You are reading a page from An Elementary Treatise on Actuarial Mathematics by Harry Freeman (1932)
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CHAPTER XIX
MEAN VALUE. THE APPLICATION OF
THE CALCULUS TO THE SOLUTION
OF QUESTIONS IN PROBABILITY
1. The ordinary arithmetical average of a number of quantities is the sum of the quantities divided by the number of the quantities. If there are n quantities 0 (a),O (a + h1), (a + h2), ... (a + hn_1), then their average or mean value is
0(a)+0(a+h1)+0(a+h2)+...+0(a+hn 1)
n Suppose that y = 0 (x) is a function of x, and that x has the n successive values a, a + h, a + zh, ... a + (n — I) h. Then the mean value of 0 (x) for these n values from x = a to x = a + (n — I) h is, as above,
0(a)+0(a+h)+0(a+zh)+...+0(a+n—ih)
n
Letb=a+nhsothat nh=b—a. h r=n-1
Then the mean value =     E 0 (a + rh)
nh r=o
h r=n-1
E 0 (a + rh). b — a r-o
If x varies continuously between a and b so that the number of values, n, tends to infinity, the mean value becomes Lt h[0(a)+0(a+h)+...+0(a+n—Ih)]

n-->oo    b — a
I    b
b — a Ja O (x) dx.
It should be noted that the mean value depends on the law governing the selected values. For example, the mean value of the ordinate of a semicircle determined by ordinates passing through equidistant points along the diameter is different from the mean value determined by taking equidistant points along the circumference. It thus appears that

MEAN VALUE 355 the mean value of a continuous function 9, (x) is not a definite quantity but a quantity varying according to the law assumed for the successive values of x.
2. This application of the integral calculus enables us to solve many problems involving mean values. The solution of these problems can generally be effected by the use of single integrals, although some of the more difficult questions necessitate the use of double integration. The two following examples illustrate the use of single integrals and, as will be seen, the solutions present little difficulty.
Example 1.
Find (i) the mean value of the ordinate, (ii) the mean value of the square of the ordinate of the curve y = a sin nx for the range x = o
7r
to x = . n
(i) We have to find the sum of the ordinates over the given range divided by the number of ordinates. Since the number of these ordinates will tend to infinity as the distance between them tends to zero, we shall have
~nydx fn a sin nxdx o    .o
M.V. _    _

[ a t-     - cosnxn
t    n    o
[xi
Jo
a    a
- - cos 7r-}--coso
n    n
n
a    a

since cos 7r = — I,
2a 7r
R    ,r
on dx    Jon
I    dx
7r
n
23-2
1

356
(ii) Similarly
MEAN VALUE AND PROBABILITY

a2 sin2 nxdx
In
.o
dx
n
a2~ (I —cos znx) dx .o
n
7
in Ia2x — ?a2 I 2n sin 2nx
0
IT n = a2 7T/7T
- = ia2.
2n/ n    2 Example 2.
A straight line of length a is divided at random into two parts. Find the mean value of the rectangle contained by the two parts.
Let x be the length of one part ; then a — x is the length of the other part. The rectangle contained by the two parts is x (a — x), and we have to integrate this function with respect to x. Also, since x may have any value from o to a, these values will be the limits of integration.
Hence,
x (a — x) dx
ladx .o
fa
J (ax — x2) dx 0
M.V.
_
[ax3 x3l a
2    3 to rx a
L _o

THE USE OF DOUBLE INTEGRALS    357
a3 a3
2    3
a
a2 6
Note. Since we are required to find the mean value of an area, the result must be of the second degree in a.
3. The use of double integrals.
Consider first a function, q (x), involving a single variable x. Then the mean value of 0 (x) between the limits x = a and x = b is
x=b    x=b
E (x)/ E I.
x=a    x=a
If there be a function i (x, y) of two variables x and y, we may write similarly
x=b y=R    x=b y=R
M.V. = E E 0 (x, y)/ E E r,
x=a y=a    x=a y=a
where x and y proceed by small but finite intervals.
If these intervals, say I/n and I/m for x and y respectively, be very small, the numerator and denominator of this fraction will be very large.
Multiply both numerator and denominator by (I/n. I/m), so that the fraction is
x=b y=R    I I x=b y=S I I
E E 0 (x, y) - -1 E E I -
x=a y=a    n m x=a ?/=a n m'
Replace the small quantities I/n and I/m by Ax and Ay, and find the limit when Ox and Ay each tend to zero. Then, since the limit of a quotient is the quotient of the limits of the numerator and denominator (provided that the limit of the denominator is not zero), we have    x=b y=3
E E 0(x,y)AyLx
    
m.v. = Lt x=a v=a    
x=b v=R
E E Ay Ax
x=a y=a
when Ax, Ay each tend to zero,
b R
I Jcb(x,y)dydx a
Ja1' Jdydx a

