# Total Probability with “Weighting”

I am reading a text and I have a somewhat "abstract" question regarding probability. It involves a technique I have never seen before. The text is: Mobile Communication Systems by Parsons, page 208-209. Another text which contains the same information but in a slightly different form is Microwave Mobile Communications by Jakes, page 406-408.

The context is signal levels - a stochastic process can be either above or below a threshold, and we want to calculate probabilities related to this. There exist 2 input signals, $r_1(t)$ and $r_2(t)$. The output signal $R(t)$, takes on the value $r_i(t)$ until it falls below a threshold A, at which time it immediately takes on the value $r_j(t)$, $i \neq j$. $R(t)$ switches only when the current branch $r_i(t)$ falls below a threshold level A. Thus, it is entirely possible for R(t) to take on values below A, if both $r_1(t)$ and $r_2(t)$ are both below A. $r_1(t)$ and $r_2(t)$ have identically independent stationary probability distributions. So, we can calculate the probability of $r_i(t)$ at any time being below a level, within a range, or above a level.

Suppose the probability $P(r_i(t) < A) = q$. Suppose the average amount of time $r_i(t)$ is above level A is $\tau_p$. The average amount of time is defined as: the total amount of time above the level divided by the number of times the signal crosses into the level. Suppose the average amount of time $r_i$ is below level A is $\tau_q$. A "successful" transition is defined as: the switch occurs such that the signal that is being switched into is already above the level A. A "unsuccessful" transition is defined as: the switch occurs such that the signal that is being switched into is below the level A (so we did not improve our situation - a higher signal level is considered better).

We also have the following conditional probability distributions: $$P(r_i(t) < B | r_i(t) > A) = x, B>A$$ $$P(r_i(t) < B | r_i(t) > A) = 0, B<A$$ $$P(r_i(t) < B | r_i(t) < A) = 1, A<B$$ $$P(r_i(t) < B | r_i(t) < A) = y, A>B$$

We are interested in calculating the probability $$P(R(t) < B)$$ The statement (paraphrased) that makes me confused is: To obtain the overall cumulative distribution, it is necessary to combine the four equations above, with a weighting to account for the relative times over which they apply.

The book goes on to say the average duration of a segment following a successful transition is $\frac{\tau_p}{2}$. The average duration of a segment following an unsuccessful transition is $\frac{\tau_q}{2} + \tau_p$. I can agree with the last two sentences.

Then, my question is, why is this statement true, exact quote from the book: "The probabilities of successful and unsuccessful transitions are (1-q) and q, respectively, and the weighting appropriate to each distribution is proportional to probability of occurrence X average duration". Why is this true? I am looking at the units here, and we are multiplying a time by a probability. Eventually the final result $P(R(t) < B)$ is the product of a 2x2 matrix (consisting of the four conditional probabilities) by a vector. The vector contains the "probability of occurrence X average duration", normalized (the vector elements sum to 1). Why is this true?

I know this question is (too) long. Thanks for any help, I will attempt to clarify if I can.

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I asked my professor, and he said it has something to do with ergodicity. I have narrowed the problem down to something like $P(R(t) < B) = P(R(t) < B | R(t) < A)unknownfactor1 + P(R(t) < B | R(t) > A)unknownfactor2$. It turns out the unknown factors "correspond" to the average length of the time segment multiplied by the probability reaching that time segment. The question is what is the mathematical definition of "correspond"? Here we are saying corresponds means probability X average_time_segment. – jrand Apr 6 '11 at 3:03