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Stat 61 - Homework #5 - SOLUTIONS (10/29/07)
1. Let X be a random variable with E(X) = 100 and Var(X) = 15. What are…
a. E ( X2 ) (Not 10000)
Var(X) = E(X2) – E(X)2, so 15 = E(X2) – 10000, so E(X2) = 10015.
b. E ( 3X + 10 )
=3 E(X) + 10 = 310
c. E (-X)
= –100
d. Standard deviation of –X ?
= 15  3.9, NOT –3.9
2. Let X be the random variable with pdf
fX ( x ) = 2x
0
if 0 ≤ x ≤ 1
otherwise.
a. What is the cdf, FX(a) ?
FX(a) =

a

0

f ( x)dx = a 2
1

if a  0
if 0  a  1
if a  1
b. Let Y be another random variable defined by Y = X2. What is its cdf,
FY (b) ?
( This is possible because if you know P( X ≤ a ) for every a, you can get
P( Y ≤ b ) for any b.)
FY(b) = P ( Y ≤ b )
(by definition of FY)
2
=P(X ≤b)
(same event)
(This is clearly zero of b < 0, so from here on
assume b ≥ 0)
= P(X≤ b )
(again, it’s the same event, at least
when
b is nonnegative)
= FX ( b )
(by definition of FX)
= b
=b
2
(or: 1 if b  1, 0 if b ≤ 0)
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c. What is the pdf, fY ?
It’s the derivative of FY…
if b<0
0

fY(b) = 1 if 0  b  1
0
if b  1

d. Compute E(Y) directly from FY. (Use the secret formula.) (See website if
necessary)
E(Y) =
=

1
y 0

y 0 1  F ( y)  dy  y  F ( y)dy
0
(1  y )dy  0
= 1/2.
e. Compute E(Y) from fY.
E(Y) =


y 
yf ( y )dy  
1
y 0
 y 1 dy
= 1/2
f. Compute E(Y) = E(X2) directly from fX. (If your answers to d-e-f don’t
agree, panic.)
E(Y) = E(Y )  E ( X 2 )  

x 
x2 f x ( x)dx  
1
x 0
x 2 (2 x)dx =1/2
3. (Sum of random variables) Let X and Y be independent random variables, with
these pmf’s:
pX(n) = (1/3) for n = 1, 2, 3; else 0
pY(n) = (1/2) for n = 1, 2; else 0
Let Z = X + Y. Using the formula
pZ (n)   pX (m) pY (n  m) ,
all m
[ = P(X=1) P(Y=n-1) + P(X=2) P(Y=n-2) + P(X=3) P(Y=n-3) etc. ]
construct the pmf for Z.
2
P ( Z = 2 ) = P( X=1)P(Y=1) = (1/3)(1/2) = 1/6
P ( Z = 3 ) = P(X=1)P(Y=2) + P(X=2)P(Y=1) = (1/3)(1/2) + (1/3)(1/2) = 2/6
P ( Z = 4 ) = P(X=2)P(Y=2) + P(X=3)P(Y=1) = etc. = 2/6
P ( Z = 5 ) = P(X=3)P(Y=2) = etc. = 1/6
(I guess there are infinitely many other terms in each of these sums,
but they are all zero)
so the pmf for Z is
z
pZ(z)
2
1/6
3
2/6
4
2/6
5
1/6
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4. Now let X and Y be independent Poisson random variables, with means X and
Y respectively. Let Z = X + Y. Using the formula from problem 3, derive the
pmf for Z.
n
n!
a mb n  m . )
m  0 m !( n  m)!
(Hint: The binomial formula is (a  b) n  
In what follows, x, y, and z are integers (being typical values of X, Y,
Z resp.).
pZ ( z )   p X ( x) pY ( z  x)
all x
z
 x
   X e  X
x 0  x !
  Y z  x  Y
e
 
   z  x !



 z

1
 e  (  X  Y )  
 X x Y z  x 
 x 0 x ! z  x  !


z!
 1  z
 e  (  X  Y )    
 X x Y z  x 
 z !   x 0 x ! z  x  !

z
1
 e  (  X  Y )     X  Y 
 z!
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This is the Poisson pmf for Z, assuming Z = X + Y. So: The sum of
two independent Poisson random variables is itself a Poisson
random variable.
5. (The mother of all Poisson examples) Do problem 4.2.10, page 287. Either actually
read the problem and do what it says, or do parts a and b below, which amount to
the same thing. Specifically, given the following distribution…
k
Observed number of
corps-years in which
k fatalities occurred
0
1
2
3
4
≥5
109
65
22
3
1
0
200
[This is a frequency table describing 200 annual reports from pre-WWI Prussian
cavalry corps, each giving a number of soldiers killed by being kicked by horses.]
It is suggested that these values represent 200 independent draws from a Poisson
distribution.
a. What is a reasonable estimate for the mean of the distribution?
One reasonable estimate is the mean of the 200 observed values,
which is
 = [ 109(0)+65(1)+22(2)+3(3)+1(4) ] / 200 = 0.61.
(As we’ll see soon, this is the method-of-moments estimator for the
mean, which happens to agree in this situation with the maximum
likelihood estimator.)
b. What would the entries in the right-hand column be if the 200 draws were
exactly distributed according to this Poisson distribution? (The entries would not
be integers.)
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Using  = 0.61:
k
0
1
2
3
4
≥5
observed
109
65
22
3
1
0
theoretical
108.67 = (200) ( 0.610 / 0! ) ( e–0.61 )
66.29
20.22
4.11
0.63
0.09
200
200
(In November we’ll address this question: Comparing the original data to the
theoretical values in part b, is it plausible that these really are draws from a
Poisson distribution? Or should we discard that theory?)
This looks to me like a very good fit --- even suspiciously good --- but
we may try a Chi-Square test later.
6. (Editing a Poisson distribution) At a certain boardwalk attraction, customers appear
according to a Poisson process at a rate of  = 15 customers/hour. So, if X is
the number of customers appearing between noon and 1pm, X has a Poisson
distribution with mean 15.
Assume that each customer wins a prize with (independent) probability 1/5.
Let Y be the number of customers winning prizes between noon and 1pm.
One way to understand Y is that if the value of X is given, then Y is binomial,
with parameters n = [value of X] and p = 1/5. This reasoning gives us:
P(Y=k)=