358    MEAN VALUE AND PROBABILITY
The following examples are illustrative of the method of applicaof double integrals.
Example 3.
A rod of length a is divided at random into three parts. Find the mean value of the sum of the squares on the three parts.
Let OP be the rod of length a. Take any point X in the rod distant x from 0, and another point Y distant
Y from X. The squares on the three
0
 """""x"'""
segments of the line will be x2, y2,
(a — x — y)2, respectively.		   _

Let X be fixed. Then y will vary
from o to a — x, so that the total values of the sum of the squares OX2, X Y2, YP2 will be
E [x2 + y2 + (a — x — y)2],
where y has every value from o to a — x.
Now let x vary. The limits of variation of x are evidently from o to a. Then
a a x
[x2 + y2 + (a — x — y)2] dydx
M.V. =
Jo . o    
a a—x
dydx o
2x2+2y2+a2—2ax—lay+2xy)dydx .o .o
a -a—.r
dydx o . o
l a 2x2y + sy3 + a2y — 2axy — aye + xy2      o—x dx
a    a—x
dx o y_Jo
a
x)3 + a2 (a — x) — 2ax (a — x)
— a (a — x)2 + x (a — x)2] dx
I (a—x)dx .o
a
which becomes, on evaluating the integral,
a4
4— 2 a2 — 2Q
><" " . J """'
+    P X a 3'
F
g. 39.
2

ILLUSTRATIVE EXAMPLES    359
It is important to note that if the sums of the required values can be obtained by considering separate sets of the values, the total sum must be found and divided by the total of the values. For example, if we could best solve a mean value problem by summing (x) for all values of x varying continuously from o to a, and t/i (x) for all values from a to b, then we must write
f ac(x)dx+ I /i(x)dx
M.V. = 0    a    
foci dx+ Jbdx
a
and not
(x) dx j b (x) dx
ra dx    I G dx .a
The fallacy in the second expression is easily seen when we consider that


`4 + B is not necessarily the same as A      + B
C D    C+D
Example 4.
Find the mean value of the distance from one corner of a square to any point in the square.
Let OABC be the square. Take any point X in the side OC distant x
from O, and draw Xili parallel to the side
Y be any point in XM distant y from X. %
The length 0 Y = 1/x2 + y2 and
XM = OX = x.    x


since Y may take up all possible positions    Fig. 40.
on the straight line XM. But x may have all values from o to a.
Therefore the sum of the distances from the corner 0 to any point in the triangle OBC
ra fx
— r0    1x2 + y2 dydx. .o
CB to meet the diagonal OB in M. Let C    / B

For a fixed value of x the sum of all values of OY, i.e. of 1x2+ y2, will be
E 1/x2 + y2, y—x    
o
0
Y
A

360    MEAN VALUE AND PROBABILITY
For the sum of the distances from 0 to any point in the square OABC we must double this. The required mean value is therefore
2 fa I -Vx2 + y2 dy dx .o .o
2 (a jxdydx o .o


I a x2 [1/2 + log (i + \/2)] dx, a /0
on integrating with respect to y and inserting the limits o and x, =3[\+log(I+"V2)].
4. Application of the calculus to probability.
When we are dealing with problems in probability where the number of cases involved depends upon magnitudes varying continuously over a given range, the method of approach is similar to that outlined above for the solution of mean value problems. The general principle is to take the quotient of the number of favourable ways by the number of possible ways, where all ways are equally likely.
The application of the integral calculus to problems in probability is best illustrated by examples; as a general rule it is sufficient to employ single integrals, although in some instances it is of advantage to use double integration. Many problems can be solved by either method, and examples of both methods are given below.
Example 5.
A line of given length is divided into three parts by two points taken at random. Find the chance that no one part is greater than the sum of the other two.
We shall adopt the method of single integration for the solution of this question.
Let one of the random positions
be at P distant x from the end A of A    IP    C    B
the line, and let AC = CB = 2a.    <•---x---->
(i) Consider the favourable cases    Fig. 41.
in which AP (= x) is less than za.
Take a point Q in the line such that PQ = a. Then, for the conditions of the problem to be satisfied, the other random point R must lie in

