Variance (Page 4/4)

Applied probability Page 4 / 4

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] = \int \int (t^{2} + 2 t u) f_{X Y} (t, u) d u d t = 3 \int_{- 1}^{0} \int_{0}^{1 + t} u (t^{2} + 2 t u) d u d t + 3 \int_{0}^{1} \int_{0}^{1 - t} u (t^{2} + 2 t u) d u d t = 1 / 10

E [Z^{2}] = 3 \int_{- 1}^{0} \int_{0}^{1 + t} u {(t^{2} + 2 t u)}^{2} d u d t + 3 \int_{0}^{1} \int_{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 \approx 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 = \{\begin{matrix} X & for max {X, Y} \leq 1 \\ 2 Y & for max {X, Y} > 1 \end{matrix}) = I_{Q} (X, Y) X + I_{Q^{c}} (X, Y) 2 Y

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

Determine $E [W]$ and $Var [W]$ .

ANALYTIC SOLUTION

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

E [W] = \frac{1}{2} \int_{0}^{1} \int_{1 - t}^{1} t d u d t + \frac{1}{2} \int_{0}^{1} \int_{1}^{1 + t} 2 u d u d t + \frac{1}{2} \int_{1}^{2} \int_{t - 1}^{3 - t} 2 u d u d t = 11 / 6 \approx 1.8333

E [W^{2}] = \frac{1}{2} \int_{0}^{1} \int_{1 - t}^{1} t^{2} d u d t + \frac{1}{2} \int_{0}^{1} \int_{1}^{1 + t} 4 u^{2} d u d t + \frac{1}{2} \int_{1}^{2} \int_{t - 1}^{3 - t} 4 u^{2} d u d t = 103 / 24

Var [W] = 103 / 24 - {(11 / 6)}^{2} = 67 / 72 \approx 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

Got questions? Get instant answers now!

A function with compound definition

f_{X Y} (t, u) = 3 on 0 \leq u \leq t^{2} \leq 1

Z = I_{Q} (X, Y) X + I_{Q^{c}} (X, Y) for Q = {(t, u) : u + t \leq 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 \int_{0}^{t_{0}} t \int_{0}^{t^{2}} d u d t + 3 \int_{t_{0}}^{1} t \int_{0}^{1 - t} d u d t + 3 \int_{t_{0}}^{1} \int_{1 - t}^{t^{2}} d u d t = \frac{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 $t_{0} = (\sqrt{5} - 1) / 2 \approx 0.6180$ , we get $Var [Z] = (2125 t_{0} - 1309) / 80 \approx 0.0540$ .

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

Got questions? Get instant answers now!

Standard deviation and the Chebyshev inequality

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

P (\frac{| X - μ |}{σ} \leq t) = P (| X - μ | \leq 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 (\frac{| X - μ |}{σ} \geq a) \leq \frac{1}{a^{2}} or P (| X - μ | \geq a σ) \leq \frac{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 $A = {| X - μ | \geq a σ} = {{(X - μ)}^{2} \geq a^{2} σ^{2}}$ . Then $a^{2} σ^{2} I_{A} \leq {(X - μ)}^{2}$ .

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

a^{2} σ^{2} P (A) \leq 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^{*} = \frac{X - μ}{σ} = \frac{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^{*} = \frac{X - μ_{X}}{σ_{X}}, Y^{*} = \frac{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] \neq 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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