 n
nk
 
  P( X  n)    k  1/ 5
k

(4 / 5) nk 

(The second part of the summand is the probability that Y=k given that X=n.)
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a. Simplify this formula, to show that Y is itself a Poisson distribution with
mean 3.
P(Y=k)=
 n

  P( X  n)    k  1/ 5
k
 
nk

(4 / 5) n  k 



 15 15    n 
k
 
e     1/ 5  (4 / 5) n  k 
nk  n !
 k 

n

1 15  n  1   1
e 15 
 n
k!
nk
 (n  k )!   5

1 15 k   3n  k 4n  k 
e (3 ) 

k!
n  k  ( n  k )! 
15n
 nk
4
...and
now
since
 (3k )(3n  k )...

n
5

When we substitute m for n-k in the last sum, we get

12m
 ,
m 0 m !
which is the Taylor series for e+12; so
3k 3
P(Y=k)=
e
k!
as desired.
(Or, if after getting off to a good start you find this problem too annoying, do
part b instead.)
b. Find a much simpler explanation of why Y should be Poisson with mean 3.
If the probability of a customer arriving in a small interval is 15
times the length of the interval in hours, independently of what
happens in other intervals, then the probability of a winner appearing
during that small interval is 3 times the length of the interval, and is
still independent of what happens in other intervals.
That means that the “winners” process is a Poisson process with
mean 3, and the result follows.
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7. (Waiting times) Buses arrive according to a Poisson process with mean (1/10) min-1.
Let W be the waiting time from time t = 0 till the arrival of the second bus.
a. What is E(W) ?
If X is the time we have to wait for the first bus, then X is
exponentially distributed with mean 10. After the first bus arrives,
we can start waiting for the second bus, and if Y is the time that
takes, then Y also has mean 10. Now W = X + Y. It doesn’t matter
whether X and Y are independent (they are) --- in any case,
E(W) = E(X) + E(Y) = 20 minutes.
Part a is of no use in solving part b.
b. Can you construct a pdf or cdf for W ?
( Good start: P(W ≤ t) = 1 – P(exactly 0 buses or exactly 1 bus between
times 0 and t). Another approach: W is the sum of two independent one-bus
waiting times.)
P ( W ≤ t ) = 1 – exp(-(1/10) t) – (1/10) t exp(-(1/10) t).
That’s the same as FW(t).
If you want the density function fW(t), just differentiate FW(t);
things cancel and you get
fW(t) = (1/100) t exp(-(1/10)t ).
This happens to be the density for (a special case of) the gamma
distribution.
8. (The first boring normal tables problem) If Z is a standard normal variable, what
are…
a. P ( -1.0 ≤ Z ≤ +1.0 ) = 0.68
b. P ( -2.0 ≤ Z ≤ +2.0 ) = 0.95
c. P ( -3.0 ≤ Z ≤ +3.0 ) = 0.997
( by the “Rule of 68 – 95 – 99.7” )
( or, use Excel’s NORMSDIST function to get 0.682689, 0.9545,
0.9973 )
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9. (The second boring normal tables problem) Let Z be a standard normal variable.
a. If P ( -a ≤ Z ≤ +a ) = 0.95, what is a ?
Find z for which Phi(z) = 0.975 --- namely, z = 1.96.
(Or: =NORMSINV(0.975) = 1.9600 )
b. If P ( -b ≤ Z ≤ +b ) = 0.99, what is b ?
Find z for which Phi(z) = 0.995 --- namely, z = 2.58
(Or: =NORMSINV(0.995) = 2.575829 )
10. (The third boring normal tables problem)
a. If X is normal with mean 500 and standard deviation 110,
what is P ( X ≥ 800 ) ?
If Z = (X-500)/110 then Z is standard normal.
P ( X ≥ 800 ) = P ( Z ≥ (800-500)/110 ) = P ( Z ≥ 2.7272 ) =
1 – (2.72) = 0.0032
b. If X is normal with mean 500 and standard deviation 120,
what is P ( X ≥ 800 ) ?
If Z = (X-500)/120 then Z is standard normal.
P ( X ≥ 800 ) = P ( Z ≥ (800-500)/120 ) = P ( Z ≥ 2.5 ) =
1–(2.5) = 0.0062
(end)
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