IDEAS THAT WE WILL NEED FROM CHAPTER 5
Chapter 5 is about the simultaneous behavior of two or more random variables. For example, we might
be concerned with the probability that a randomly selected person will have a body mass index at most
35 AND a systolic blood pressure at most 140. Body mass index (Y1) is one random variable, and
systolic blood pressure (Y2) is another. The values 35 and 140 are arbitrary. As we slide these values
around, we can create a function called the joint (cumulative) distribution function:
𝐹𝑌1, 𝑌2 (y1,y2) = P(Y1  y1 AND Y2  y2)
Much of chapter 5 is devoted to methods for calculating and manipulating these functions.
Fortunately for us, we will only need to be able to calculate the means and standard deviations for linear
combinations of multiple variables. That is, we will frequently be concerned with functions of the form
W  a1Y1  a2Y2  ...  akYk  b
Suppose that each Yi has expected value denoted by i . What can we say about E (W ) ?
RESULT #1: E (W )  a11  a2 2  ...  ak k  b  a1E (Y1 )  a2 E (Y2 )  ...  ak E (Yk )  b
While I am deliberately avoiding developing the machinery for a proof of this important result, it is not
surprising that expectations distribute over linear operations on multiple variables in the same way that
they distribute over linear functions of single variables.
Example 1. You are scoring college applicants on the basis of their High School GPA (Y1) and their
total SAT score (Y2). The scoring function you use is S  10Y1  Y2 / 40 .
If applicants have a mean High School GPA of 3.1 and a mean total SAT of 1100, what is the mean for
the values of S?
The mean is 10*3.1 + 1100/40 = 58.5
Example 2. The mean time to service a rental car when it is returned is 25 minutes. 10 cars are
returned during the day. What is the expected total time to service all of them?
Intuitively, you know that this must be 25*10 = 250 minutes. Formally, let Y1 be the time to
service the first car, Y2 the time to service the second, etc.
total  T  Y1  Y2  ...  Y10
Then E(T) = 25 + 25 + ….+ 25 = 10*25 = 250
In addition to getting the mean, we also need to know the standard deviation. This works out reasonably
easily IF we can assume the variables are independent.
Definition: Let the random variable 𝑌1 have c.d.f. 𝐹𝑌1 , and let the random variable 𝑌2 have c.d.f. 𝐹𝑌2 .
Let the two random variables together have the joint c.d.f. 𝐹𝑌1 ,𝑌2 . The two random variables are said to
be independent if, for all 𝑦1 𝑎𝑛𝑑 𝑦2 , we have 𝐹𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = 𝐹𝑌1 (𝑦1 )𝐹𝑌2 (𝑦2 ). If 𝑌1 and 𝑌2 are not
independent, they are said to be dependent.
Note: If the two random variables are continuous, then independence also implies that the joint p.d.f. is
factorable into the product of the individual (marginal) p.d.f.’s.
If the two random variables are discrete, then independence implies that the joint p.m.f. is factorable into
the product of the individual (marginal) p.m.f.’s.
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Example: Consider the joint p.d.f. for two continuous r.v.’s 𝑌1 and 𝑌2 given by
1
1
𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = 2𝜋𝜎2 𝑒𝑥𝑝 {− 2𝜎2 [(𝑦1 − 𝜇1 )2 + (𝑦2 − 𝜇2 )2 ]} , for −∞ < 𝑦1 < +∞, and
−∞ < 𝑦2 < +∞. This joint p.d.f. may be factored into a product of two normal p.d.f.’s with different
means but the same standard deviation. Thus the two r.v.’s 𝑌1 and 𝑌2 are independent.
Example: p. 253, 5.52
Example: Consider the joint p.m.f. for two discrete r.v.’s 𝑌1 and 𝑌2 given by
1
𝑝𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = 36, for (𝑦1 , 𝑦2 ) ∈ {1, 2, 3, 4, 5, 6}, or 𝑝𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = 0, otherwise. This joint p.m.f.
may be factored into the product of
1
𝑝𝑌1 (𝑦1 ) = 6, for (𝑦1 ) ∈ {1, 2, 3, 4, 5, 6}, or 𝑝𝑌1 , (𝑦1 ) = 0, otherwise; and
1
𝑝𝑌2 (𝑦2 ) = 6, for (𝑦2 ) ∈ {1, 2, 3, 4, 5, 6}, or 𝑝𝑌2 , (𝑦2 ) = 0, otherwise. Thus the two r.v.’s 𝑌1 and 𝑌2 are
independent.
