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A jointly distributed pair ( Example 14 From "mathematical expectation; general random variables")

Suppose the pair { X , Y } has joint density f X Y ( t , u ) = 3 u on the triangular region bounded by u = 0 , u = 1 + t , u = 1 - t . Let Z = g ( X , Y ) = X 2 + 2 X Y .

Determine E [ Z ] and Var [ Z ] .

ANALYTIC SOLUTION

E [ Z ] = ( t 2 + 2 t u ) f X Y ( t , u ) d u d t = 3 - 1 0 0 1 + t u ( t 2 + 2 t u ) d u d t + 3 0 1 0 1 - t u ( t 2 + 2 t u ) d u d t = 1 / 10
E [ Z 2 ] = 3 - 1 0 0 1 + t u ( t 2 + 2 t u ) 2 d u d t + 3 0 1 0 1 - t u ( t 2 + 2 t u ) 2 d u d t = 3 / 35
Var [ Z ] = E [ Z 2 ] - E 2 [ Z ] = 53 / 700 0 . 0757

APPROXIMATION

tuappr Enter matrix [a b]of X-range endpoints [-1 1] Enter matrix [c d]of Y-range endpoints [0 1] Enter number of X approximation points 400Enter number of Y approximation points 200 Enter expression for joint density 3*u.*(u<=min(1+t,1-t)) Use array operations on X, Y, PX, PY, t, u, and PG = t.^2 + 2*t.*u; % g(X,Y) = X^2 + 2XY EG = total(G.*P) % E[g(X,Y)]EG = 0.1006 % Theoretical value = 1/10 VG = total(G.^2.*P) - EG^2VG = 0.0765 % Theoretical value 53/700 = 0.0757 [Z,PZ]= csort(G,P); % Distribution for Z EZ = Z*PZ' % E[Z]from distribution EZ = 0.1006VZ = Z.^2*PZ' - EZ^2 VZ = 0.0765
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A function with compound definition ( Example 15 From "mathematical expectation; general random variables")

The pair { X , Y } has joint density f X Y ( t , u ) = 1 / 2 on the square region bounded by u = 1 + t , u = 1 - t , u = 3 - t , and u = t - 1 .

W = X for max { X , Y } 1 2 Y for max { X , Y } > 1 = I Q ( X , Y ) X + I Q c ( X , Y ) 2 Y

where Q = { ( t , u ) : max { t , u } 1 } = { ( t , u ) : t 1 , u 1 } .

Determine E [ W ] and Var [ W ] .

ANALYTIC SOLUTION

The intersection of the region Q and the square is the set for which 0 t 1 and 1 - t u 1 . Reference to Figure 11.3.2 shows three regions of integration.

E [ W ] = 1 2 0 1 1 - t 1 t d u d t + 1 2 0 1 1 1 + t 2 u d u d t + 1 2 1 2 t - 1 3 - t 2 u d u d t = 11 / 6 1 . 8333
E [ W 2 ] = 1 2 0 1 1 - t 1 t 2 d u d t + 1 2 0 1 1 1 + t 4 u 2 d u d t + 1 2 1 2 t - 1 3 - t 4 u 2 d u d t = 103 / 24
Var [ W ] = 103 / 24 - ( 11 / 6 ) 2 = 67 / 72 0 . 9306
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 2] Enter number of X approximation points 200Enter number of Y approximation points 200 Enter expression for joint density ((u<=min(t+1,3-t))&... (u$gt;=max(1-t,t-1)))/2Use array operations on X, Y, PX, PY, t, u, and P M = max(t,u)<=1; G = t.*M + 2*u.*(1 - M); % Z = g(X,Y)EG = total(G.*P) % E[g(X,Y)] EG = 1.8340 % Theoretical 11/6 = 1.8333VG = total(G.^2.*P) - EG^2 VG = 0.9368% Theoretical 67/72 = 0.9306 [Z,PZ]= csort(G,P); % Distribution for Z EZ = Z*PZ' % E[Z]from distribution EZ = 1.8340VZ = (Z.^2)*PZ' - EZ^2 VZ = 0.9368
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A function with compound definition

f X Y ( t , u ) = 3 on 0 u t 2 1
Z = I Q ( X , Y ) X + I Q c ( X , Y ) for Q = { ( t , u ) : u + t 1 }

The value t 0 where the line u = 1 - t and the curve u = t 2 meet satisfies t 0 2 = 1 - t 0 .

E [ Z ] = 3 0 t 0 t 0 t 2 d u d t + 3 t 0 1 t 0 1 - t d u d t + 3 t 0 1 1 - t t 2 d u d t = 3 4 ( 5 t 0 - 2 )

For E [ Z 2 ] replace t by t 2 in the integrands to get E [ Z 2 ] = ( 25 t 0 - 1 ) / 20 .

Using t 0 = ( 5 - 1 ) / 2 0 . 6180 , we get Var [ Z ] = ( 2125 t 0 - 1309 ) / 80 0 . 0540 .

