more discrete random variable solved problems
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7242019 More Discrete Random Variable Solved Problems
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325 Solved Problems
More about Discrete Random Variables
Problem 1
Let be a discrete random variable with the following PMF
Find and plot the CDF of
Solution
The CDF is defined by We have
Problem 2
Let be a discrete random variable with the following PMF
a Find b Find Var
c If find
Solutiona
b We can use Var Thus we need to find Using LOTUS we have
Thus we have
X
( 983160 ) = P
X
⎧
⎩
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
0 3
0 2
0 3
0 2
0
f o r 983160 = 3
f o r 983160 = 5
f o r 983160 = 8
f o r 983160 = 1 0
o t h e r w i s e
X
( 983160 ) = P ( X le 983160 ) F
X
( 983160 ) = F
X
⎧
⎩
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
0
( 3 ) = 0 3 P
X
( 3 ) + ( 5 ) = 0 5 P
X
P
X
( 3 ) + ( 5 ) + ( 8 ) = 0 8 P
X
P
X
P
X
1
f o r 983160 lt 3
f o r 3 le 983160 lt 5
f o r 5 le 983160 lt 8
f o r 8 le 983160 lt 1 0
f o r 983160 ge 1 0
X
( 983147 ) = P
X
⎧
⎩
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
0 1
0 4
0 3
0 2
0
f o r 983147 = 0
f o r 983147 = 1
f o r 983147 = 2
f o r 983147 = 3
o t h e r w i s e
E X
( X )
Y = ( X minus 2 )
2
E Y
E X
= ( ) sum
isin 983160
983147
R
X
983160
983147
P
X
983160
983147
= 0 ( 0 1 ) + 1 ( 0 4 ) + 2 ( 0 3 ) + 3 ( 0 2 )
= 1 6
( X ) = E minus ( E X = E minus ( 1 6 X
2
)
2
X
2
)
2
E X
2
E = ( 0 1 ) + ( 0 4 ) + ( 0 3 ) + ( 0 2 ) = 3 4 X
2
0
2
1
2
2
2
3
2
2
7242019 More Discrete Random Variable Solved Problems
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c Again using LOTUS we have
Problem 3
Let be a discrete random variable with PMF
Define Find the PMF of
Solution
First note that Thus
Thus
Problem 4
Let Find
Solution
The PMF of is given by
where Thus
V a r ( X ) = ( 3 4 ) minus ( 1 6 = 0 8 4 )
2
E ( X minus 2 = ( 0 minus 2 ( 0 1 ) + ( 1 minus 2 ( 0 4 ) + ( 2 minus 2 ( 0 3 ) + ( 3 minus 2 ( 0 2 ) = 1 )
2
)
2
)
2
)
2
)
2
X
( 983147 ) = P
X
⎧
⎩
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
0 2
0 2
0 3
0 3
0
f o r 983147 = 0
f o r 983147 = 1
f o r 983147 = 2
f o r 983147 = 3
o t h e r w i s e
Y = X ( X minus 1 ) ( X minus 2 ) Y
= 983160 ( 983160 minus 1 ) ( 983160 minus 2 ) | 983160 isin 0 1 2 3 = 0 6 R
Y
( 0 ) P
Y
= P ( Y = 0 ) = P ( ( X = 0 ) o r ( X = 1 ) o r ( X = 2 ) )
= ( 0 ) + ( 1 ) + ( 2 ) P
X
P
X
P
X
= 0 7
( 6 ) P
Y
= P ( X = 3 ) = 0 3
( 983147 ) = P
Y
⎧
⎩
⎨
⎪
⎪
0 7
0 3
0
f o r 983147 = 0
f o r 983147 = 6
o t h e r w i s e
X sim G 983141 983151 983149 983141 983156 983154 983145 c ( 983152 ) E 983131 983133
1
2
X
X
( 983147 ) = 983163 P
X
983152 983153
983147 minus 1
0
f o r 983147 = 1 2 3
o t h e r w i s e
983153 = 1 minus 983152
E 983131 983133
1
2
X
= ( 983147 ) sum
infin
983147 = 1
1
2
983147
P
X
= 983152 sum
infin
983147 = 1
1
2
983147
983153
983147 minus 1
=
983152
2
sum
infin
983147 = 1
( 983081
983153
2
983147 minus 1
=
983152
2
1
1 minus
983153
2
=
983152
1 + 983152
7242019 More Discrete Random Variable Solved Problems
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Problem 5
If find
Solution
The PMF of is given by
where
Finding directly seems to be very complicated So lets try to see if we can find an easier way to find
