11.1 Mathematical expectation; general random variables (Page 5/7)

Applied probability Page 5 / 7

Remark . It is easy to show that the function $λ (\cdot)$ is nondecreasing. This fact is used in establishing Jensen's inequality for conditional expectation.

The product rule for expectations of independent random variables

Product rule for simple random variables

Consider an independent pair ${X, Y}$ of simple random variables

X = \sum_{i = 1}^{n} t_{i} I_{A_{i}} Y = \sum_{j = 1}^{m} u_{j} I_{B_{j}} (both in canonical form)

We know that each pair ${A_{i}, B_{j}}$ is independent, so that $P (A_{i} B_{j}) = P (A_{i}) P (B_{j})$ . Consider the product $X Y$ . According to the pattern described after Example 9 from "Mathematical Expectation: Simple Random Variables."

X Y = \sum_{i = 1}^{n} t_{i} I_{A_{i}} \sum_{j = 1}^{m} u_{j} I_{B_{j}} = \sum_{i = 1}^{n} \sum_{j = 1}^{m} t_{i} u_{j} I_{A_{i} B_{j}}

The latter double sum is a primitive form, so that

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

Thus the product rule holds for independent simple random variables.

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Approximating simple functions for an independent pair

Suppose ${X, Y}$ is an independent pair, with an approximating simple pair ${X_{s}, Y_{s}}$ . As functions of X and Y , respectively, the pair ${X_{s}, Y_{s}}$ is independent. According to [link] , above, the product rule $E [X_{s} Y_{s}] = E [X_{s}] E [Y_{s}]$ must hold.

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Product rule for an independent pair

For $X \geq 0, Y \geq 0$ , there exist nondecreasing sequences ${X_{n} : 1 \leq n}$ and ${Y_{n} : 1 \leq n}$ of simple random variables increasing to X and Y , respectively. The sequence ${X_{n} Y_{n} : 1 \leq n}$ is also a nondecreasing sequence of simple random variables, increasing to $X Y$ . By the monotone convergence theorem (MC)

E [X_{n}] ↗ E [X], E [Y_{n}] ↗ E [Y], and E [X_{n} Y_{n}] ↗ E [X Y]

Since $E [X_{n} Y_{n}] = E [X_{n}] E [Y_{n}]$ for each n , we conclude $E [X Y] = E [X] E [Y]$

In the general case,

X Y = (X^{+} - X^{-}) (Y^{+} - Y^{-}) = X^{+} Y^{+} - X^{+} Y^{-} - X^{-} Y^{+} + X^{-} Y^{-}

Application of the product rule to each nonnegative pair and the use of linearity gives the product rule for the pair ${X, Y}$

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Remark . It should be apparent that the product rule can be extended to any finite independent class.

The joint distribution of three random variables

The class ${X, Y, Z}$ is independent, with the marginal distributions shown below. Let $W = g (X, Y, Z) = 3 X^{2} + 2 X Y - 3 X Y Z$ . Determine $E [W]$ .

X = 0:4;
Y = 1:2:7;Z = 0:3:12;
PX = 0.1*[1 3 2 3 1];
PY = 0.1*[2 2 3 3];
PZ = 0.1*[2 2 1 3 2];

icalc3                            % Setup for joint dbn for{X,Y,Z}
Enter row matrix of X-values  XEnter row matrix of Y-values  Y
Enter row matrix of Z-values  ZEnter X probabilities  PX
Enter Y probabilities  PYEnter Z probabilities  PZ
Use array operations on matrices X, Y, Z,PX, PY, PZ, t, u, v, and P
EX = X*PX'% E[X]
EX =    2EX2 = (X.^2)*PX'              % E[X^2]
EX2 =   5.4000EY =  Y*PY'                   % E[Y]
EY =    4.4000EZ = Z*PZ'                    % E[Z]
EZ =    6.3000G = 3*t.^2 + 2*t.*u - 3*t.*u.*v;  % W = g(X,Y,Z) = 3X^2 + 2XY - 3XYZ
EG = total(G.*P)              % E[g(X,Y,Z)]EG = -132.5200
[W,PW]= csort(G,P);          % Distribution for W = g(X,Y,Z)
EW = W*PW'                    % E[W]EW = -132.5200
ew = 3*EX2 + 2*EX*EY - 3*EX*EY*EZ % Use of linearity and product ruleew = -132.5200

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A function with a compound definition: truncated exponential

Suppose $X \sim$ exponential (0.3). Let

Z = \{\begin{matrix} X^{2} & for X \leq 4 \\ 16 & for X > 4 \end{matrix}) = I_{[0, 4]} (X) X^{2} + I_{(4, \infty]} (X) 16

Determine $E [Z]$ .

ANALYTIC SOLUTION

E [g (X)] = \int g (t) f_{X} (t) d t = \int_{0}^{\infty} I_{[0, 4]} (t) t^{2} 0.3 e^{- 0.3 t} d t + 16 E [I_{(4, \infty]} (X)]

= \int_{0}^{4} t^{2} 0.3 e^{- 0.3 t} d t + 16 P (X > 4) \approx 7.4972 (by Maple)

APPROXIMATION

To obtain a simple aproximation, we must approximate the exponential by a bounded random variable. Since $P (X > 50) = e^{- 15} \approx 3 \cdot 10^{- 7}$ we may safely truncate X at 50.

tappr
Enter matrix [a b]of x-range endpoints  [0 50]
Enter number of x approximation points  1000Enter density as a function of t  0.3*exp(-0.3*t)
Use row matrices X and PX as in the simple caseM = X<= 4;
G = M.*X.^2 + 16*(1 - M);  % g(X)EG = G*PX'                 % E[g(X)]
EG =  7.4972[Z,PZ] = csort(G,PX);      % Distribution for Z = g(X)EZ = Z*PZ'                 % E[Z] from distributionEZ =  7.4972

Because of the large number of approximation points, the results agree quite closely with the theoretical value.

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

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