[EMAIL PROTECTED] (driver_writer) wrote in message news:<[EMAIL PROTECTED]>... > In my readings, I've come across unbounded discrete probability > functions, e.g. > > f(x) = (1/2)^x, x = 1,2,3... > = 0 elsewhere > > How does one take the Expected value or a r.v. X, for an unbounded (or > in this case bounded on the left side) p.f? > > My guess is you have to see if the series converges or diverges. If it > converges, you take the sum and then ?
It's not clear which sum you intend taking there. You don't sum the p.f. - you already know it sums to 1 if it's a valid p.f. You just take the usual form of the expectation, which you should be able to look up, or google on "discrete expectation|expected" or something similar (the first few links seem to have it right). (and yes, the formula for the expectation will be an infinite series) So you find the sum of that series for the expectation (if it converges, which is not always the case, so yes, you need to make sure of that). > Also, in order to take the mode of the distribution, for the > continuous pdf you can find the maximum by taking the derivative, > setting it to 0 and solving. How would you find the mode for a > discrete p.f like the one above? You find which discrete value has largest f. i.e. find m such that f(m)>=f(x), for all x. There are a variety of ways to demonstrate that you have found such a value (such as showing that f(m)-f(x) >= 0 for all x, or that f(x)/f(m)<=1 for all x, or you may be able to show that f(x-1)/f(x)<1 for x<m and the equivalent on the other side of the mode, that sort of thing). Glen . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