In particular a powerful tool that we have is linearity of expectation Can we write as the sum of
simpler random variables To do so lets remember the random experiment behind the hypergeometric
distribution You have a bag that contains blue marbles and red marbles You choose marblesat random (without replacement) and let be the number of blue marbles in your sample In particular lets
define the indicator random variables as follows
Then we can write
Thus
To find we note that for any particular all marbles are equally likely to be chosen This is
because of symmetry no marble is more likely to be chosen than the th marble as any other marblesTherefore
We conclude
Thus we have
Problem 6
In Example 314 we showed that if then We found this by writing as the sum
of random variables Now find directly using Hint Use
Solution
X sim H 983161 983152 983141 983154 983143 983141 983151 983149 983141 983156 983154 983145 c ( b 983154 983147 ) E X
X
( 983160 ) = P
X
⎧
⎩
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
( 983081 ( )
b
983160
983154
983147 minus
983160
( 983081
b + 983154
983147
0
f o r 983160 isin R
X
o t h e r w i s e
= m a x ( 0 983147 minus 983154 ) m a x ( 0 983147 minus 983154 ) + 1 m a x ( 0 983147 minus 983154 ) + 2 m i n ( 983147 b ) R
X
E X
E X X
X
983145
b 983154 983147 le b + 983154
X
X
983145
= 983163 X
983145
1
0
i f t h e 983145 t h c h o s e n m a r b l e i s b l u e
o t h e r w i s e
X = + + ⋯ + X
1
X
2
X
983147
E X = E + E + ⋯ + E X
1
X
2
X
983147
P ( = 1 ) X
983145
X
983145
983145
P ( = 1 ) = f o r a l l 983145 isin 1 2 ⋯ 983147 X
983145
b
b + 983154
E X
983145
= 0 sdot 983152 ( = 0 ) + 1 sdot P ( = 1 ) X
983145
X
983145
=
b
b + 983154
E X =
983147 b
b + 983154
X sim B 983145 983150 983151 983149 983145 a 983148 ( 983150 983152 ) E X = 983150 983152 X
983150 B 983141 983154 983150 983151 983157 983148 983148 983145 ( 983152 ) E X E X = ( ) sum
isin 983160
983147
R
X
983160
983147
P
X
983160
983147
983147 ( ) = 983150 ( 983081
983150
983147
983150 minus 1
983147 minus 1
( 983081
7242019 More Discrete Random Variable Solved Problems
httpslidepdfcomreaderfullmore-discrete-random-variable-solved-problems 45
First note that we can prove by the following combinatorial interpretation Suppose that
from a group of students we would like to choose a committee of students one of whom is chosen to bethe committee chair We can do this
1 by choosing people first (in ways) and then choosing one of them to be the chair ( ways) or
2 by choosing the chair first ( possibilities and then choosing students from the remaining
students (in ways))
Thus we conclude
Now lets find for
Note that the last line is true because the is equal to for a
random variable that has distribution hence it is equal to
Problem 7
Let be a discrete random variable with Prove
Solution Note that
Thus
Problem 8
If find Var
Solution
Problem 9
983147 ( ) = 983150 ( 983081
983150
983147
983150 minus 1
983147 minus 1
983150 983147
983147 ( )
983150
983147
983147
983150 983147 minus 1
983150 minus 1 ( 983081
983150 minus 1
983147 minus 1
983147 983080 983081 = 983150 983080 983081
983150
983147
983150 minus 1
983147 minus 1
E X X sim B 983145 983150 983151 983149 983145 a 983148 ( 983150 983152 )
