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CM141A – Probability and Statistics I Solutions
to exercise sheet 4
1. We know that E X [ ]= 1 and Var X ( )= 5
a. First, using the various properties we studied in class we can simplify the expression
E ⎡⎣( 2 + X
2
= E ⎡
4 + 4 X + X
2
⎤ = + 44 E X [ ]+ E X ⎡
2
We obviously know that E X [ ]=1, but what is E ⎡
X
2
This can be determined from the knowledge of both the mean and the variance. Note that
5 = Var X ( ) = E X ⎡
2
− E X [ ]
2
= E X
2
2
⇒ E X ⎡
2
= 6 And thus:
E ⎡( 2 + X )
2
⎦
⎤ = = +4 4 E X [ ]+ E X
2
b. Var ( 4 + 3 X ) = E
⎡( 4 + 3 X )
2
⎦
⎤− E [ 4 + 3 X ]
2
= E ⎡
16 + 24 X + 9 X
2
−( 4 + 3 E X [ ])
2
= 16 + 24 E X [ ] + 9 E X ⎡
2
−( 16 + 24 E X [ ] + 9 E X [ ]
2
= 9 ( E X ⎡
2
− E X [ ]
2
) = 9 Var X () = 9 × 5 = 45
- a. Below is a table comparing the direct evaluation of n! with the result of the
Stirling approximation n!~ 2 πn
n 12+
e
−n
n n!
π
nn 12+ e−n
(n!− 2 πnn 12+ e−n)/ n!
This formula is results from something called an asymptotic expansion, which is similar to a
Taylor expansion around infinity. In fact the above-cited formula is simply the first term in
the expansion, and one can improve over it by taking more terms, e.g.
n!~ 2 πnn 12+ e−n ⎡⎢ 1 +121n + 2881n 2 −
51840139n 3 − ⎤⎥⎦ ⎣
(2k)! 2 π( 2 k
2k+ 12
e−2k 2 2 π
2 k+12 2k 12k + e−2k 1
b. k!k!2 2
k
~ ( 2 π
kk 12+ e− ) 2 2
2 πk2k+ 1 e−2k 2 2k
πk k 2k
- If we assume that the dice are fair and act independently of each other, then the number
of times that the number bet appears is a binomial random variable with parameters (3, ).
Hence, letting X denote the player’s winnings in the game we have
P X (=− = 1 ) ⎛ ⎞⎜ ⎟ 03 ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 16
0
3
1
2
P X (= = 1 ) ⎜ ⎟⎝ ⎠ 1 ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 66 = 216
P X (= 2 )=⎜ ⎟⎝ ⎠ 2 ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 6 6 = 216
3
0
P X (= 3 )=⎜ ⎟⎝ ⎠ 3 ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 6 6 = 216
In order to determine whether or not this is a fair game for the player, let us calculate E [ X ].
From the preceding probabilities we obtain
(− 1 )× 125 +1 75×+ 2 15×+3 1×
E X [ ]=
= −
Hence, in the long run, the player will lose 17 units per every 216 games he plays.
P Y p i.
i = 10 i = 10 ⎝ ⎠ i
Not very high, but can happen.
- Let X be a Binomial random variable with parameters n and p ( X ~ B n p ( , )).
n + 1
Show that E ⎡⎢⎣ X 1 + 1 ⎤⎥⎦= 1 − −( n ( 1 + 1 p )) p
Proof:
E ⎡⎢⎣ X 1 + 1 ⎤⎥⎦=∑ i = n 0 ( i + 11 )⎛ ⎞⎝ ⎠⎜ ⎟ ni p qi n i − =∑ i = n 0 ( i + 11 ) i n !( n −! i )!
