Continuations

Last week in Uncertainty Wednesday, I introduced the example of the PSA test for prostate cancer. I provided all the probabilities for elementary events assuming a 50-year old male. We then did some simple addition to figure out the probability that I would receive a high PSA level signal which turned out to be P(H) = 0.091302. The crucial question then is (quoting from the previous post)What we are looking for is P(B | H) this is read as the probability of state B *conditional* on signal H b...

Last time in Uncertainty Wednesday, I announced that we would look at the PSA Test Example using absolute numbers instead of probabilities. Now you may recall that we defined probability as how likely something is to happen. And in introducing the PSA Test Example, I provided the probabilities for the elementary events for a 50-year old male as follows: P({AL}) = P(healthy *and* low PSA) = 0.907179 P({AH}) = P(healthy *and* high PSA) = 0.089721 P({BL}) = P(cancer *and* low PSA) = 0.001519 P({...

OK, so technically it is Thursday but as it turns out I had kind of an epic day yesterday and so Uncertainty Wednesday is a day late (thus introducing some uncertainty about when you will get your next dose of Uncertainty Wednesday). Today I will continue with the PSA Test Example. Last time I had shown the probability of having cancer conditional on a positive test, which was P(B | H) = 0.017316. We also saw that the reason this probability, while significantly higher than the unconditional ...

After a 2 week hiatus for travel here is another Uncertainty Wednesday. After going through probability I had promised a concrete example. As a 50 year old male one that is close to home is prostate cancer. The system in question is the prostate and simplistically it can be in two states (A) healthy and (B) with cancer. I am saying simplistic because there are various severities of prostate cancer in terms of aggressiveness and whether or not the cancer is still contained in the prostate. Thi...

And we are back! Sorry for the, ahem, uncertainty about when Uncertainty Wednesday would appear. I decided to work less this year on vacation and that included not blogging. Today we will wrap up the Zoltar Example. In Part 3, I first introduced the idea that even a very simple explanation in the form of the equation B + G = N (reminder B is a bad fortune, G a good one and N the total number of fortunes) let’s us start to reason about the uncertainty involved. In particular, I discussed why a...

Last Uncertainty Wednesday, I set up the first example which I called Zoltar as a machine thatonly prints out two different fortunes: it either prints “You will have a Good day tomorrow” or “You will have a Bad day tomorrow.” For brevity we will refer to these as G and B respectively. And I asked how to use the Uncertainty Wednesday framework to think about uncertainty here. This requires us to do two things. We have to think about what our observations are (and how those have clear limits) a...

I was going to start formalizing the concepts from Uncertainty Wednesday but I got a reader suggestion to do this in the context of a coin flipping example. This is what pretty much every introductory text uses because, well, coin flips are actually used in practice to decide things (especially in sports) and also coins have only two sides (heads and tails). It will be interesting to see if I can show how the Uncertainty Wednesday framework makes coin flipping more interesting than the way it...

In Uncertainty Wednesday today we continue our analysis of the Zoltar example, which I introduced two weeks ago. We came up with the super simple expression B + G = N, where B is the number of Bad fortunes, G is the number of Good fortunes and N is the number of all fortunes. Now while this may still strike you as simply an accounting identity, let’s consider the case of N = 1. Because these have to be integers, it must be the case that either (B = 1 and G = 0) or (B = 0 and G = 1). Why do th...