E X = 983147 ( ) sum
983150
983147 = 0
983150
983147
983152
983147
983153
983150 minus 983147
= 983147 ( ) sum
983150
983147 = 1
983150
983147
983152
983147
983153
983150 minus 983147
= 983150 ( 983081 sum
983150
983147 = 1
983150 minus 1
983147 minus 1
983152
983147
983153
983150 minus 983147
= 983150 983152 ( 983081 sum
983150
983147 = 1
983150 minus 1
983147 minus 1
983152
983147 minus 1
983153
983150 minus 983147
= 983150 983152 ( 983081 sum
983150 minus 1
983148 = 0
983150 minus 1
983148
983152
983148
983153
( 983150 minus 1 ) minus 983148
= 983150 983152
( 983081 sum
983150 minus 1
983148 = 0
983150 minus 1
983148
983152
983148
983153
( 983150 minus 1 ) minus 983148
( 983148 ) sum
983150 minus 1
983148 = 0
P
Y
Y B 983145 983150 983151 983149 983145 a 983148 ( 983150 minus 1 983152 ) 1
X sub 0 1 2 R
X
E X = P ( X gt 983147 ) sum
983147 = 0
infin
P ( X gt 0 ) = ( 1 ) + ( 2 ) + ( 3 ) + ( 4 ) + ⋯ P
X
P
X
P
X
P
X
P ( X gt 1 ) = ( 2 ) + ( 3 ) + ( 4 ) + ⋯ P
X
P
X
P
X
P ( X gt 2 ) = ( 3 ) + ( 4 ) + ( 5 ) + ⋯ P
X
P
X
P
X
P ( X gt 983147 ) sum
infin
983147 = 0
= P ( X gt 0 ) + P ( X gt 1 ) + P ( X gt 2 ) +
= ( 1 ) + 2 ( 2 ) + 3 ( 3 ) + 4 ( 4 ) + P
X
P
X
P
X
P
X
= E X
X sim P 983151 983145 983155 983155 983151 983150 ( λ ) ( X )
( 2 minus ) = 6
7242019 More Discrete Random Variable Solved Problems
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Let and be two independent random variables Suppose that we know Var and Var
Find Var and Var
Solution
Lets first make sure we understand what Var and Var mean They are Var and
Var where the random variables and are defined as and
Since and are independent random variables then and are independent random variables
Also and are independent random variables Thus by using Equation 37 we can write
By solving for Var and Var we obtain Var and Var
larr previousnext rarr
X Y ( 2 X minus Y ) = 6
( X + 2 Y ) = 9 ( X ) ( Y )
( 2 X minus Y ) ( X + 2 Y ) ( Z )
( W ) Z W Z = 2 X minus Y W = X + 2 Y
X Y 2 X minus Y
X 2 Y
V a r ( 2 X minus Y ) = V a r ( 2 X ) + V a r ( - Y ) = 4 V a r ( X ) + V a r ( Y ) = 6
V a r ( X + 2 Y ) = V a r ( X ) + V a r ( 2 Y ) = V a r ( X ) + 4 V a r ( Y ) = 9
( X ) ( Y ) ( X ) = 1 ( Y ) = 2
7242019 More Discrete Random Variable Solved Problems
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c Again using LOTUS we have
Problem 3
Let be a discrete random variable with PMF
Define Find the PMF of
Solution
First note that Thus
Thus
Problem 4
Let Find
Solution
The PMF of is given by
where Thus
V a r ( X ) = ( 3 4 ) minus ( 1 6 = 0 8 4 )
2
E ( X minus 2 = ( 0 minus 2 ( 0 1 ) + ( 1 minus 2 ( 0 4 ) + ( 2 minus 2 ( 0 3 ) + ( 3 minus 2 ( 0 2 ) = 1 )
2
)
2
)
2
)
2
)
2
X
( 983147 ) = P
X
⎧
⎩
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
0 2
0 2
0 3
0 3
0
f o r 983147 = 0
f o r 983147 = 1
f o r 983147 = 2
f o r 983147 = 3
o t h e r w i s e
Y = X ( X minus 1 ) ( X minus 2 ) Y
= 983160 ( 983160 minus 1 ) ( 983160 minus 2 ) | 983160 isin 0 1 2 3 = 0 6 R
Y
( 0 ) P
Y
= P ( Y = 0 ) = P ( ( X = 0 ) o r ( X = 1 ) o r ( X = 2 ) )
= ( 0 ) + ( 1 ) + ( 2 ) P
X
P
X
P
X
= 0 7
( 6 ) P
Y
= P ( X = 3 ) = 0 3
( 983147 ) = P
Y
⎧
⎩
⎨
⎪
⎪
0 7
0 3
0
f o r 983147 = 0
f o r 983147 = 6
o t h e r w i s e