APPLICATION OF CALCULUS TO PROBABILITY 361 CQ (otherwise PR or RB will be greater than half the line and consequently greater than the sum of the other two parts AP, RB or AP, PR).
Now P lies in the small part dx between distances x + dx and x from A, and since P has been taken at random, the chance that it falls
in this small part is dx a
The chance that R lies in CQ is
CQ AQ—AC x+PQ — AC x
    a =    a    a    a' since PO and AC are each 2a.
Therefore the total chance that P falls in dx and R in CQ is
fia x dx
Io a' a'
the limits of x being o and 2a.
(ii) Consider the favourable cases in
    which AP (= x) is greater than 2a. A    Q    C P
Then, as above, R must lie in QC,    x
    the chance of which is    Fig. 42.
QC PQ—CP a—x+2a a—x
    a =    a =    a    = a
Therefore, since the limits of x are now 2a and a, the chance that P falls in dx and R in CO is
.a a — x dx
a    a '
The total chance required
_I(lax dx+~aa—xdx
—JO a. a    .ia a    a
= a2 Lzx2ioa + a-9 Lax — 2x2]aa
2
= a2 1 2 4 + a2 (a2 2a2 2a2 -+
-
=.
Example 6.
Two points are selected at random on a line of length a. What is the probability that none of the three sections into which the line is thus divided is less than ia?
B

362    MEAN VALUE AND PROBABILITY
As an alternative this question will be solved by the use of double integrals.
Let AB be the given line divided    I,    Q
into four equal parts at C, D, E. Now A     C    D    E    B if P and O be the random points, then
to satisfy the required conditions    Fig. 43•
neither can be in AC or EB.
Let P be at distance x from A. Then the limits of x will evidently be 4a and aa. Let Q be at distance y from A ; then Q can take up any position from ea along the line from P to the point E. As the origin is at A, the limits of y will be 4a + x and 4a.
Again, when P is fixed, all possible positions of Q will be from P to B, i.e. y can vary from x to a. Obviously x can vary from o to a. Then by the unitary definition, the required chance is
a 3    a    3
2 4a    2    4a
fa    dydx    dx
    4    4(+x    4    4+r_
J a f a dydx    a [y]a dx
0 x    o    x a
fa2 [4a—fa— x] dx 4
Ia(a—x)dx 0 a
l2
[lax — lax — 2x2J
a 4
[ax — ix21
_ 18
on inserting the limits and simplifying.
Note. Each of the above two problems can be solved by the different methods here demonstrated. The two questions are exactly similar, and with the necessary alterations in the limits precisely the same working can be applied.
Example 7.
Two independent events, A and B, must each happen once and once only in the future. The chances of their happening in the interval from
a 0

    ILLUSTRATIVE EXAMPLES    363


t to t + dt are proportionate to atdt and btdt respectively, where t is the time elapsed and a and b are constants (positive fractions). Find the chance that the two events happen in the order AB.
Let the chance that A happens at the moment of time dt be kat dt and the chance that B happens be lbtdt. Then, since the events must happen, we have
-co
Jo katdt = z and    lbtdt =I. . o
This gives    k = — log a and 1= — log b.
Now the chance that B happens between now and time t from now
t
=    lbxdx. .o
Therefore the chance that B has not happened by that time
= — t lbxdx.
0
The chance that A happens at the moment of time dt = katdt,
and the chance that A happens at that moment, B not having happened,
t
= (I —    lbxdx) katdt. .o
Therefore total chance that the events happen in the order AB
=    (I — t lbxdx katdt,
o    .o    ~ which becomes
log a
log a + log b
on substituting for k and 1 and evaluating the integrals.
Example S.
In a certain year A and B were in London for one period only in each case, A for one-third of a year, B for one-quarter of a year. Asthat in the case of A any one period of one-third of a year and in the case of B any one period of one-quarter of a year is as likely as any other period, find the probabilities that
(i) A was in London the whole of the time that B was;
 364    MEAN VALUE AND PROBABILITY
(1) The chance that A arrived in London at point of time between t and t + dt from the beginning of the year is dt since he must have arrived in the first two-thirds of the year.
The chance that B arrived in the one month permissible = 13= = i 4 9 Therefore the chance that A was in London the whole of the time
2
that B was = 13 dt =
.0 9 s    9
(z) The chance that A and B were not in London together = (a) the chance that B arrived after A had left + (b) the chance that B left before A arrived.
For (a) B must have arrived between times t + i and , i.e. in the space of time i' — t. The limits of t are o and -.
5
I'; — t dt
Therefore    chance = 12
Jo    4    = 25
144
For (b) B must have arrived between the times o and t — i, and the chance of this event, namely that B left before A arrived
2
~3t—1dt    25
. 1    t    3    1.1..1. . 4
Total chance under this head = -g.
(3) This is evidently the complement of (2) and is — 72
"6 Or 7 41 ~ .
 GEOMETRICAL SOLUTIONS    365
simpler manner by the use of geometry than by having recourse to the methods of the calculus.
Consider for instance Example 6 above :
If the straight line be divided at random into three parts x, y, a — x — y, the following must hold to satisfy the conditions of the problem :
x+y<a; x,y,a—x—yeach >4a. We have    x > 4a
y> ia)
and    a — x — y>la,
giving    x + y < 4a.
Fig. 44.
Draw the straight lines LM, x + y = a and NP, x + y = 4a and complete the diagram as shown:
OM = a = OL; OA = 4a; OB = 4a; AP = 2a.
Then the above conditions may be illustrated thus :
x> IIa means that the points must be to the right of AD; y > IIa means that the points must be above BC;
x + y < 'i a means that the points must be below the line NP. The only points satisfying all these conditions are contained in the shaded area.

