2

and fZ (z) = 0 otherwise.