X sim G 983141 983151 983149 983141 983156 983154 983145 c ( 983152 ) E 983131 983133
1
2
X
X
( 983147 ) = 983163 P
X
983152 983153
983147 minus 1
0
f o r 983147 = 1 2 3
o t h e r w i s e
983153 = 1 minus 983152
E 983131 983133
1
2
X
= ( 983147 ) sum
infin
983147 = 1
1
2
983147
P
X
= 983152 sum
infin
983147 = 1
1
2
983147
983153
983147 minus 1
=
983152
2
sum
infin
983147 = 1
( 983081
983153
2
983147 minus 1
=
983152
2
1
1 minus
983153
2
=
983152
1 + 983152
7242019 More Discrete Random Variable Solved Problems
httpslidepdfcomreaderfullmore-discrete-random-variable-solved-problems 35
Problem 5
If find
Solution
The PMF of is given by
where
Finding directly seems to be very complicated So lets try to see if we can find an easier way to find
In particular a powerful tool that we have is linearity of expectation Can we write as the sum of
simpler random variables To do so lets remember the random experiment behind the hypergeometric
distribution You have a bag that contains blue marbles and red marbles You choose marblesat random (without replacement) and let be the number of blue marbles in your sample In particular lets
define the indicator random variables as follows
Then we can write
Thus
To find we note that for any particular all marbles are equally likely to be chosen This is
because of symmetry no marble is more likely to be chosen than the th marble as any other marblesTherefore
We conclude
Thus we have
Problem 6
In Example 314 we showed that if then We found this by writing as the sum
of random variables Now find directly using Hint Use
Solution
X sim H 983161 983152 983141 983154 983143 983141 983151 983149 983141 983156 983154 983145 c ( b 983154 983147 ) E X
X
( 983160 ) = P
X
⎧
⎩
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
( 983081 ( )
b
983160
983154
983147 minus
983160
( 983081
b + 983154
983147
0
f o r 983160 isin R
X
o t h e r w i s e
= m a x ( 0 983147 minus 983154 ) m a x ( 0 983147 minus 983154 ) + 1 m a x ( 0 983147 minus 983154 ) + 2 m i n ( 983147 b ) R
X
E X
E X X
X
983145
b 983154 983147 le b + 983154
X
X
983145
= 983163 X
983145
1
0
i f t h e 983145 t h c h o s e n m a r b l e i s b l u e
o t h e r w i s e
X = + + ⋯ + X
1
X
2
X
983147
E X = E + E + ⋯ + E X
1
X
2
X
983147
P ( = 1 ) X
983145
X
983145
983145
P ( = 1 ) = f o r a l l 983145 isin 1 2 ⋯ 983147 X
983145
b
b + 983154
E X
983145
= 0 sdot 983152 ( = 0 ) + 1 sdot P ( = 1 ) X
983145
X
983145
=
b
b + 983154
E X =
983147 b
b + 983154
X sim B 983145 983150 983151 983149 983145 a 983148 ( 983150 983152 ) E X = 983150 983152 X
983150 B 983141 983154 983150 983151 983157 983148 983148 983145 ( 983152 ) E X E X = ( ) sum
isin 983160
983147
R
X
983160
983147
P
X
983160
983147
983147 ( ) = 983150 ( 983081
983150
983147
983150 minus 1
983147 minus 1
( 983081
7242019 More Discrete Random Variable Solved Problems
httpslidepdfcomreaderfullmore-discrete-random-variable-solved-problems 45
First note that we can prove by the following combinatorial interpretation Suppose that
from a group of students we would like to choose a committee of students one of whom is chosen to bethe committee chair We can do this
1 by choosing people first (in ways) and then choosing one of them to be the chair ( ways) or
2 by choosing the chair first ( possibilities and then choosing students from the remaining