APPROXIMATION

% Theoretical values t0 = (sqrt(5) - 1)/2t0 = 0.6180 EZ = (3/4)*(5*t0 -2)EZ = 0.8176 EZ2 = (25*t0 - 1)/20EZ2 = 0.7225 VZ = (2125*t0 - 1309)/80VZ = 0.0540 tuapprEnter matrix [a b] of X-range endpoints [0 1]Enter matrix [c d] of Y-range endpoints [0 1]Enter number of X approximation points 200 Enter number of Y approximation points 200Enter expression for joint density 3*(u<= t.^2) Use array operations on X, Y, t, u, and PG = (t+u<= 1).*t + (t+u>1); EG = total(G.*P)EG = 0.8169 % Theoretical = 0.8176VG = total(G.^2.*P) - EG^2 VG = 0.0540 % Theoretical = 0.0540[Z,PZ] = csort(G,P);EZ = Z*PZ' EZ = 0.8169VZ = (Z.^2)*PZ' - EZ^2 VZ = 0.0540
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Standard deviation and the Chebyshev inequality

In Example 5 from "Functions of a Random Variable," we show that if X N ( μ , σ 2 ) then Z = X - μ σ N ( 0 , 1 ) . Also, E [ X ] = μ and Var [ X ] = σ 2 . Thus

P | X - μ | σ t = P ( | X - μ | t σ ) = 2 Φ ( t ) - 1

For the normal distribution, the standard deviation σ seems to be a natural measure of the variation away from the mean.

For a general distribution with mean μ and variance σ 2 , we have the

Chebyshev inequality

P | X - μ | σ a 1 a 2 or P ( | X - μ | a σ ) 1 a 2

In this general case, the standard deviation appears as a measure of the variation from the mean value. This inequality is useful in many theoretical applications as wellas some practical ones. However, since it must hold for any distribution which has a variance, the bound is not a particularly tight. It may be instructive to comparethe bound on the probability given by the Chebyshev inequality with the actual probability for the normal distribution.

t = 1:0.5:3; p = 2*(1 - gaussian(0,1,t));c = ones(1,length(t))./(t.^2); r = c./p;h = [' t Chebyshev Prob Ratio'];m = [t;c;p;r]';disp(h) t Chebyshev Prob Ratiodisp(m) 1.0000 1.0000 0.3173 3.15151.5000 0.4444 0.1336 3.3263 2.0000 0.2500 0.0455 5.49452.5000 0.1600 0.0124 12.8831 3.0000 0.1111 0.0027 41.1554

DERIVATION OF THE CHEBYSHEV INEQUALITY

Let A = { | X - μ | a σ } = { ( X - μ ) 2 a 2 σ 2 } . Then a 2 σ 2 I A ( X - μ ) 2 .

Upon taking expectations of both sides and using monotonicity, we have

a 2 σ 2 P ( A ) E [ ( X - μ ) 2 ] = σ 2

from which the Chebyshev inequality follows immediately.

We consider three concepts which are useful in many situations.

Definition . A random variable X is centered iff E [ X ] = 0 .

X ' = X - μ is always centered.

Definition . A random variable X is standardized iff E [ X ] = 0 and Var [ X ] = 1 .

X * = X - μ σ = X ' σ is standardized

Definition . A pair { X , Y } of random variables is uncorrelated iff

E [ X Y ] - E [ X ] E [ Y ] = 0

It is always possible to derive an uncorrelated pair as a function of a pair { X , Y } , both of which have finite variances. Consider

U = ( X * + Y * ) V = ( X * - Y * ) , where X * = X - μ X σ X , Y * = Y - μ Y σ Y

Now E [ U ] = E [ V ] = 0 and

E [ U V ] = E ( X * + Y * ) ( X * - Y * ) ] = E [ ( X * ) 2 ] - E [ ( Y * ) 2 ] = 1 - 1 = 0

so the pair is uncorrelated.

Determining an uncorrelated pair

We use the distribution for Examples Example 10 from "Mathematical Expectation: Simple Random Variables" and [link] , for which

E [ X Y ] - E [ X ] E [ Y ] 0
jdemo1 jcalcEnter 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 PEX = total(t.*P) EX = 0.6420EY = total(u.*P) EY = 0.0783EXY = total(t.*u.*P) EXY = -0.1130c = EXY - EX*EY c = -0.1633 % {X,Y} not uncorrelated VX = total(t.^2.*P) - EX^2 VX = 3.3016VY = total(u.^2.*P) - EY^2 VY = 3.6566SX = sqrt(VX) SX = 1.8170SY = sqrt(VY) SY = 1.9122x = (t - EX)/SX; % Standardized random variables y = (u - EY)/SY;uu = x + y; % Uncorrelated random variables vv = x - y;EUV = total(uu.*vv.*P) % Check for uncorrelated condition EUV = 9.9755e-06 % Differs from zero because of roundoff
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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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