MAT 470, Worksheet 4
Solutions
1. For a random variable Y with a geometric distribution, the moment generating function is
given by
pet
m(t ) 
, where q  1  p .
1  qet
a). Demonstrate how the derivative of the moment generating function may be used to
determine the expected value, E(Y) = 1/ p.
Evaluate the derivative at t = 0.
1  qet  pet  pet (qet )

m(t ) 
and so
t 2
1

qe


E (Y )  m(0) 
1  q  p  p(q)  p
2
2
1  q 
1  q 

p 1

p2 p
b). Determine E(Y) and V(Y) for a random variable Y with a moment generating function
given.
m(t ) 
5 t
5et
8e

and so Y has a geometric distribution with p = 5/8, q = 3/8.
8  3et 1  83 et
Therefore, E(Y) = 1/p = 8/5 and V(Y) = q / p2 = 24/25.
c). Determine E(Y) and V(Y) for a random variable Y with a moment generating function
given.
m(t )   0.2  0.8et  is for a binomial distribution with n = 25 and p = 0.8.
25
Therefore, E(Y) = np = 20 and V(Y) = npq = 4.
d). Determine E(Y) and V(Y) for a random variable Y with a moment generating function
given.
15
15
 3  2e3t 
 3  2et 
t 15
m(t )  
 . Consider instead mX (t )  
  (0.6  0.4e )
 5 
 5 
15
 3  2e(3t ) 
which is binomial with n = 15 and p = 0.4. Note mX (3t )  
  m(t ) .
5


Therefore, Y = 3X and so we find E(Y) = 3E(X) = 3np = 18
and V(Y) = 9V(X) = 9npq = 32.4.
2.
For a discrete random variable Y with moment generating function given by
10
 et  4 
m(t )  

 5 
is binomial with n = 10, p = 1/5, q = 4/5
a). Determine the expected value of Y, E( Y ) = np = 2
b). Determine the expected value E( 3Y 2 + 5 ) = 3E( Y 2 ) + 5 = 3(28/5) + 5 = 21.8
where V(Y) = npq = 8/5 and so E( Y2 ) = V(Y) + (E(Y))2 = 8/5 + 4 = 28/5
c). Determine the variance V( 4Y + 10 ) = 16V( Y ) = 16 (8/5) = 25.6
3.
The Poisson probability distribution with  = 3, has moment generating function
m(t )  e3e 3 .
t
Its first and second derivatives are given by
m(t )  3et e3e 3 and m(t )  e3e 3 9e2t  3et  .
t
t
a). Determine the second moment, E (Y )  m(0)  (e
2
b). Compute the third derivative,
33
)(9  3)  12

m(t )  e3e 3 (18e2t  3et )  (9e2t  3et ) e3e 3 (3et )
t
t
c). Determine the third moment, E (Y )  m(0)  (18  3)  (9  3)(3)  57
3
4.
Consider a discrete random variable with the following probability distribution.
x
0
2
4
6
p(x) 0.4 0.3 0.2 0.1
a). Construct the moment generating function m( t ) for this random variable.
m( t ) = E( etx ) = e0(0.4) + e2t(0.3) + e4t(0.2) + e6t(0.1)
= 0.4 + 0.3e2t + 0.2e4t + 0.1e6t
b). Using the derivative, compute the expected value E(X) = m’(0) =
m’( t ) = 0.6e2t + 0.8e4t + 0.6e6t and so E(X) = m’( 0 ) = 0.6 + 0.8 + 0.6 = 2.0.
c). Compute the expected value using the definition,
E ( X )   x p( x)  0(0.4) + 2(0.3) + 4(0.2) + 6(0.1) = 2.0, as above.
x

5.
Let Y be a random variable with mean 16. Based on Tchebysheff’s theorem,
P(| Y   |  k ) 
1
.
k2
a). Find the value of k such that we’re guaranteed P(| Y  16 |  k )  0.16
Here, let k2 = 1/.16 and so k = 2.5.
b). Suppose this is actually a Poisson distribution with mean 16. Determine the interval
| Y  16 |  k , using the value of k from part “a”.
For Poisson, V(Y) = E(Y) = 16 and so the standard deviation is 4.
The interval 16 – 2.5 < Y < 16 + 2.5 is the interval ( 6, 26 ).
6.
Let Y be a binomial random variable with n = 100 and p = 0.03. Based on Tchebysheff’s
4
, but this is just an upper bound. For this particular
9
distribution, determine the actual probability P(| Y   |  1.5 ).
Since V(Y) = npq = 2.91 and  = E(Y) = np = 3, the interval   1.5  Y    1.5
theorem P(| Y   |  1.5 ) 
is the interval 3  1.5 2.91  Y  3  1.5 2.91 or simply (0.44, 5.56). For this
binomial distribution, P( 0.44 < Y < 5.56) = P(Y = 1) + … + P( Y = 5) = 0.8716.
Therefore, the complement P(| Y   |  1.5 ) = 1 – 0.8716 = 0.1284 is indeed
less than 4/9, as promised.