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Calculate directly the generating function
Calculate directly the generating function
A projection bulb has life (in hours) represented by
Simple random variable X has distribution
Setting
Use the moment generating function to obtain the variances for the following distributions
Exponential
The pair
The pair
Use the moment generating function for the symmetric triangular
distribution on
in the section "Three Basic Transforms".
Let
Random variable X has moment generating function
The pair
Random variable X has moment generating function
By recognizing forms and using rules of combinations, determine
Random variable X has moment generating function
By recognizing forms and using rules of combinations, determine
Suppose the class
The distribution is
Catalogue of Useful Matlab Filesc = [-3 2 4 0];
P = 0.1*[3 5 2];
canonic
Enter row vector of coefficients c
Enter row vector of minterm probabilities minprob(P)
Use row matrices X and PX for calculations
Call for XDBN to view the distribution
P1 = [0.7 0.3];
P2 = [0.5 0.5];
P3 = [0.8 0.2];
X1 = [0 -3];
X2 = [0 2];
X3 = [0 4];
[x,px] = mgsum3(X1,X2,X3,P1,P2,P3);
disp([X;PX;x;px]')
-3.0000 0.1200 -3.0000 0.1200
-1.0000 0.1200 -1.0000 0.1200
0 0.2800 0 0.2800
1.0000 0.0300 1.0000 0.0300
2.0000 0.2800 2.0000 0.2800
3.0000 0.0300 3.0000 0.0300
4.0000 0.0700 4.0000 0.0700
6.0000 0.0700 6.0000 0.0700
Suppose the pair
Binomial iff both have same p, as shown below.
Suppose the pair
Always Poisson, as the argument below shows.
However,
Suppose the pair
Suppose the pair
x = 0:6;
px = ibinom(6,0.51,x);
[Z,PZ] = mgsum(2*x,4*x,px,px);
disp([Z(1:5);PZ(1:5)]')
0 0.0002 % Cannot be binomial, since odd values missing
2.0000 0.0012
4.0000 0.0043
6.0000 0.0118
8.0000 0.0259
- - - - - - - -
Suppose the pair
Let
X = 0:5;
Y = 0:7;
PX = ibinom(5,0.33,X);
PY = ibinom(7,0.47,Y);
G = 3*X.^2 - 2*X;
H = 2*Y.^2 + Y + 3;
[Z,PZ] = mgsum(G,H,PX,PY);
icalc
Enter row matrix of X-values X
Enter row matrix of Y-values Y
Enter X probabilities PX
Enter Y probabilities PY
Use array operations on matrices X, Y, PX, PY, t, u, and P
M = 3*t.^2 - 2*t + 2*u.^2 + u + 3;
[z,pz] = csort(M,P);
e = max(abs(pz - PZ)) % Comparison of p values
e = 0
Suppose the pair
X = 0:8;
Y = [-1.3 -0.5 1.3 2.2 3.5];
PX = ibinom(8,0.39,X);
PY = (1/5)*ones(1,5);
U = 3*X.^2 - 2*X + 1;
V = Y.^3 + 2*Y - 3;
[Z,PZ] = mgsum(U,V,PX,PY);
icalc
Enter row matrix of X-values X
Enter row matrix of Y-values Y
Enter X probabilities PX
Enter Y probabilities PY
Use array operations on matrices X, Y, PX, PY, t, u, and P
M = 3*t.^2 - 2*t + 1 + u.^3 + 2*u - 3;
[z,pz] = csort(M,P);
e = max(abs(pz - PZ))
e = 0
If X is a nonnegative integer-valued random variable, express the generating function as a power series.
Since power series may be differentiated term by term
Let
Setting
Use derivatives of
To simplify writing use
Use moment generating functions to show that variances add for the sum or difference of independent random variables.
To simplify writing, set
Setting
Taking the difference gives
The pair
Use the central limit theorem to show that for large enough sample size (usually 20 or more), the sample average
is approximately
By the central limit theorem, An is approximately normal, with the mean and variance above.
A population has standard deviation approximately three. It is desired to determine the sample size n needed to ensure that with probability 0.95 the sample average will be within 0.5 of the mean value.
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This work is licensed by Paul E Pfeiffer under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.
Last edited by Daniel Williamson on Sep 17, 2009 4:37 pm -0500.