Example: p. 251, 5.45
The most important consequence of independence is that the variances add in a sensible way. If the
variables were not independent, then the variance of the linear combination would also include terms
involving covariances of pairs of the Y’s.
RESULT #2: If the variables are independent, V (W )  a12V (Y1 )  a22V (Y2 )  ....  ak2V (Yk ) .
Notice that the additive constant, the ‘b’ term, disappears, just as it did for the single variable case in
chapters 3 and 4.
Important Note: Random sampling implies that the measurements taken on separate members of the
random sample will be independent random variables.
Whether or not variables are independent can be an interesting question. For example, we are fairly sure
that body mass index is NOT independent of blood pressure – those people with extremely high BMI are
more likely to have high blood pressures. Similarly, High School GPA is not independent of total SAT.
Notice that this does not mean that those with high HS GPA must have high total SAT, but only that
they are more likely to have high total SAT. On the other hand, the HS GPA’s of two randomly selected
applicants are independent because the results of the first applicant will not give any information about
the results of the second applicant.
Example 2, continued. Suppose that the standard deviation for a single car’s service time is 6 minutes.
Also, assume it is reasonable that the times for different cars are independent.
a. What is the standard deviation for T = total time to service the 10 cars?
Solution:
Each car has a time with variance  2  62  36 . Hence, the variance for T is
V (T )  V (Y1 )  V (Y2 )  ...  V (Y10 )  10  36  360
and the standard deviation is
360  18.97
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b. Suppose that T  T /10 is the mean time in the sample of 10 cars. What is the expected value and
standard deviation for T ?
Using the basic results from chapters 3 and 4, we know E (T )  E (T ) /10  25
And V (T )  V (T ) /102  3.6
The standard deviation for T is
3.6  1.897
It is extremely important to notice that the standard deviation for T was much smaller than the
standard deviation for any single car. This fundamental property of averages is the principle that
makes statistical inference possible!
Problem A. Suppose that the time between service calls on a copy machine is exponentially distributed
with a mean  . Let T be the sample mean for records of 36 independent times.
a. Express E (T ) and V (T ) as expressions using  .
b. Suppose the sample mean is 20.8 days. Provide a common sense estimate of  . Assuming that the
distribution of T is nearly bell-shaped, is it likely that the true value of  differs from your estimate by
more than 10 days?
The behavior of the the sample mean ( T ) is of great concern to people doing practical statistics. Does it
provide a stable estimate of the population mean ? Results #1 and #2 yield the following:
RESULT #3. Suppose that Y1 , Y2 ,..., Yn are independent random variables all having the same mean 
and variance  2 . Then Y has mean  and variance  2 / n .
proof: Using result #1, E (Y )  E (Y1 / n  Y2 / n  ...  Yn / n)   / n   / n  ...   / n  
Using result #2 and the assumption of independence,
V (Y )  V (Y1 / n  Y2 / n  ...  Yn / n)   2 / n2   2 / n2  ....   2 / n2   2 / n
Notice that we only needed independence for the variance calculation.
MOMENT GENERATING FUNCTIONS (This is actually covered in Chapter 6.)
Another useful property of independent random variables is that the expectation distributes over
PRODUCTS. Notice that expectations always distribute over +/-, but it takes the special case of
independent to allow them to distribute over multiplication!
RESULT #4: If the variables are independent, then
E ( g1 (Y1 )  g2 (Y2 )  ...  gk (Yk ))  E( g1 (Y1 ))  E( g2 (Y2 ))  ...  E( g k (Yk ))
There are many useful consequences of this, and we will mostly be concerned with the implications for
the moment generating function of the linear combination W.
E(etW )  E(exp  a jYj  b)  E(eta1Y1  eta2Y2  ...  etakYk  etB )  etb  E (e
ta jY j
)  etb  m j (a jt )
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That is, the moment generating functions will multiply!
Example 3. Suppose that Y1 , Y2 ,..., Yn are independent random variables having the exponential
distribution with mean  .
a. What is the distribution of T  Y1  Y2  ...  Yn ?
b. What is the distribution of T  T / n ?
Example 4. Suppose that Y1 is a binomial variable with parameters n1 , p and Y2 is an independent
binomial variable with parameters n2 , p . Note that the values of p are the same, but not necessarily the
values of n . What is the distribution of T  Y1  Y2 ? Would the same result hold if the values of p were
allowed to differ?
Problem B. Suppose that Y1 is a normally distributed variable with parameters 1 , 12 and Y2 is an
independent normally distributed variable with parameters 2 ,  22 . What is the distribution of
T  Y1  Y2 ?
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