students (in ways))
Thus we conclude
Now lets find for
Note that the last line is true because the is equal to for a
random variable that has distribution hence it is equal to
Problem 7
Let be a discrete random variable with Prove
Solution Note that
Thus
Problem 8
If find Var
Solution
Problem 9
983147 ( ) = 983150 ( 983081
983150
983147
983150 minus 1
983147 minus 1
983150 983147
983147 ( )
983150
983147
983147
983150 983147 minus 1
983150 minus 1 ( 983081
983150 minus 1
983147 minus 1
983147 983080 983081 = 983150 983080 983081
983150
983147
983150 minus 1
983147 minus 1
E X X sim B 983145 983150 983151 983149 983145 a 983148 ( 983150 983152 )
E X = 983147 ( ) sum
983150
983147 = 0
983150
983147
983152
983147
983153
983150 minus 983147
= 983147 ( ) sum
983150
983147 = 1
983150
983147
983152
983147
983153
983150 minus 983147
= 983150 ( 983081 sum
983150
983147 = 1
983150 minus 1
983147 minus 1
983152
983147
983153
983150 minus 983147
= 983150 983152 ( 983081 sum
983150
983147 = 1
983150 minus 1
983147 minus 1
983152
983147 minus 1
983153
983150 minus 983147
= 983150 983152 ( 983081 sum
983150 minus 1
983148 = 0
983150 minus 1
983148
983152
983148
983153
( 983150 minus 1 ) minus 983148
= 983150 983152
( 983081 sum
983150 minus 1
983148 = 0
983150 minus 1
983148
983152
983148
983153
( 983150 minus 1 ) minus 983148
( 983148 ) sum
983150 minus 1
983148 = 0
P
Y
Y B 983145 983150 983151 983149 983145 a 983148 ( 983150 minus 1 983152 ) 1
X sub 0 1 2 R
X
E X = P ( X gt 983147 ) sum
983147 = 0
infin
P ( X gt 0 ) = ( 1 ) + ( 2 ) + ( 3 ) + ( 4 ) + ⋯ P
X
P
X
P
X
P
X
P ( X gt 1 ) = ( 2 ) + ( 3 ) + ( 4 ) + ⋯ P
X
P
X
P
X
P ( X gt 2 ) = ( 3 ) + ( 4 ) + ( 5 ) + ⋯ P
X
P
X
P
X
P ( X gt 983147 ) sum
infin
983147 = 0
= P ( X gt 0 ) + P ( X gt 1 ) + P ( X gt 2 ) +
= ( 1 ) + 2 ( 2 ) + 3 ( 3 ) + 4 ( 4 ) + P
X
P
X
P
X
P
X
= E X
X sim P 983151 983145 983155 983155 983151 983150 ( λ ) ( X )
( 2 minus ) = 6
7242019 More Discrete Random Variable Solved Problems
httpslidepdfcomreaderfullmore-discrete-random-variable-solved-problems 55
Let and be two independent random variables Suppose that we know Var and Var
Find Var and Var
Solution
Lets first make sure we understand what Var and Var mean They are Var and
Var where the random variables and are defined as and
Since and are independent random variables then and are independent random variables
Also and are independent random variables Thus by using Equation 37 we can write
By solving for Var and Var we obtain Var and Var
larr previousnext rarr
X Y ( 2 X minus Y ) = 6
( X + 2 Y ) = 9 ( X ) ( Y )
( 2 X minus Y ) ( X + 2 Y ) ( Z )
( W ) Z W Z = 2 X minus Y W = X + 2 Y
X Y 2 X minus Y
X 2 Y
V a r ( 2 X minus Y ) = V a r ( 2 X ) + V a r ( - Y ) = 4 V a r ( X ) + V a r ( Y ) = 6
V a r ( X + 2 Y ) = V a r ( X ) + V a r ( 2 Y ) = V a r ( X ) + 4 V a r ( Y ) = 9
( X ) ( Y ) ( X ) = 1 ( Y ) = 2
7242019 More Discrete Random Variable Solved Problems
httpslidepdfcomreaderfullmore-discrete-random-variable-solved-problems 35
Problem 5
If find
Solution
The PMF of is given by
where
Finding directly seems to be very complicated So lets try to see if we can find an easier way to find
In particular a powerful tool that we have is linearity of expectation Can we write as the sum of
