Mathematical expectation: simple random variables (Page 4/4)

Applied probability Page 4 / 4

E [X + Y] = \sum_{i = 1}^{n} \sum_{j = 1}^{m} (t_{i} + u_{j}) P (A_{i} B_{j}) = \sum_{i = 1}^{n} \sum_{j = 1}^{m} t_{i} P (A_{i} B_{j}) + \sum_{i = 1}^{n} \sum_{j = 1}^{m} u_{j} P (A_{i} B_{j})

= \sum_{i = 1}^{n} t_{i} \sum_{j = 1}^{m} P (A_{i} B_{j}) + \sum_{j = 1}^{m} u_{j} \sum_{i = 1}^{n} P (A_{i} B_{j})

We note that for each i and for each j

P (A_{i}) = \sum_{j = 1}^{m} P (A_{i} B_{j}) and P (B_{j}) = \sum_{i = 1}^{n} P (A_{i} B_{j})

Hence, we may write

E [X + Y] = \sum_{i = 1}^{n} t_{i} P (A_{i}) + \sum_{j = 1}^{m} u_{j} P (B_{j}) = E [X] + E [Y]

Now $a X$ and $b Y$ are simple if X and Y are, so that with the aid of [link] we have

E [a X + b Y] = E [a X] + E [b Y] = a E [X] + b E [Y]

If $X, Y, Z$ are simple, then so are $a X + b Y$ , and $c Z$ . It follows that

E [a X + b Y + c Z] = E [a X + b Y] + c E [Z] = a E [X] + b E [Y] + c E [Z]

By an inductive argument, this pattern may be extended to a linear combination of any finite number of simple random variables. Thus we may assert

Linearity . The expectation of a linear combination of a finite number of simple random variables is that linear combination of the expectations of the individual random variables.

— $□$

Expectation of a simple random variable in affine form

As a direct consequence of linearity, whenever simple random variable X is in affine form, then

E [X] = E [c_{0} + \sum_{i = 1}^{n} c_{i} I_{E_{i}}] = c_{0} + \sum_{i = 1}^{n} c_{i} P (E_{i})

Thus, the defining expression holds for any affine combination of indicator functions , whether in canonical form or not.

Binomial distribution $(n, p)$

This random variable appears as the number of successes in n Bernoulli trials with probability p of success on each component trial. It is naturally expressed in affine form

X = \sum_{i = 1}^{n} I_{E_{i}} so that E [X] = \sum_{i = 1}^{n} p = n p

Alternately, in canonical form

X = \sum_{k = 0}^{n} k I_{A_{k n}}, with p_{k} = P (A_{k n}) = P (X = k) = C (n, k) p^{k} q^{n - k}, q = 1 - p

so that

E [X] = \sum_{k = 0}^{n} k C (n, k) p^{k} q^{n - k}, q = 1 - p

Some algebraic tricks may be used to show that the second form sums to $n p$ , but there is no need of that. The computation for the affine form is much simpler.

Got questions? Get instant answers now!

Expected winnings

A bettor places three bets at $2.00 each. The first bet pays $10.00 with probability 0.15, the second pays $8.00 with probability 0.20, and the third pays $20.00 with probability 0.10.What is the expected gain?

SOLUTION

The net gain may be expressed

X = 10 I_{A} + 8 I_{B} + 20 I_{C} - 6, with P (A) = 0.15, P (B) = 0.20, P (C) = 0.10

Then

E [X] = 10 \cdot 0.15 + 8 \cdot 0.20 + 20 \cdot 0.10 - 6 = - 0.90

These calculations may be done in MATLAB as follows:

c = [10 8 20 -6];p = [0.15 0.20 0.10 1.00]; % Constant a = aI_(Omega), with P(Omega) = 1
E = c*p'E =  -0.9000

Got questions? Get instant answers now!

Functions of simple random variables

If X is in a primitive form (including canonical form) and g is a real function defined on the range of X , then

Z = g (X) = \sum_{j = 1}^{m} g (c_{j}) I_{C_{j}} a primitive form

so that

E [Z] = E [g (X)] = \sum_{j = 1}^{m} g (c_{j}) P (C_{j})

Alternately, we may use csort to determine the distribution for Z and work with that distribution.

Caution . If X is in affine form (but not a primitive form)

X = c_{0} + \sum_{j = 1}^{m} c_{j} I_{E_{j}} then g (X) \neq g (c_{0}) + \sum_{j = 1}^{m} g (c_{j}) I_{E_{j}}

so that

E [g (X)] \neq g (c_{0}) + \sum_{j = 1}^{m} g (c_{j}) P (E_{j})

Expectation of a function of X

Suppose X in a primitive form is

X = - 3 I_{C_{1}} - I_{C_{2}} + 2 I_{C_{3}} - 3 I_{C_{4}} + 4 I_{C_{5}} - I_{C_{6}} + I_{C_{7}} + 2 I_{C_{8}} + 3 I_{C_{9}} + 2 I_{C_{10}}

with probabilities $P (C_{i}) = 0.08, 0.11, 0.06, 0.13, 0.05, 0.08, 0.12, 0.07, 0.14, 0.16$ .

