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This is from Bayesian Data Analysis - Andrew Gelman 2nd edition, example on pg.11.

If theta = 1 = mother is carrier, theta = 0 = mother is non carrier and y1, y2, y3 = 3 sons, where y1 is either 1 = affected or 0 = unaffected.

After calculating p(theta = 1 | y1, y2 = 0) = .2, if the mother has a third son who is unaffected, Why is the new likelihood .5? The book says before, With 2 sons,

But with the third son who is unaffected, likelihood changes to .5?

I get how the posterior becomes new prior, but I don't get why new likelihood p(y1, y2, y3 = 0 | theta = .2) = .5

Hope I made the question clear. Thanks.

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  • $\begingroup$ Three comments: it looks to me as though $\theta$ can equal either $0$ or $1$, but you have $\theta = 0.2$ in your final line. Also, since you haven't written out the likelihood function anywhere, it's not possible to answer your question as written, since no-one will be able to figure out why it equals what it does if they don't know what it is. Third, please use mathjax for formatting, it will help readability immensely. $\endgroup$
    – jbowman
    Commented Jul 1, 2018 at 23:03
  • $\begingroup$ The new likelihood only involves the third son because the first two sons have been incorporated into the new prior. $\endgroup$
    – mef
    Commented Jul 1, 2018 at 23:50

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$P(\theta =1|Y_1=0, Y_2=0, Y_3=0)$ $$ = \dfrac{P(Y_1=0, Y_2=0, Y_3=0|\theta=1)P(\theta=1)}{P(Y_1=0,Y_2=0, Y_3=0)}$$ $$=\frac{P(Y_3=0|Y_2=0,Y_1=0,\theta=1)P(Y_1=0,Y_2=0|\theta=1)P(\theta=1)}{P(Y_3=0|Y_2=0,Y_1=0)P(Y_2=0,Y_1=0)}$$ $$=\frac{P(Y_3=0|Y_2=0,Y_1=0, \theta=1)P(\theta=1|Y_2=0,Y_1=0)}{P(Y_3=0|Y_2=0,Y_1=0)}$$ $$=\frac{P(Y_3=0|\theta=1)P(\theta=1|Y_2=0,Y_1=0)}{P(Y_3=0| Y_2=0,Y_1=0)}$$

The following held because the events $(Y_1=0, Y_2=0)$ and $Y_3=0$ are independent (one child having the trait doesn't influence the other ones) $$P(Y_3=0|Y_2=0,Y_1=0,\theta=1) = P(Y_3=0|\theta=1)$$

And I believe $P(Y_3=0|\theta=1)=0.5$

This is actually exactly what I was asking in Apply Bayes rule sequentially

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  • $\begingroup$ Ah ok. so the likelihood is independent of previous "iteration", and so it only considers y3. Thanks. Baye's theorem is really efficient it seems. $\endgroup$ Commented Jul 3, 2018 at 5:54

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