simpler random variables To do so lets remember the random experiment behind the hypergeometric
distribution You have a bag that contains blue marbles and red marbles You choose marblesat random (without replacement) and let be the number of blue marbles in your sample In particular lets
define the indicator random variables as follows
Then we can write
Thus
To find we note that for any particular all marbles are equally likely to be chosen This is
because of symmetry no marble is more likely to be chosen than the th marble as any other marblesTherefore
We conclude
Thus we have
Problem 6
In Example 314 we showed that if then We found this by writing as the sum
of random variables Now find directly using Hint Use
Solution
X sim H 983161 983152 983141 983154 983143 983141 983151 983149 983141 983156 983154 983145 c ( b 983154 983147 ) E X
X
( 983160 ) = P
X
⎧
⎩
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
( 983081 ( )
b
983160
983154
983147 minus
983160
( 983081
b + 983154
983147
0
f o r 983160 isin R
X
o t h e r w i s e
= m a x ( 0 983147 minus 983154 ) m a x ( 0 983147 minus 983154 ) + 1 m a x ( 0 983147 minus 983154 ) + 2 m i n ( 983147 b ) R
X
E X
E X X
X
983145
b 983154 983147 le b + 983154
X
X
983145
= 983163 X
983145
1
0
i f t h e 983145 t h c h o s e n m a r b l e i s b l u e
o t h e r w i s e
X = + + ⋯ + X
1
X
2
X
983147
E X = E + E + ⋯ + E X
1
X
2
X
983147
P ( = 1 ) X
983145
X
983145
983145
P ( = 1 ) = f o r a l l 983145 isin 1 2 ⋯ 983147 X
983145
b
b + 983154
E X
983145
= 0 sdot 983152 ( = 0 ) + 1 sdot P ( = 1 ) X
983145
X
983145
=
b
b + 983154
E X =
983147 b
b + 983154
X sim B 983145 983150 983151 983149 983145 a 983148 ( 983150 983152 ) E X = 983150 983152 X
983150 B 983141 983154 983150 983151 983157 983148 983148 983145 ( 983152 ) E X E X = ( ) sum
isin 983160
983147
R
X
983160
983147
P
X
983160
983147
983147 ( ) = 983150 ( 983081
983150
983147
983150 minus 1
983147 minus 1
( 983081
7242019 More Discrete Random Variable Solved Problems
httpslidepdfcomreaderfullmore-discrete-random-variable-solved-problems 45
First note that we can prove by the following combinatorial interpretation Suppose that
from a group of students we would like to choose a committee of students one of whom is chosen to bethe committee chair We can do this
1 by choosing people first (in ways) and then choosing one of them to be the chair ( ways) or
2 by choosing the chair first ( possibilities and then choosing students from the remaining
students (in ways))
Thus we conclude
Now lets find for
Note that the last line is true because the is equal to for a
random variable that has distribution hence it is equal to
Problem 7
Let be a discrete random variable with Prove
Solution Note that
Thus
Problem 8
If find Var
Solution
Problem 9
983147 ( ) = 983150 ( 983081
983150
983147
983150 minus 1
983147 minus 1
983150 983147
983147 ( )
983150
983147
983147
983150 983147 minus 1
983150 minus 1 ( 983081
983150 minus 1
983147 minus 1
983147 983080 983081 = 983150 983080 983081
983150
983147
983150 minus 1
983147 minus 1
E X X sim B 983145 983150 983151 983149 983145 a 983148 ( 983150 983152 )
E X = 983147 ( ) sum
983150
983147 = 0
983150
983147
983152
983147