Let $g (t) = t^{2} + 2 t$ . Determine $E [g (X)]$ .

c = [-3 -1 2 -3 4 -1 1 2 3 2];            % Original coefficientspc = 0.01*[8 11 6 13 5 8 12 7 14 16];     % Probabilities for C_jG = c.^2 + 2*c                            % g(c_j)
G =  3  -1   8   3  24  -1   3   8  15   8EG = G*pc'                            % Direct computation
EG =  6.4200[Z,PZ] = csort(G,pc);                 % Distribution for Z = g(X)disp([Z;PZ]')                         % Optional display-1.0000    0.1900
    3.0000    0.33008.0000    0.2900
   15.0000    0.140024.0000    0.0500
EZ = Z*PZ'                            % E[Z]from distribution for Z
EZ =  6.4200

Got questions? Get instant answers now!

A similar approach can be made to a function of a pair of simple random variables, provided the joint distribution is available. Suppose $X = \sum_{i = 1}^{n} t_{i} I_{A_{i}}$ and $Y = \sum_{j = 1}^{m} u_{j} I_{B_{j}}$ (both in canonical form). Then

Z = g (X, Y) = \sum_{i = 1}^{n} \sum_{j = 1}^{m} g (t_{i}, u_{j}) I_{A_{i} B_{j}}

The $A_{i} B_{j}$ form a partition, so Z is in a primitive form. We have the same two alternative possibilities: (1) direct calculation from values of $g (t_{i}, u_{j})$ and corresponding probabilities $P (A_{i} B_{j}) = P (X = t_{i}, Y = u_{j})$ , or (2) use of csort to obtain the distribution for Z .

Expectation for $Z = g (X, Y)$

We use the joint distribution in file jdemo1.m and let $g (t, u) = t^{2} + 2 t u - 3 u$ . To set up for calculations, we use jcalc.

% file jdemo1.m
X = [-2.37 -1.93 -0.47 -0.11  0  0.57 1.22 2.15 2.97 3.74];
Y = [-3.06 -1.44 -1.21  0.07 0.88 1.77 2.01 2.84];
P = 0.0001*[ 53   8 167 170 184  18  67 122  18  12;11  13 143 221 241 153  87 125 122 185;
            165 129 226 185  89 215  40  77  93 187;165 163 205  64  60  66 118 239  67 201;
            227   2 128  12 238 106 218 120 222  30;93  93  22 179 175 186 221  65 129   4;
            126  16 159  80 183 116  15  22 113 167;198 101 101 154 158  58 220 230 228 211];

jdemo1                   % Call for data
jcalc                    % Set upEnter JOINT PROBABILITIES (as on the plane)  P
Enter row matrix of VALUES of X  XEnter row matrix of VALUES of Y  Y
 Use array operations on matrices X, Y, PX, PY, t, u, and PG = t.^2 + 2*t.*u - 3*u; % Calculation of matrix of [g(t_i, u_j)]
EG = total(G.*P)         % Direct calculation of expectationEG =  3.2529
[Z,PZ]= csort(G,P);     % Determination of distribution for Z
EZ = Z*PZ'               % E[Z]from distribution
EZ =  3.2529

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Applied probability' conversation and receive update notifications?

Ask

	College physics By OpenStax Read Online Course
	Fluid Mechanics MCQ By Stephanie Redfern Start Quiz
©flickr: U.S.	Biology Chapter 8 By Michael Sag Start Exam
	9 Psychology MCQ 2011 2 Exam By John Gabrieli Start Exam
	31 Biology 31 Soil and Plant Nutrition MCQ By OpenStax Start Quiz
	6 Arts Society: Theater 6 By Jonathan Long Start Quiz
	23 AP Key Terms 23 The Digestive System By OpenStax Start Key Terms
	8 AP 08 Appendicular Skeleton Essay By OpenStax Start Flashcards
	1 Lec:1 Descriptive Epidemiology By Janet Forrester Start Quiz
	2 Endocrinology Essay By Rohini Ajay Start Test

Mathematical expectation: simple random variables (Page 4/4)

Binomial distribution ( n , p )

Expected winnings

Expectation of a function of X

Expectation for Z = g ( X , Y )

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Binomial distribution $(n, p)$

Expectation for $Z = g (X, Y)$