A property of expectation---additive

One of the motivations to write this article is this question:

http://acm.hdu.edu.cn/showproblem.php?pid=5194

This is the first question of the BESTCODER35 competition. Thought for a long while only violent methods. Attach the answer given by the author of the original question:

The title means:

n a black ball, m a white ball. Now don't put back the ball until there is no ball to take. Take out the black Note 1. The white Note 0. Expectation of the number of "01" occurrences of the entire sequence

The additive of expectation is mentioned in the original solution. Woo, this time has been forgotten. Today, when you are ready to review the probability of a certain knowledge point, by the way review this nature.

This property is:

E (x+y) =e (X) +e (Y);

The proof is as follows:

Note that x, Y independence is not required in the certification process.

This is what I want to say the nature ~.

Back to the topic:

In this topic, we can define the random variable XI as:

Then there are:

x=x1+x2+x3+ ...

and, E (xi) =p (xi=1) = (m/m+n) (n/m+n-1)

Then E (X) =e (x1+x2+x3+ ... X (m+n-1))

=e (1) +e (2) +...+e (m+n-1)

=mn/(M+n)

The "Probability Theory and Mathematical statistics" of the Higher Education Society there is an example in the teacher's book, which is also attached here for comparison.

Note that the emphasis on independence in this topic is to calculate probabilities. The desired additive does not need to be independent.