983153
983150 minus 983147
= 983147 ( ) sum
983150
983147 = 1
983150
983147
983152
983147
983153
983150 minus 983147
= 983150 ( 983081 sum
983150
983147 = 1
983150 minus 1
983147 minus 1
983152
983147
983153
983150 minus 983147
= 983150 983152 ( 983081 sum
983150
983147 = 1
983150 minus 1
983147 minus 1
983152
983147 minus 1
983153
983150 minus 983147
= 983150 983152 ( 983081 sum
983150 minus 1
983148 = 0
983150 minus 1
983148
983152
983148
983153
( 983150 minus 1 ) minus 983148
= 983150 983152
( 983081 sum
983150 minus 1
983148 = 0
983150 minus 1
983148
983152
983148
983153
( 983150 minus 1 ) minus 983148
( 983148 ) sum
983150 minus 1
983148 = 0
P
Y
Y B 983145 983150 983151 983149 983145 a 983148 ( 983150 minus 1 983152 ) 1
X sub 0 1 2 R
X
E X = P ( X gt 983147 ) sum
983147 = 0
infin
P ( X gt 0 ) = ( 1 ) + ( 2 ) + ( 3 ) + ( 4 ) + ⋯ P
X
P
X
P
X
P
X
P ( X gt 1 ) = ( 2 ) + ( 3 ) + ( 4 ) + ⋯ P
X
P
X
P
X
P ( X gt 2 ) = ( 3 ) + ( 4 ) + ( 5 ) + ⋯ P
X
P
X
P
X
P ( X gt 983147 ) sum
infin
983147 = 0
= P ( X gt 0 ) + P ( X gt 1 ) + P ( X gt 2 ) +
= ( 1 ) + 2 ( 2 ) + 3 ( 3 ) + 4 ( 4 ) + P
X
P
X
P
X
P
X
= E X
X sim P 983151 983145 983155 983155 983151 983150 ( λ ) ( X )
( 2 minus ) = 6
7242019 More Discrete Random Variable Solved Problems
httpslidepdfcomreaderfullmore-discrete-random-variable-solved-problems 55
Let and be two independent random variables Suppose that we know Var and Var
Find Var and Var
Solution
Lets first make sure we understand what Var and Var mean They are Var and
Var where the random variables and are defined as and
Since and are independent random variables then and are independent random variables
Also and are independent random variables Thus by using Equation 37 we can write
By solving for Var and Var we obtain Var and Var
larr previousnext rarr
X Y ( 2 X minus Y ) = 6
( X + 2 Y ) = 9 ( X ) ( Y )
( 2 X minus Y ) ( X + 2 Y ) ( Z )
( W ) Z W Z = 2 X minus Y W = X + 2 Y
X Y 2 X minus Y
X 2 Y
V a r ( 2 X minus Y ) = V a r ( 2 X ) + V a r ( - Y ) = 4 V a r ( X ) + V a r ( Y ) = 6
V a r ( X + 2 Y ) = V a r ( X ) + V a r ( 2 Y ) = V a r ( X ) + 4 V a r ( Y ) = 9
( X ) ( Y ) ( X ) = 1 ( Y ) = 2
7242019 More Discrete Random Variable Solved Problems
httpslidepdfcomreaderfullmore-discrete-random-variable-solved-problems 45
First note that we can prove by the following combinatorial interpretation Suppose that
from a group of students we would like to choose a committee of students one of whom is chosen to bethe committee chair We can do this
1 by choosing people first (in ways) and then choosing one of them to be the chair ( ways) or
2 by choosing the chair first ( possibilities and then choosing students from the remaining
students (in ways))
Thus we conclude
Now lets find for
Note that the last line is true because the is equal to for a
random variable that has distribution hence it is equal to
Problem 7
Let be a discrete random variable with Prove
Solution Note that
Thus
Problem 8
If find Var
Solution
Problem 9
983147 ( ) = 983150 ( 983081
983150
983147
983150 minus 1
983147 minus 1
983150 983147
983147 ( )
983150
983147
983147
983150 983147 minus 1
983150 minus 1 ( 983081
983150 minus 1
983147 minus 1
983147 983080 983081 = 983150 983080 983081
983150
983147
983150 minus 1
