Does having your first child not be color-blind change the odds that the next ones will be? I'm graduated now, and swear I remember this exact kind of problem coming up in my Bayesian statistics class, but I can't remember what the answer was.
The Problem
So my wife's brother is red/green colorblind (xY), but her father is not(XY), and her mother is not, so her mother must be a carrier(Xx). Red/green color-blindness is recessive on the X Chromosome(x), which means in order to have the phenotype (actually have it), all of your X chromosomes must have the gene(xx or xY). This means women must get it from both their mother and father, while men must get it from their mother.
As a result, my wife has a 50% chance of being a carrier(Xx or XX), while all of my sons have a 25% chance of being red/green colorblind(XY or xY).
What I'm trying to figure out is if my first son ends up NOT being red-green colorblind (XY), does that change the probability that my wife is a carrier, since I've now observed one datapoint that refutes that possibility? If so, by how much?
My Thoughts
I think that Frequentists would say that it remains a 50% chance due to them being Independent events, but again, I can't recall exactly how it worked. I may be spacing it, but if my memory serves, Bayesians would include the original assumption(50%) as essentially a datapoint, but would subjectively give it more or less weight. I'm wondering if there's an objective way to handle this.
Of course, I can always calculate the odds of her being a carrier and how unlikely it is that she would have X sons without color-blindness (simple Geometric Distribution), but that doesn't tell me anything about her original probability.
 A: Here is my simple attempt.
We have prior of your wife being a carrier $\pi(Y = 1) = 0.5$ and the probability of your first son being colourblind $p(X = 0|Y = 1) = 0.5,\  p(X = 0|Y = 0) = 1$. Now using simple Bayes rule we have:
$$\pi(Y = 1|X = 0) = \frac{p(X = 0| Y = 1)\pi(Y = 1)}{p(X = 0)} = \frac{p(X = 0| Y = 1)\pi(Y = 1)}{p(X = 0| Y = 1)\pi(Y = 1) + p(X = 0| Y = 0)\pi(Y = 0)} = \frac{\frac{1}{2} * \frac{1}{2}}{\frac{1}{2} * \frac{1}{2} + 1 * \frac{1}{2}} = \frac{1}{3}$$
A: I do not think that Frequentist's would say that (the 50% probability). Imagine the ridiculousness of the statement if you extend it to ten first sons without color blindness.

*

*A Bayesian statement would go something like assigning probabilities to the different cases. And then you have P(mother = Xx) and P(mother = XX) changing from estimates of the odds ratio prior to the experiment, $\frac{P(XX)}{P(xX)} = 1$, to odds ratio posterior to the experiment $\frac{P(XX|obs)}{P(xX|obs)} = \frac{P(XX)}{P(xX)}  \frac{P(obs|XX)}{P(obs|xX)} = 2$. A frequentist could also make a 95% confidence interval which woudl both include the mother having xX and XX, another not so useful approach in this case.


*A frequentist would more like create a statement that has a certain success rate. For instance, do not reject the hypothesis that the mother has xX, since the p-value is only 0.5, (which means the approach, if it would have been rejected, fails 50% of the time when the mother has xX after all, this does not mean that the probability for a particular experiment is 50%). In the case of such small numbers, and for a single point estimate, the frequentist approach seems less useful.
The frequentist might also do something like create a 95% confidence
interval, which would in this case include both the xX and XX,
another not so useful approach in this case.
Of course this 'a frequentists would...' is a bit ridiculous. No statistician is frequentist or bayesian. Only the methods have this property, and frequentists methods, as this example shows, have a disadvantage for such small experiments, with few observations.
