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I'm currently trying to learn Bayesian statistics and am working through the book "Bayesian Data Analysis". At the end of chapter 1 is a question asking to derive a conditional probability. I can't figure out the answer even though there is a printed solution here (question 1.3).

I'm not going to post the exact text of the question due to fear of copyright issues but it is roughly as follows:

You inherit a trait from your parents based on your genes

| Gene 1 | Gene 2 | Trait |   
|---     |---     |---|---|  
| X      | X      | A     |    
| X      | Y      | B     |   
| Y      | X      | B     |      
| Y      | Y      | B     |    

In particular you are classed as a heterozygote if you have either XY or YX. A proportion $p^2$ of people have trait A and $2p(1-p)$ are heterozygote carriers of a single recessive allele where $0 < p < 1$. There is a 50/50 chance on either gene being passed on from parents to children.

Show that probability of a child being heterozygote given that they are B and their parents are both B is $2p/(1+2p)$

For my attempt so far I have the following:

For ease I've used the following notation
H = child is a heterozygotes
B = child has trait B
P = parents have trait B

$$ \begin{align} P( H | B , P ) &= \frac{P( H , B , P)}{P(B,P)} \\ \\ &= \frac{P( B | H , P ) P(H , P ) }{ P(B | P) P(P) } \\ \\ &= \frac{1 * P(H | P ) P(P) }{ P(B | P) P(P) } \\ \\ &= \frac{P(H | P )}{ P(B | P) } \\ \end{align} $$

Also I calculate that the proportion of the population who are YY should be $(1-p)^2$.

From here I am stuck on where to go; my guess would be to use the law of total probability to calculate each conditional probability but I get lost whenever I attempt the algebra. Additionally when I look at the above linked solution I don't appear to be even on the right line. Any help would be appreciated.

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You're right so far, so you can certainly start from trying to evaluate the final expression you wrote.

As you know, scenario $P$ entails three possibilities that are mutually exclusive. They are:

  1. two heterozygous parents
  2. two parents of type YY
  3. one YY and one heterozygous parent

Let's call those scenarios $P_1,$ $P_2,$ and $P_3$ respectively. Then, $$ P(P) = P(P_1) + P(P_2) + P(P_3). $$

The law of total probability states, $$ P(H \mid P) = P(H,P_1 \mid P) + P(H,P_2 \mid P) + P(H,P_3 \mid P), $$ which we could rewrite as, $$ P(H \mid P) = \frac{1}{P(P)} \left[P(H \mid P_1) P(P_1) + P(H \mid P_2) P(P_2) + P(H \mid P_3) P(P_3)\right]. $$ Here, I used $P(H, P_1 \mid P) = P(H \mid P_1, P) P(P_1 \mid P) = P(H \mid P_1) P(P_1) / P(P).$

You can use the same logic for $P(B \mid P).$

If you let $q = 2 p (1-p)$ and $w = (1-p)^2$ then you can show that $P(P_1) = q^2,$ $P(P_2) = w^2,$ and $P(P_3) = 2 q w.$ Then you can work out $P(H \mid P_i)$ and $P(B \mid P_i)$ with the assumption that there's a $1/2$ chance of getting either gene from each parent. (For example, $P(B \mid P_1) = 3/4.$ If both parents are heterozygous, then there's only a $1/4$ chance of the child getting both X genes, which is the only way the child can have trait A.)

I think it's easier if you write $\frac{P(H \mid P)}{P(B \mid P)}$ in terms of $q$s and $w$s as I defined above, then convert them back into $p$s once you have the expression written out.

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  • $\begingroup$ That you for this it makes complete sense ! $\endgroup$ – gowerc Jan 12 '18 at 23:40
  • $\begingroup$ As a side note how simple / hard is a question like this. For being the third question in a Bayesian text book it seemed pretty steep to me ! Is this a sign of things to come for Bayesian analysis ? $\endgroup$ – gowerc Jan 12 '18 at 23:42

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