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Then, by the law of total probability (CE1b) ,
If there is a conditional density such that
then
A careful, measure-theoretic treatment shows that it may not be true that is a distribution function for all t in the range of X . However, in applications, this is seldom a problem. Modeling assumptions often start with such a family of distribution functions or density functions.
As in [link] , suppose exponential , where the parameter u is the value of a parameter random variable H . If the parameter random variable uniform , determine the distribution function F X .
SOLUTON
As in [link] , take the assumption on the conditional distribution to mean
Then
By the law of total probability
Differentiation with respect to t yields the expression for
The following example uses a discrete conditional distribution and marginal distribution to obtain the joint distribution for the pair.
A number N is chosen by a random selection from the integers from 1 through 20 (say by drawing a card from a box). A pair of dice is thrown N times. Let S be the number of “matches” (i.e., both ones, both twos, etc.). Determine the jointdistribution for .
SOLUTION
uniform on the integers 1 through 20. for . Since there are 36 pairs of numbers for the two dice and six possible matches, theprobability of a match on any throw is 1/6. Since the i throws of the dice constitute a Bernoulli sequence with probability 1/6 of a success (a match), we have S conditionally binomial , given . For any pair , ,
Now . so that
The following MATLAB procedure calculates the joint probabilities and arranges them “as on the plane.”
% file randbern.m
p = input('Enter the probability of success ');N = input('Enter VALUES of N ');
PN = input('Enter PROBABILITIES for N ');n = length(N);
m = max(N);S = 0:m;
P = zeros(n,m+1);for i = 1:n
P(i,1:N(i)+1) = PN(i)*ibinom(N(i),p,0:N(i));end
PS = sum(P);P = rot90(P);
disp('Joint distribution N, S, P, and marginal PS')randbern % Call for the procedure
Enter the probability of success 1/6Enter VALUES of N 1:20
Enter PROBABILITIES for N 0.05*ones(1,20)Joint distribution N, S, P, and marginal PS
ES = S*PS'ES = 1.7500 % Agrees with the theoretical value We introduce the regression problem in the treatment of linear regression. Here we are concerned with more general regression. A pair of real random variables has a joint distribution. A value is observed. We desire a rule for obtaining the “best” estimate of the corresponding value . If is the actual value and is the estimate, then is the error of estimate. The best estimation rule (function) is taken to be that for which the average square of the error is a minimum. That is, we seek a function r such that
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