366    MEAN VALUE AND PROBABILITY
Similarly, for the total possible positions governed by the conx + y < a, it will be seen that all possible points lie in the triangle LOM.
The required chance is therefore
Shaded area = 2 (4a)2 __3a2 =    as before.
LOM    2a2    2a2    11~
6. We will conclude this chapter by solving a further problem by integral calculus and by plane geometry. The alternative solutions by integral calculus are given in order to show that there may be more than one method of approaching the question, and the geosolution is an excellent example of the application of elementary methods to a seemingly difficult problem.
Example 9.
X starts between 2.30 and 3 o'clock to walk at a uniform rate of 4 miles per hour from A through B and C to D. Y starts from D in the reverse direction between 2 and 3 o'clock and walks at a uniform rate of 3 miles per hour. From A to B is 2 miles, B to C i mile, C to D 3 miles. What is the chance that they meet between B and C assuming that between the given limits any time of starting is equally likely?
Method (i).
The chance that Y leaves D between t and t + dt past 2 is 60 . He
arrives at C at t minutes past 3, and at B at t + 20 minutes past 3. The chance that he meets X between B and C is that X has not reached C by t minutes past 3, i.e. that Y has not started after t + 15 minutes past z and that X has reached B by t + 20 minutes past 3.
If X has reached B by t + 20 minutes past 3, he must have started by t + 5o minutes past 2.
In order to meet, X must have started between (t + 15) and (t + 50) minutes past z ; the earliest time cannot be before 2.30 and the latest after 3.
Therefore if t is lo or less, X can start at any time between 30 and t + 50;
chance =     t + 20
30
If t is io to 15, X can start anywhere from 30 to 6o;
chance = 3S.

ILLUSTRATIVE EXAMPLES    367 If t is 15 to 45, X can start anywhere from t + 15 to 6o;
chance = 45 — t. 30
Required probability
I0 t+2odt+ (1530dt+ 4545—tdt
f o    30 6o Jlo 30 6o 15 30 6o
= ae (on evaluating the integrals).

Method (ii).
The chance that they miss is the sum of the chances that Y starts
too early and Y starts too late.
 368    MEAN VALUE AND PROBABILITY
Draw BB1 perpendicular to AD meeting DA3 and AZD2 in b1, b$ respectively, and CC1 perpendicular to AD meeting A1D1 and A2D2
                                         
                                         
 
50
                                      
                                 
                     
~~~t
            
MMWIM
    
                                         
               
.0%15
      
43
                
                                         
         
.la.
      
V 2.-.
  
c2
  
'`
    
~~
        
                           
           
          EsammPill     i~                     l~i
           
      ii°hiii    
2.30
                                    
                                   
A
                  
"
                
  2 %  
j
            
`HI!1
    
                         

Fig. 45.
in c1, c2 respectively. Then the space of time over which X and Y can meet between the points B and C on their journey is represented by
the area btib1Mcic2.
EXAMPLES rq

r. There are m posts in a straight line at equal distances of a yard apart. A man starts from any one and walks to any other; prove that the average distance which he will travel after doing this at random a great many times is i (m + r) yards.
The required chance = area b2 bl Mcl cz area KLMN
By simple geometrical methods the value of this is found to be

EXAMPLES    369
 370    MEAN VALUE AND PROBABILITY
the other at any time between i and 2. All times within these limits are equally likely. Find the chance that they are not out together.
17. Find the chance that the roots of the equation x2 — zax + b = o are real, where a and b are positive proper fractions chosen at random.
i8. The point 111 is the centre of a line LMN of length 4a. Two points P, Q on the line are chosen at random. Find the chance that the sum of the two distances MP and MQ is greater than a.