983147 minus 1
E X X sim B 983145 983150 983151 983149 983145 a 983148 ( 983150 983152 )
E X = 983147 ( ) sum
983150
983147 = 0
983150
983147
983152
983147
983153
983150 minus 983147
= 983147 ( ) sum
983150
983147 = 1
983150
983147
983152
983147
983153
983150 minus 983147
= 983150 ( 983081 sum
983150
983147 = 1
983150 minus 1
983147 minus 1
983152
983147
983153
983150 minus 983147
= 983150 983152 ( 983081 sum
983150
983147 = 1
983150 minus 1
983147 minus 1
983152
983147 minus 1
983153
983150 minus 983147
= 983150 983152 ( 983081 sum
983150 minus 1
983148 = 0
983150 minus 1
983148
983152
983148
983153
( 983150 minus 1 ) minus 983148
= 983150 983152
( 983081 sum
983150 minus 1
983148 = 0
983150 minus 1
983148
983152
983148
983153
( 983150 minus 1 ) minus 983148
( 983148 ) sum
983150 minus 1
983148 = 0
P
Y
Y B 983145 983150 983151 983149 983145 a 983148 ( 983150 minus 1 983152 ) 1
X sub 0 1 2 R
X
E X = P ( X gt 983147 ) sum
983147 = 0
infin
P ( X gt 0 ) = ( 1 ) + ( 2 ) + ( 3 ) + ( 4 ) + ⋯ P
X
P
X
P
X
P
X
P ( X gt 1 ) = ( 2 ) + ( 3 ) + ( 4 ) + ⋯ P
X
P
X
P
X
P ( X gt 2 ) = ( 3 ) + ( 4 ) + ( 5 ) + ⋯ P
X
P
X
P
X
P ( X gt 983147 ) sum
infin
983147 = 0
= P ( X gt 0 ) + P ( X gt 1 ) + P ( X gt 2 ) +
= ( 1 ) + 2 ( 2 ) + 3 ( 3 ) + 4 ( 4 ) + P
X
P
X
P
X
P
X
= E X
X sim P 983151 983145 983155 983155 983151 983150 ( λ ) ( X )
( 2 minus ) = 6
7242019 More Discrete Random Variable Solved Problems
httpslidepdfcomreaderfullmore-discrete-random-variable-solved-problems 55
Let and be two independent random variables Suppose that we know Var and Var
Find Var and Var
Solution
Lets first make sure we understand what Var and Var mean They are Var and
Var where the random variables and are defined as and
Since and are independent random variables then and are independent random variables
Also and are independent random variables Thus by using Equation 37 we can write
By solving for Var and Var we obtain Var and Var
larr previousnext rarr
X Y ( 2 X minus Y ) = 6
( X + 2 Y ) = 9 ( X ) ( Y )
( 2 X minus Y ) ( X + 2 Y ) ( Z )
( W ) Z W Z = 2 X minus Y W = X + 2 Y
X Y 2 X minus Y
X 2 Y
V a r ( 2 X minus Y ) = V a r ( 2 X ) + V a r ( - Y ) = 4 V a r ( X ) + V a r ( Y ) = 6
V a r ( X + 2 Y ) = V a r ( X ) + V a r ( 2 Y ) = V a r ( X ) + 4 V a r ( Y ) = 9
( X ) ( Y ) ( X ) = 1 ( Y ) = 2
7242019 More Discrete Random Variable Solved Problems
httpslidepdfcomreaderfullmore-discrete-random-variable-solved-problems 55
Let and be two independent random variables Suppose that we know Var and Var
Find Var and Var
Solution
Lets first make sure we understand what Var and Var mean They are Var and
Var where the random variables and are defined as and
Since and are independent random variables then and are independent random variables
Also and are independent random variables Thus by using Equation 37 we can write
By solving for Var and Var we obtain Var and Var
larr previousnext rarr
X Y ( 2 X minus Y ) = 6
( X + 2 Y ) = 9 ( X ) ( Y )
( 2 X minus Y ) ( X + 2 Y ) ( Z )
( W ) Z W Z = 2 X minus Y W = X + 2 Y
X Y 2 X minus Y
X 2 Y
V a r ( 2 X minus Y ) = V a r ( 2 X ) + V a r ( - Y ) = 4 V a r ( X ) + V a r ( Y ) = 6
V a r ( X + 2 Y ) = V a r ( X ) + V a r ( 2 Y ) = V a r ( X ) + 4 V a r ( Y ) = 9
( X ) ( Y ) ( X ) = 1 ( Y ) = 2