p qin i − =∑ i = n 0 ( i +1 !) ( n! n − i )! nn ++ 11 p qin i −
∑
=
n
( i +(1! n ) (+1! n )− i )! pp
i + 1
n i − ( n + 11 ) p
i
=
n
0
n
1
q ( n + − + 1 ) ( i 1 ) n + 1 i 0
q
=
∑
i
1 n 1 ⎛ n + 1 ⎞
j i = + 1 ( n + 1 ) p ∑ j += 1 ⎝⎜ j ⎠⎟ p qj ( n + − 1 ) j = ( n + 11 ) p ⎣⎢⎡∑ nj =+ 10 ⎛⎝⎜ n + j
1 ⎠⎟⎞ p qj ( n + − 1 ) j −⎝⎜⎛ n 0 + 1 ⎠⎟⎞ p q 0 ( n + −1 0) ⎤⎥⎦
= 1 ⎡⎢∑ n =+ 1 ⎛⎜ n + j 1 ⎞⎟⎠ p qj ( n + − 1 ) j − −( 1 p ) n + 1 ⎤⎥⎦
( n + 1 ) p ⎣ j 0 ⎝
Note that
∑
n
j =
0
n +
j
j ( n + − 1 ) j
1 as this is the normalization condition for a Binomial RV
p q =
with parameters n+1 and p. Therefore
= n + 1 p ⎢∑ n =+ 1 ⎜⎛ n + j 1 ⎞⎠⎟ p qj ( n + − 1 ) j −
−( 1 p ) n + 1 ⎦⎥⎤= 1 − −( n ( 1 + 1 p )) pn + 1 QED
[ ]
n1 n
2 2
9 9
= =
=π++ + + + = π× = π
E ⎡⎢⎣ X 1 + 1 ⎤⎥⎦= ( 1 ) ⎡⎣ j 0 ⎝
- a. The probability mass function of X, a uniform discrete random variable between a
and b ( a b < ), is simply
P X (= = i ) ⎧⎨ b a − +1 1 a ≤ ≤ i b for i = a a , +1,…, b
0 otherwise
Since the sample space contains b a − + 1 events, which are all equiprobable.
b. μ= X = = iP X = =
i
b 1 a 1 = i = b 1 a 1 2
( a + b )( b − + a 1
( a + b ).
i a − + i a − + ⎦
Note that this is simply the mean of a and b.
b a
2
a b 2 ab + 2 b
2
)−
( a + b )
⎦
⎤
⎥
( b − a )( b − + a 2 )
And thus σ= = ( 2 a − + +
- Let R be the radius of the sphere. The R is a Uniformly distributed discrete RV over
{1,2,3,4,5,6}. The Expected volume of perfume that interests us is therefore simply:
6 6
V = 43 πR 3 = π 43 R 3 = π 43 ∑n 3 × = π 1692 ∑n 3 =
- Let Y = # of holes drilled until a productive well is found.
Given the information in the problem, we know that Y G (.2) (geometrically distributed).
a. The probability that the 3
rd
hole drilled is the first to yield a productive well is just
P Y ( = 3 )= q p
2
2
× =.2 .128.
b. The probability that he will fail to find a productive well within 10 drills is
q =
2
2
VarX X X
σ
( ) ( )
2 2 2
2 2 2 2
b b
ia ia
X iPX i i
b a
b a a a b ab b a a b ab b
= =
∑ ∑
⇒ Var Y ( )= Y 2 − Y 2 = 2 p − 2 p −⎛ ⎞⎝ ⎠⎜ ⎟ 1 p = 1 − p 2 p
b. For a positive integer n : P Y > n =∑ q
y
−
1
p = p ∑∞ q
y
−
1
p 1 q
−
nq = p qpn = qn QED
c. Let m and n be any positive integers. Thus
P Y ( > + m n Y > n )= P ⎡⎣( Y > + P Ym ( n >) n )( Y > n )⎦⎤ ( Y m n > +=) (⊃ Y n > )
P Y ( P Y (> + m > n ) n ) And using the result of part b we get
P Y ( > + m n ) q
m n + m
P Y ( > m ) QED
P Y ( > + mn Y > n )= =
P Y ( > n )
q
n = q =
This property is called the memoryless property of the geometric distribution since it says
that the only thing that matters is not when we actually started the experiment but rather the
last reported event of which we know there wasn’t a success.
The reason is that the trails are independent, and so knowing that there was no success up to
the n
th
event means that the first success will not depend at all on the trails which preceded
this one.
- As there are 12 fish in the pond, with 3 large ones, the probability of catching a large
fish is p = 12
3
1
4
. Therefore, the number of fish we catch until we get a large one, denoted
here by Y, is Geometrically distributed with p = (if you are confused, try to see the analogy
between this and flipping a coin until you get a Tail). Since Y ~ G( ) , we immediately
know that its expected value is E Y[ ]=
1
p
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