<< Chapter < Page | Chapter >> Page > |
The class is independent, with respective probabilities
0.43, 0.53, 0.46, 0.37, 0.45, 0.39. Let
cx = [6 13 -8 0];cy = [-3 4 1 -7];px = 0.01*[43 53 46 100];py = 0.01*[37 45 39 100];EX = dot(cx,px)
EX = 5.7900EY = dot(cy,py)
EY = -5.9200VX = sum(cx.^2.*px.*(1-px))
VX = 66.8191VY = sum(cy.^2.*py.*(1-py))
VY = 6.2958EZ = 3*EY - 2*EX
EZ = -29.3400VZ = 9*VY + 4*VX
VZ = 323.9386
Consider . The class has minterm
probabilities (data are in m-file npr12_10.m )
npr12_10 Data are in cx, cy, pmx and pmy
canonicEnter row vector of coefficients cx
Enter row vector of minterm probabilities pmxUse row matrices X and PX for calculations
Call for XDBN to view the distributionEX = dot(X,PX)
EX = -1.2200VX = dot(X.^2,PX) - EX^2
VX = 18.0253G = 2*X.^2 - 3*X + 2;
[W,PW]= csort(G,PX);
EW = dot(W,PW)EW = 44.6874
VW = dot(W.^2,PW) - EW^2VW = 2.8659e+03
Consider a second random variable in addition to that in [link] . The class has minterm probabilities (in mfile npr12_10.m )
The pair is independent.
(Continuation of [link] )
[Y,PY] = canonicf(cy,pmy);EY = dot(Y,PY)
EY = 19.2000VY = dot(Y.^2,PY) - EY^2
VY = 178.3600icalc
Enter row matrix of X-values XEnter row matrix of Y-values Y
Enter X probabilities PXEnter Y probabilities PY
Use array operations on matrices X, Y, PX, PY, t, u, and PH = t.^2 + 2*t.*u - u;
[Z,PZ]= csort(H,P);
EZ = dot(Z,PZ)EZ = -46.5343VZ = dot(Z.^2,PZ) - EZ^2
VZ = 3.7165e+04
Suppose the pair is independent, with gamma (3,0.1) and
Poisson (13). Let . Determine and .
gamma (3, 0.1) implies and . Poisson (13) implies . Then
The pair is jointly distributed with the following parameters:
Determine .
EX = 3;
EY = 4;EXY = 15;
EX2 = 11;VY = 5;
VX = EX2 - EX^2VX = 2
CV = EXY - EX*EYCV = 3
VZ = 9*VX + 4*VY - 6*2*CVVZ = 2
The class is independent, with respective probabilities
Let
px = 0.01*[47 33 46 100];py = 0.01*[27 41 37 100];cx = [8 11 -7 0];cy = [-3 5 1 -3];ex = dot(cx,px)
ex = 4.1700ey = dot(cy,py)
ey = -1.3900vx = sum(cx.^2.*px.*(1 - px))
vx = 54.8671vy = sum(cy.^2.*py.*(1-py))
vy = 8.0545[X,PX] = canonicf(cx,minprob(px(1:3)));[Y,PY] = canonicf(cy,minprob(py(1:3)));icalc
Enter row matrix of X-values XEnter row matrix of Y-values Y
Enter X probabilities PXEnter Y probabilities PY
Use array operations on matrices X, Y, PX, PY, t, u, and PEX = dot(X,PX)
EX = 4.1700EY = dot(Y,PY)
EY = -1.3900VX = dot(X.^2,PX) - EX^2
VX = 54.8671VY = dot(Y.^2,PY) - EY^2
VY = 8.0545EZ = 3*EY - 2*EX
EZ = -12.5100VZ = 9*VY + 4*VX
VZ = 291.9589
Notification Switch
Would you like to follow the 'Applied probability' conversation and receive update notifications?